$int_{0}^{infty} frac{e^{-x} sin(x)}{x} dx$ Evaluate Integral












25












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Compute the following integral:



$$int_{0}^{infty} frac{e^{-x} sin(x)}{x} dx$$



Any hint, suggestion is welcome.






calculus integration limits improper-integrals







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  • $begingroup$
    this looks like something that could be done with complex contour integration. en.wikipedia.org/wiki/Methods_of_contour_integration
    $endgroup$
    – Dennis Gulko
    Jun 18 '12 at 10:08










  • $begingroup$
    http://www.wolframalpha.com/input/?i=Integrate[%28e^-x*Sin[x]%2Fx%29%2C{x%2C0%2CInfinity} Ans = Pi/4=0.785398. In case you don't want to use limits and make it indefinite than use "Integrate[(e^-x*Sin[x]/x)]". This is not step by step solution obviously but may be helpful to you.
    $endgroup$
    – Rorschach
    Jun 18 '12 at 10:17








  • 7




    $begingroup$
    Consider $I(p)=int_{0}^{infty}frac{e^{-px}sin(x)}{x}dx$,then $frac{d}{dp}I(p)=-int_{0}^{infty}e^{-px}sin(x)dx$.$I(p)$ tends to zero when $p$ tends to infinity.Solve this differential equation and work out $I(1)$.That is the answer.
    $endgroup$
    – y zh
    Jun 18 '12 at 10:21












  • $begingroup$
    @yzhao Would you consider writing that as an answer?
    $endgroup$
    – anon
    Jun 18 '12 at 10:38
















25












$begingroup$


Compute the following integral:



$$int_{0}^{infty} frac{e^{-x} sin(x)}{x} dx$$



Any hint, suggestion is welcome.






calculus integration limits improper-integrals







share|cite|improve this question











$endgroup$












  • $begingroup$
    this looks like something that could be done with complex contour integration. en.wikipedia.org/wiki/Methods_of_contour_integration
    $endgroup$
    – Dennis Gulko
    Jun 18 '12 at 10:08










  • $begingroup$
    http://www.wolframalpha.com/input/?i=Integrate[%28e^-x*Sin[x]%2Fx%29%2C{x%2C0%2CInfinity} Ans = Pi/4=0.785398. In case you don't want to use limits and make it indefinite than use "Integrate[(e^-x*Sin[x]/x)]". This is not step by step solution obviously but may be helpful to you.
    $endgroup$
    – Rorschach
    Jun 18 '12 at 10:17








  • 7




    $begingroup$
    Consider $I(p)=int_{0}^{infty}frac{e^{-px}sin(x)}{x}dx$,then $frac{d}{dp}I(p)=-int_{0}^{infty}e^{-px}sin(x)dx$.$I(p)$ tends to zero when $p$ tends to infinity.Solve this differential equation and work out $I(1)$.That is the answer.
    $endgroup$
    – y zh
    Jun 18 '12 at 10:21












  • $begingroup$
    @yzhao Would you consider writing that as an answer?
    $endgroup$
    – anon
    Jun 18 '12 at 10:38














25












25








25


14



$begingroup$


Compute the following integral:



$$int_{0}^{infty} frac{e^{-x} sin(x)}{x} dx$$



Any hint, suggestion is welcome.






calculus integration limits improper-integrals







share|cite|improve this question











$endgroup$




Compute the following integral:



$$int_{0}^{infty} frac{e^{-x} sin(x)}{x} dx$$



Any hint, suggestion is welcome.






calculus integration limits improper-integrals




calculus integration limits improper-integrals






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share|cite|improve this question













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edited Jan 12 at 6:49









Abcd

3,02821235




3,02821235










asked Jun 18 '12 at 10:02









user 1357113user 1357113

22.4k877226




22.4k877226












  • $begingroup$
    this looks like something that could be done with complex contour integration. en.wikipedia.org/wiki/Methods_of_contour_integration
    $endgroup$
    – Dennis Gulko
    Jun 18 '12 at 10:08










  • $begingroup$
    http://www.wolframalpha.com/input/?i=Integrate[%28e^-x*Sin[x]%2Fx%29%2C{x%2C0%2CInfinity} Ans = Pi/4=0.785398. In case you don't want to use limits and make it indefinite than use "Integrate[(e^-x*Sin[x]/x)]". This is not step by step solution obviously but may be helpful to you.
    $endgroup$
    – Rorschach
    Jun 18 '12 at 10:17








  • 7




    $begingroup$
    Consider $I(p)=int_{0}^{infty}frac{e^{-px}sin(x)}{x}dx$,then $frac{d}{dp}I(p)=-int_{0}^{infty}e^{-px}sin(x)dx$.$I(p)$ tends to zero when $p$ tends to infinity.Solve this differential equation and work out $I(1)$.That is the answer.
    $endgroup$
    – y zh
    Jun 18 '12 at 10:21












  • $begingroup$
    @yzhao Would you consider writing that as an answer?
    $endgroup$
    – anon
    Jun 18 '12 at 10:38


















  • $begingroup$
    this looks like something that could be done with complex contour integration. en.wikipedia.org/wiki/Methods_of_contour_integration
    $endgroup$
    – Dennis Gulko
    Jun 18 '12 at 10:08










  • $begingroup$
    http://www.wolframalpha.com/input/?i=Integrate[%28e^-x*Sin[x]%2Fx%29%2C{x%2C0%2CInfinity} Ans = Pi/4=0.785398. In case you don't want to use limits and make it indefinite than use "Integrate[(e^-x*Sin[x]/x)]". This is not step by step solution obviously but may be helpful to you.
    $endgroup$
    – Rorschach
    Jun 18 '12 at 10:17








  • 7




    $begingroup$
    Consider $I(p)=int_{0}^{infty}frac{e^{-px}sin(x)}{x}dx$,then $frac{d}{dp}I(p)=-int_{0}^{infty}e^{-px}sin(x)dx$.$I(p)$ tends to zero when $p$ tends to infinity.Solve this differential equation and work out $I(1)$.That is the answer.
    $endgroup$
    – y zh
    Jun 18 '12 at 10:21












  • $begingroup$
    @yzhao Would you consider writing that as an answer?
    $endgroup$
    – anon
    Jun 18 '12 at 10:38
















$begingroup$
this looks like something that could be done with complex contour integration. en.wikipedia.org/wiki/Methods_of_contour_integration
$endgroup$
– Dennis Gulko
Jun 18 '12 at 10:08




$begingroup$
this looks like something that could be done with complex contour integration. en.wikipedia.org/wiki/Methods_of_contour_integration
$endgroup$
– Dennis Gulko
Jun 18 '12 at 10:08












$begingroup$
http://www.wolframalpha.com/input/?i=Integrate[%28e^-x*Sin[x]%2Fx%29%2C{x%2C0%2CInfinity} Ans = Pi/4=0.785398. In case you don't want to use limits and make it indefinite than use "Integrate[(e^-x*Sin[x]/x)]". This is not step by step solution obviously but may be helpful to you.
$endgroup$
– Rorschach
Jun 18 '12 at 10:17






$begingroup$
http://www.wolframalpha.com/input/?i=Integrate[%28e^-x*Sin[x]%2Fx%29%2C{x%2C0%2CInfinity} Ans = Pi/4=0.785398. In case you don't want to use limits and make it indefinite than use "Integrate[(e^-x*Sin[x]/x)]". This is not step by step solution obviously but may be helpful to you.
$endgroup$
– Rorschach
Jun 18 '12 at 10:17






7




7




$begingroup$
Consider $I(p)=int_{0}^{infty}frac{e^{-px}sin(x)}{x}dx$,then $frac{d}{dp}I(p)=-int_{0}^{infty}e^{-px}sin(x)dx$.$I(p)$ tends to zero when $p$ tends to infinity.Solve this differential equation and work out $I(1)$.That is the answer.
$endgroup$
– y zh
Jun 18 '12 at 10:21






$begingroup$
Consider $I(p)=int_{0}^{infty}frac{e^{-px}sin(x)}{x}dx$,then $frac{d}{dp}I(p)=-int_{0}^{infty}e^{-px}sin(x)dx$.$I(p)$ tends to zero when $p$ tends to infinity.Solve this differential equation and work out $I(1)$.That is the answer.
$endgroup$
– y zh
Jun 18 '12 at 10:21














$begingroup$
@yzhao Would you consider writing that as an answer?
$endgroup$
– anon
Jun 18 '12 at 10:38




$begingroup$
@yzhao Would you consider writing that as an answer?
$endgroup$
– anon
Jun 18 '12 at 10:38










8 Answers
8






active

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votes


















43












$begingroup$

Yet a different approach: parametric integration. Let
$$
F(lambda)=int_{0}^{infty} frac{e^{-lambda x} sin(x)}{x},dx,qquadlambda>0.
$$
Then
$$
F'(lambda)=-int_{0}^{infty} e^{-lambda x} sin(x),dx=-frac{1}{1+lambda^2}.
$$
Integrating and taking into account that $lim_{lambdatoinfty}F(lambda)=0$ we have
$$
F(lambda)=fracpi2-arctanlambda
$$
and
$$
int_{0}^{infty} frac{e^{-x} sin(x)}{x},dx=F(1)=fracpi4.
$$






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  • $begingroup$
    @Chris:I'm sorry for just noticing you comment,but Julian gives a detailed description of my idea.
    $endgroup$
    – y zh
    Jun 18 '12 at 11:21












  • $begingroup$
    Nice answer (+1). I enjoy seeing a variety of solutions!
    $endgroup$
    – robjohn♦
    Jun 18 '12 at 11:44



















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Using Laplace Transform,
$$mathcal{L}(sin(x)) = frac{1}{s^2 + 1}$$
$$mathcal{L}left(frac{sin(x)}{x}right) = int_r^infty frac{1}{s^2 + 1} ds = frac{pi}{2} - arctan(r)$$
Therefore,
$$int_0^infty e^{-rx} frac{sin(x)}{x} dx = frac{pi}{2} - arctan(r)$$
Substituting r = 1,
$$int_0^infty e^{-x} frac{sin(x)}{x} dx = frac{pi}{4}$$






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  • $begingroup$
    thanks for your delicate solution!
    $endgroup$
    – user 1357113
    Jun 18 '12 at 10:57



















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Another approach:
$$begin{eqnarray*}
int_{0}^{infty} dx, frac{e^{-x} sin(x)}{x}
&=& int_{0}^{infty}dx, frac{e^{-x}}{x} sum_{k=0}^infty frac{(-1)^k x^{2k+1}}{(2k+1)!} \
&=& sum_{k=0}^infty frac{(-1)^k}{(2k+1)!}
int_{0}^{infty}dx, x^{2k} e^{-x} \
&=& sum_{k=0}^infty frac{(-1)^k}{(2k+1)!}(2k)! \
&=& sum_{k=0}^infty frac{(-1)^k}{2k+1}
hspace{5ex} textrm{(Leibniz series for $pi$)}\
&=& frac{pi}{4}.
end{eqnarray*}$$






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  • $begingroup$
    this is another magic shot! Nice job! Thanks! :-)
    $endgroup$
    – user 1357113
    Jun 19 '12 at 5:52












  • $begingroup$
    @Chris: Thanks, Chris. Another good question!
    $endgroup$
    – user26872
    Jun 19 '12 at 7:40



















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Write this as
$$
lim_{epsilonto0}int_epsilon^{1/epsilon}frac{e^{-(1-i)x}-e^{-(1+i)x}}{2ix},mathrm{d}xtag{1}
$$
and then consider the path integral
$$
frac{1}{2i}int_{gamma_epsilon} e^{-z},frac{mathrm{d}z}{z}tag{2}
$$
where $gamma_epsilon$ comes in along the line $(1+i)x$, makes a quarter circle clockwise along $|z|=epsilon$, goes out along the line $(1-i)x$ and then back a quarter circle counter-clockwise along $|z|=1/epsilon$. There are no poles inside this path, so the integral in $(2)$ is $0$.



The part along $|z|=1/epsilon$ dies away exponentially as $epsilonto0$. The two parts along the lines sum to our integral, $(1)$, and the part along $|z|=epsilon$ tends to $frac14$ of the integral of $frac{1}{2iz}$ clockwise around the origin; that is, $-pi/4$. Since the sum of these parts is $0$, the limit in $(1)$ must be $pi/4$. That is,
$$
int_0^inftyfrac{e^{-x}sin(x)}{x}mathrm{d}x=frac{pi}{4}tag{3}
$$






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  • $begingroup$
    @robjohn that's a nice solution! But I think that you should also justify that principal value of the integral equals integral itself.
    $endgroup$
    – qoqosz
    Jun 18 '12 at 11:08










  • $begingroup$
    @qoqosz: I am not using a principal value since $frac{sin(x)}{x}$ is a nice bounded function. I use the $epsilon$ and $1/epsilon$ to connect with the contour integral. Perhaps I am missing your concern.
    $endgroup$
    – robjohn♦
    Jun 18 '12 at 11:17










  • $begingroup$
    @robjohn $frac{x}{1+x^2}$ is also bounded but when integrating over $mathbb{R}$ it exists only in P.V. sense. Well, maybe it's not a 'nice' function ;).
    $endgroup$
    – qoqosz
    Jun 18 '12 at 11:21










  • $begingroup$
    @qoqosz: $e^{-x}frac{sin(x)}{x}$ is a nice bounded function that dies away exponentially. It converges absolutely over $[0,infty)$. In fact, all of the integrals above converge absolutely. What am I missing?
    $endgroup$
    – robjohn♦
    Jun 18 '12 at 11:31






  • 1




    $begingroup$
    @qoqosz: Ah, I just saw the improper-integral tag. In the Riemann integral sense, it is improper because the domain of integration is not bounded, but it is not improper in the Lebesgue sense.
    $endgroup$
    – robjohn♦
    Jun 18 '12 at 11:36





















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If you know a bit about Fourier theory. You could Parseval's theorem
$$int !dx ,f(x) g(x)^* = int !dxi,hat f(xi) hat g(xi)^* $$
with $f(x) = sin(x)/x$, $g(x) = Theta(x) e^{-x}$ and $hat{f}$, $hat{g}$ their Fourier transforms and $Theta(x)$ the Heaviside step function.



Hint: $hat{f}(xi) = tfrac12sqrt{frac{pi}{2}} [Theta(1-xi) + Theta(1+xi)] =sqrt{frac{pi}{2}} mathop{rm rect}(xi) $.






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  • $begingroup$
    @Anon: Sorry for not defining: $Theta(x)$ is the Heaviside step function.
    $endgroup$
    – Fabian
    Jun 18 '12 at 10:24





















5












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Let



$$f(z) = frac{e^{-z+iz}}{z}$$



and let $C$ be the contour that travels along $0$ to $R$, makes a quarter of a circle around to $iR$ and back to $0$, properly indented around $0$ with a quarter circle of radius $delta$ to avoid the pole.



Contour



As $R to infty$, the integral over rounded part of the contour tends to $0$ and the part around $0$ tends to $-ifrac{pi}{2}$ (N.B. this is $-ifrac{pi}{2}$ of the residue at $z=0$) as $delta to 0$. Then by Cauchy's theorem:



$$
0=oint_C f(z),dz =\
int_0^infty frac{e^{-z+iz}}{z},dz -int_0^{infty} frac{e^{-iz-z}}{z},dz - ifrac{pi}{2}
$$



And upon taking imaginary parts and solving:



$$
frac{pi}{4}=int_0^infty frac{e^{-z}sin(z)}{z},dx
$$






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  • $begingroup$
    hehe (+1). Glad to have the opportunity to learn more on complex analysis. :-)
    $endgroup$
    – user 1357113
    Jul 4 '13 at 16:06










  • $begingroup$
    @Chris'swisesister I thought you may be interested in a slightly different contour for variety
    $endgroup$
    – Argon
    Jul 4 '13 at 16:06










  • $begingroup$
    sure. I'm always interested in various ways. Thanks! :-)
    $endgroup$
    – user 1357113
    Jul 4 '13 at 16:07










  • $begingroup$
    Could you also add the contour (a picture) in case you have one ready?
    $endgroup$
    – user 1357113
    Jul 4 '13 at 16:09












  • $begingroup$
    @Chris'swisesister I could try and draw one, I will update this soon.
    $endgroup$
    – Argon
    Jul 4 '13 at 16:11



















3












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$$int_0^{infty} frac{e^{-x}sin x}{x},dx=int_0^{infty}int_0^{infty} e^{-x}sin x ,e^{-xy},dy,dx=int_0^{infty} int_0^{infty} e^{-x(1+y)}sin x,dx,dy$$
From integration by parts or otherwise, one can show that:
$$int_0^{infty}e^{-x(1+y)}sin x,dx=frac{1}{1+(1+y)^2}$$
Hence
$$int_0^{infty} int_0^{infty} e^{-x(1+y)}sin x,dx,dy=int_0^{infty} frac{dy}{1+(1+y)^2}=left(arctan(1+y)right|_0^{infty}=boxed{dfrac{pi}{4}}$$






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  • $begingroup$
    Neat, but isn't that more or less the same as Julián Aguirre's answer?
    $endgroup$
    – IAmNoOne
    May 9 '14 at 0:12



















2












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$newcommand{+}{^{dagger}}%
newcommand{angles}[1]{leftlangle #1 rightrangle}%
newcommand{braces}[1]{leftlbrace #1 rightrbrace}%
newcommand{bracks}[1]{leftlbrack #1 rightrbrack}%
newcommand{ceil}[1]{,leftlceil #1 rightrceil,}%
newcommand{dd}{{rm d}}%
newcommand{down}{downarrow}%
newcommand{ds}[1]{displaystyle{#1}}%
newcommand{equalby}[1]{{#1 atop {= atop vphantom{huge A}}}}%
newcommand{expo}[1]{,{rm e}^{#1},}%
newcommand{fermi}{,{rm f}}%
newcommand{floor}[1]{,leftlfloor #1 rightrfloor,}%
newcommand{half}{{1 over 2}}%
newcommand{ic}{{rm i}}%
newcommand{iff}{Longleftrightarrow}
newcommand{imp}{Longrightarrow}%
newcommand{isdiv}{,left.rightvert,}%
newcommand{ket}[1]{leftvert #1rightrangle}%
newcommand{ol}[1]{overline{#1}}%
newcommand{pars}[1]{left( #1 right)}%
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{pp}{{cal P}}%
newcommand{root}[2]{,sqrt[#1]{,#2,},}%
newcommand{sech}{,{rm sech}}%
newcommand{sgn}{,{rm sgn}}%
newcommand{totald}[3]{frac{{rm d}^{#1} #2}{{rm d} #3^{#1}}}
newcommand{ul}[1]{underline{#1}}%
newcommand{verts}[1]{leftvert, #1 ,rightvert}$
begin{align}
&color{#00f}{largeint_{0}^{infty}{expo{-x}sinpars{x} over x},dd x}
=int_{0}^{infty}expo{-x}pars{halfint_{-1}^{1}expo{ic kx},dd k}
=halfint_{-1}^{1}int_{0}^{infty}expo{pars{ic k - 1}x},dd k
\[3mm]&=halfint_{-1}^{1}{1 over 1 - ic k},dd k
=int_{0}^{1}{dd k over 1 + k^{2}} = arctanpars{1}
=color{#00f}{Large{pi over 4}}
end{align}






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    8 Answers
    8






    active

    oldest

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    8 Answers
    8






    active

    oldest

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    active

    oldest

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    active

    oldest

    votes









    43












    $begingroup$

    Yet a different approach: parametric integration. Let
    $$
    F(lambda)=int_{0}^{infty} frac{e^{-lambda x} sin(x)}{x},dx,qquadlambda>0.
    $$
    Then
    $$
    F'(lambda)=-int_{0}^{infty} e^{-lambda x} sin(x),dx=-frac{1}{1+lambda^2}.
    $$
    Integrating and taking into account that $lim_{lambdatoinfty}F(lambda)=0$ we have
    $$
    F(lambda)=fracpi2-arctanlambda
    $$
    and
    $$
    int_{0}^{infty} frac{e^{-x} sin(x)}{x},dx=F(1)=fracpi4.
    $$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      @Chris:I'm sorry for just noticing you comment,but Julian gives a detailed description of my idea.
      $endgroup$
      – y zh
      Jun 18 '12 at 11:21












    • $begingroup$
      Nice answer (+1). I enjoy seeing a variety of solutions!
      $endgroup$
      – robjohn♦
      Jun 18 '12 at 11:44
















    43












    $begingroup$

    Yet a different approach: parametric integration. Let
    $$
    F(lambda)=int_{0}^{infty} frac{e^{-lambda x} sin(x)}{x},dx,qquadlambda>0.
    $$
    Then
    $$
    F'(lambda)=-int_{0}^{infty} e^{-lambda x} sin(x),dx=-frac{1}{1+lambda^2}.
    $$
    Integrating and taking into account that $lim_{lambdatoinfty}F(lambda)=0$ we have
    $$
    F(lambda)=fracpi2-arctanlambda
    $$
    and
    $$
    int_{0}^{infty} frac{e^{-x} sin(x)}{x},dx=F(1)=fracpi4.
    $$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      @Chris:I'm sorry for just noticing you comment,but Julian gives a detailed description of my idea.
      $endgroup$
      – y zh
      Jun 18 '12 at 11:21












    • $begingroup$
      Nice answer (+1). I enjoy seeing a variety of solutions!
      $endgroup$
      – robjohn♦
      Jun 18 '12 at 11:44














    43












    43








    43





    $begingroup$

    Yet a different approach: parametric integration. Let
    $$
    F(lambda)=int_{0}^{infty} frac{e^{-lambda x} sin(x)}{x},dx,qquadlambda>0.
    $$
    Then
    $$
    F'(lambda)=-int_{0}^{infty} e^{-lambda x} sin(x),dx=-frac{1}{1+lambda^2}.
    $$
    Integrating and taking into account that $lim_{lambdatoinfty}F(lambda)=0$ we have
    $$
    F(lambda)=fracpi2-arctanlambda
    $$
    and
    $$
    int_{0}^{infty} frac{e^{-x} sin(x)}{x},dx=F(1)=fracpi4.
    $$






    share|cite|improve this answer









    $endgroup$



    Yet a different approach: parametric integration. Let
    $$
    F(lambda)=int_{0}^{infty} frac{e^{-lambda x} sin(x)}{x},dx,qquadlambda>0.
    $$
    Then
    $$
    F'(lambda)=-int_{0}^{infty} e^{-lambda x} sin(x),dx=-frac{1}{1+lambda^2}.
    $$
    Integrating and taking into account that $lim_{lambdatoinfty}F(lambda)=0$ we have
    $$
    F(lambda)=fracpi2-arctanlambda
    $$
    and
    $$
    int_{0}^{infty} frac{e^{-x} sin(x)}{x},dx=F(1)=fracpi4.
    $$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jun 18 '12 at 11:07









    Julián AguirreJulián Aguirre

    68.2k24094




    68.2k24094












    • $begingroup$
      @Chris:I'm sorry for just noticing you comment,but Julian gives a detailed description of my idea.
      $endgroup$
      – y zh
      Jun 18 '12 at 11:21












    • $begingroup$
      Nice answer (+1). I enjoy seeing a variety of solutions!
      $endgroup$
      – robjohn♦
      Jun 18 '12 at 11:44


















    • $begingroup$
      @Chris:I'm sorry for just noticing you comment,but Julian gives a detailed description of my idea.
      $endgroup$
      – y zh
      Jun 18 '12 at 11:21












    • $begingroup$
      Nice answer (+1). I enjoy seeing a variety of solutions!
      $endgroup$
      – robjohn♦
      Jun 18 '12 at 11:44
















    $begingroup$
    @Chris:I'm sorry for just noticing you comment,but Julian gives a detailed description of my idea.
    $endgroup$
    – y zh
    Jun 18 '12 at 11:21






    $begingroup$
    @Chris:I'm sorry for just noticing you comment,but Julian gives a detailed description of my idea.
    $endgroup$
    – y zh
    Jun 18 '12 at 11:21














    $begingroup$
    Nice answer (+1). I enjoy seeing a variety of solutions!
    $endgroup$
    – robjohn♦
    Jun 18 '12 at 11:44




    $begingroup$
    Nice answer (+1). I enjoy seeing a variety of solutions!
    $endgroup$
    – robjohn♦
    Jun 18 '12 at 11:44











    19












    $begingroup$

    Using Laplace Transform,
    $$mathcal{L}(sin(x)) = frac{1}{s^2 + 1}$$
    $$mathcal{L}left(frac{sin(x)}{x}right) = int_r^infty frac{1}{s^2 + 1} ds = frac{pi}{2} - arctan(r)$$
    Therefore,
    $$int_0^infty e^{-rx} frac{sin(x)}{x} dx = frac{pi}{2} - arctan(r)$$
    Substituting r = 1,
    $$int_0^infty e^{-x} frac{sin(x)}{x} dx = frac{pi}{4}$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      thanks for your delicate solution!
      $endgroup$
      – user 1357113
      Jun 18 '12 at 10:57
















    19












    $begingroup$

    Using Laplace Transform,
    $$mathcal{L}(sin(x)) = frac{1}{s^2 + 1}$$
    $$mathcal{L}left(frac{sin(x)}{x}right) = int_r^infty frac{1}{s^2 + 1} ds = frac{pi}{2} - arctan(r)$$
    Therefore,
    $$int_0^infty e^{-rx} frac{sin(x)}{x} dx = frac{pi}{2} - arctan(r)$$
    Substituting r = 1,
    $$int_0^infty e^{-x} frac{sin(x)}{x} dx = frac{pi}{4}$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      thanks for your delicate solution!
      $endgroup$
      – user 1357113
      Jun 18 '12 at 10:57














    19












    19








    19





    $begingroup$

    Using Laplace Transform,
    $$mathcal{L}(sin(x)) = frac{1}{s^2 + 1}$$
    $$mathcal{L}left(frac{sin(x)}{x}right) = int_r^infty frac{1}{s^2 + 1} ds = frac{pi}{2} - arctan(r)$$
    Therefore,
    $$int_0^infty e^{-rx} frac{sin(x)}{x} dx = frac{pi}{2} - arctan(r)$$
    Substituting r = 1,
    $$int_0^infty e^{-x} frac{sin(x)}{x} dx = frac{pi}{4}$$






    share|cite|improve this answer











    $endgroup$



    Using Laplace Transform,
    $$mathcal{L}(sin(x)) = frac{1}{s^2 + 1}$$
    $$mathcal{L}left(frac{sin(x)}{x}right) = int_r^infty frac{1}{s^2 + 1} ds = frac{pi}{2} - arctan(r)$$
    Therefore,
    $$int_0^infty e^{-rx} frac{sin(x)}{x} dx = frac{pi}{2} - arctan(r)$$
    Substituting r = 1,
    $$int_0^infty e^{-x} frac{sin(x)}{x} dx = frac{pi}{4}$$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jun 18 '12 at 10:55

























    answered Jun 18 '12 at 10:41









    TenaliRamanTenaliRaman

    3,1641223




    3,1641223












    • $begingroup$
      thanks for your delicate solution!
      $endgroup$
      – user 1357113
      Jun 18 '12 at 10:57


















    • $begingroup$
      thanks for your delicate solution!
      $endgroup$
      – user 1357113
      Jun 18 '12 at 10:57
















    $begingroup$
    thanks for your delicate solution!
    $endgroup$
    – user 1357113
    Jun 18 '12 at 10:57




    $begingroup$
    thanks for your delicate solution!
    $endgroup$
    – user 1357113
    Jun 18 '12 at 10:57











    15












    $begingroup$

    Another approach:
    $$begin{eqnarray*}
    int_{0}^{infty} dx, frac{e^{-x} sin(x)}{x}
    &=& int_{0}^{infty}dx, frac{e^{-x}}{x} sum_{k=0}^infty frac{(-1)^k x^{2k+1}}{(2k+1)!} \
    &=& sum_{k=0}^infty frac{(-1)^k}{(2k+1)!}
    int_{0}^{infty}dx, x^{2k} e^{-x} \
    &=& sum_{k=0}^infty frac{(-1)^k}{(2k+1)!}(2k)! \
    &=& sum_{k=0}^infty frac{(-1)^k}{2k+1}
    hspace{5ex} textrm{(Leibniz series for $pi$)}\
    &=& frac{pi}{4}.
    end{eqnarray*}$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      this is another magic shot! Nice job! Thanks! :-)
      $endgroup$
      – user 1357113
      Jun 19 '12 at 5:52












    • $begingroup$
      @Chris: Thanks, Chris. Another good question!
      $endgroup$
      – user26872
      Jun 19 '12 at 7:40
















    15












    $begingroup$

    Another approach:
    $$begin{eqnarray*}
    int_{0}^{infty} dx, frac{e^{-x} sin(x)}{x}
    &=& int_{0}^{infty}dx, frac{e^{-x}}{x} sum_{k=0}^infty frac{(-1)^k x^{2k+1}}{(2k+1)!} \
    &=& sum_{k=0}^infty frac{(-1)^k}{(2k+1)!}
    int_{0}^{infty}dx, x^{2k} e^{-x} \
    &=& sum_{k=0}^infty frac{(-1)^k}{(2k+1)!}(2k)! \
    &=& sum_{k=0}^infty frac{(-1)^k}{2k+1}
    hspace{5ex} textrm{(Leibniz series for $pi$)}\
    &=& frac{pi}{4}.
    end{eqnarray*}$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      this is another magic shot! Nice job! Thanks! :-)
      $endgroup$
      – user 1357113
      Jun 19 '12 at 5:52












    • $begingroup$
      @Chris: Thanks, Chris. Another good question!
      $endgroup$
      – user26872
      Jun 19 '12 at 7:40














    15












    15








    15





    $begingroup$

    Another approach:
    $$begin{eqnarray*}
    int_{0}^{infty} dx, frac{e^{-x} sin(x)}{x}
    &=& int_{0}^{infty}dx, frac{e^{-x}}{x} sum_{k=0}^infty frac{(-1)^k x^{2k+1}}{(2k+1)!} \
    &=& sum_{k=0}^infty frac{(-1)^k}{(2k+1)!}
    int_{0}^{infty}dx, x^{2k} e^{-x} \
    &=& sum_{k=0}^infty frac{(-1)^k}{(2k+1)!}(2k)! \
    &=& sum_{k=0}^infty frac{(-1)^k}{2k+1}
    hspace{5ex} textrm{(Leibniz series for $pi$)}\
    &=& frac{pi}{4}.
    end{eqnarray*}$$






    share|cite|improve this answer









    $endgroup$



    Another approach:
    $$begin{eqnarray*}
    int_{0}^{infty} dx, frac{e^{-x} sin(x)}{x}
    &=& int_{0}^{infty}dx, frac{e^{-x}}{x} sum_{k=0}^infty frac{(-1)^k x^{2k+1}}{(2k+1)!} \
    &=& sum_{k=0}^infty frac{(-1)^k}{(2k+1)!}
    int_{0}^{infty}dx, x^{2k} e^{-x} \
    &=& sum_{k=0}^infty frac{(-1)^k}{(2k+1)!}(2k)! \
    &=& sum_{k=0}^infty frac{(-1)^k}{2k+1}
    hspace{5ex} textrm{(Leibniz series for $pi$)}\
    &=& frac{pi}{4}.
    end{eqnarray*}$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jun 19 '12 at 0:13









    user26872user26872

    14.9k22773




    14.9k22773












    • $begingroup$
      this is another magic shot! Nice job! Thanks! :-)
      $endgroup$
      – user 1357113
      Jun 19 '12 at 5:52












    • $begingroup$
      @Chris: Thanks, Chris. Another good question!
      $endgroup$
      – user26872
      Jun 19 '12 at 7:40


















    • $begingroup$
      this is another magic shot! Nice job! Thanks! :-)
      $endgroup$
      – user 1357113
      Jun 19 '12 at 5:52












    • $begingroup$
      @Chris: Thanks, Chris. Another good question!
      $endgroup$
      – user26872
      Jun 19 '12 at 7:40
















    $begingroup$
    this is another magic shot! Nice job! Thanks! :-)
    $endgroup$
    – user 1357113
    Jun 19 '12 at 5:52






    $begingroup$
    this is another magic shot! Nice job! Thanks! :-)
    $endgroup$
    – user 1357113
    Jun 19 '12 at 5:52














    $begingroup$
    @Chris: Thanks, Chris. Another good question!
    $endgroup$
    – user26872
    Jun 19 '12 at 7:40




    $begingroup$
    @Chris: Thanks, Chris. Another good question!
    $endgroup$
    – user26872
    Jun 19 '12 at 7:40











    13












    $begingroup$

    Write this as
    $$
    lim_{epsilonto0}int_epsilon^{1/epsilon}frac{e^{-(1-i)x}-e^{-(1+i)x}}{2ix},mathrm{d}xtag{1}
    $$
    and then consider the path integral
    $$
    frac{1}{2i}int_{gamma_epsilon} e^{-z},frac{mathrm{d}z}{z}tag{2}
    $$
    where $gamma_epsilon$ comes in along the line $(1+i)x$, makes a quarter circle clockwise along $|z|=epsilon$, goes out along the line $(1-i)x$ and then back a quarter circle counter-clockwise along $|z|=1/epsilon$. There are no poles inside this path, so the integral in $(2)$ is $0$.



    The part along $|z|=1/epsilon$ dies away exponentially as $epsilonto0$. The two parts along the lines sum to our integral, $(1)$, and the part along $|z|=epsilon$ tends to $frac14$ of the integral of $frac{1}{2iz}$ clockwise around the origin; that is, $-pi/4$. Since the sum of these parts is $0$, the limit in $(1)$ must be $pi/4$. That is,
    $$
    int_0^inftyfrac{e^{-x}sin(x)}{x}mathrm{d}x=frac{pi}{4}tag{3}
    $$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      @robjohn that's a nice solution! But I think that you should also justify that principal value of the integral equals integral itself.
      $endgroup$
      – qoqosz
      Jun 18 '12 at 11:08










    • $begingroup$
      @qoqosz: I am not using a principal value since $frac{sin(x)}{x}$ is a nice bounded function. I use the $epsilon$ and $1/epsilon$ to connect with the contour integral. Perhaps I am missing your concern.
      $endgroup$
      – robjohn♦
      Jun 18 '12 at 11:17










    • $begingroup$
      @robjohn $frac{x}{1+x^2}$ is also bounded but when integrating over $mathbb{R}$ it exists only in P.V. sense. Well, maybe it's not a 'nice' function ;).
      $endgroup$
      – qoqosz
      Jun 18 '12 at 11:21










    • $begingroup$
      @qoqosz: $e^{-x}frac{sin(x)}{x}$ is a nice bounded function that dies away exponentially. It converges absolutely over $[0,infty)$. In fact, all of the integrals above converge absolutely. What am I missing?
      $endgroup$
      – robjohn♦
      Jun 18 '12 at 11:31






    • 1




      $begingroup$
      @qoqosz: Ah, I just saw the improper-integral tag. In the Riemann integral sense, it is improper because the domain of integration is not bounded, but it is not improper in the Lebesgue sense.
      $endgroup$
      – robjohn♦
      Jun 18 '12 at 11:36


















    13












    $begingroup$

    Write this as
    $$
    lim_{epsilonto0}int_epsilon^{1/epsilon}frac{e^{-(1-i)x}-e^{-(1+i)x}}{2ix},mathrm{d}xtag{1}
    $$
    and then consider the path integral
    $$
    frac{1}{2i}int_{gamma_epsilon} e^{-z},frac{mathrm{d}z}{z}tag{2}
    $$
    where $gamma_epsilon$ comes in along the line $(1+i)x$, makes a quarter circle clockwise along $|z|=epsilon$, goes out along the line $(1-i)x$ and then back a quarter circle counter-clockwise along $|z|=1/epsilon$. There are no poles inside this path, so the integral in $(2)$ is $0$.



    The part along $|z|=1/epsilon$ dies away exponentially as $epsilonto0$. The two parts along the lines sum to our integral, $(1)$, and the part along $|z|=epsilon$ tends to $frac14$ of the integral of $frac{1}{2iz}$ clockwise around the origin; that is, $-pi/4$. Since the sum of these parts is $0$, the limit in $(1)$ must be $pi/4$. That is,
    $$
    int_0^inftyfrac{e^{-x}sin(x)}{x}mathrm{d}x=frac{pi}{4}tag{3}
    $$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      @robjohn that's a nice solution! But I think that you should also justify that principal value of the integral equals integral itself.
      $endgroup$
      – qoqosz
      Jun 18 '12 at 11:08










    • $begingroup$
      @qoqosz: I am not using a principal value since $frac{sin(x)}{x}$ is a nice bounded function. I use the $epsilon$ and $1/epsilon$ to connect with the contour integral. Perhaps I am missing your concern.
      $endgroup$
      – robjohn♦
      Jun 18 '12 at 11:17










    • $begingroup$
      @robjohn $frac{x}{1+x^2}$ is also bounded but when integrating over $mathbb{R}$ it exists only in P.V. sense. Well, maybe it's not a 'nice' function ;).
      $endgroup$
      – qoqosz
      Jun 18 '12 at 11:21










    • $begingroup$
      @qoqosz: $e^{-x}frac{sin(x)}{x}$ is a nice bounded function that dies away exponentially. It converges absolutely over $[0,infty)$. In fact, all of the integrals above converge absolutely. What am I missing?
      $endgroup$
      – robjohn♦
      Jun 18 '12 at 11:31






    • 1




      $begingroup$
      @qoqosz: Ah, I just saw the improper-integral tag. In the Riemann integral sense, it is improper because the domain of integration is not bounded, but it is not improper in the Lebesgue sense.
      $endgroup$
      – robjohn♦
      Jun 18 '12 at 11:36
















    13












    13








    13





    $begingroup$

    Write this as
    $$
    lim_{epsilonto0}int_epsilon^{1/epsilon}frac{e^{-(1-i)x}-e^{-(1+i)x}}{2ix},mathrm{d}xtag{1}
    $$
    and then consider the path integral
    $$
    frac{1}{2i}int_{gamma_epsilon} e^{-z},frac{mathrm{d}z}{z}tag{2}
    $$
    where $gamma_epsilon$ comes in along the line $(1+i)x$, makes a quarter circle clockwise along $|z|=epsilon$, goes out along the line $(1-i)x$ and then back a quarter circle counter-clockwise along $|z|=1/epsilon$. There are no poles inside this path, so the integral in $(2)$ is $0$.



    The part along $|z|=1/epsilon$ dies away exponentially as $epsilonto0$. The two parts along the lines sum to our integral, $(1)$, and the part along $|z|=epsilon$ tends to $frac14$ of the integral of $frac{1}{2iz}$ clockwise around the origin; that is, $-pi/4$. Since the sum of these parts is $0$, the limit in $(1)$ must be $pi/4$. That is,
    $$
    int_0^inftyfrac{e^{-x}sin(x)}{x}mathrm{d}x=frac{pi}{4}tag{3}
    $$






    share|cite|improve this answer









    $endgroup$



    Write this as
    $$
    lim_{epsilonto0}int_epsilon^{1/epsilon}frac{e^{-(1-i)x}-e^{-(1+i)x}}{2ix},mathrm{d}xtag{1}
    $$
    and then consider the path integral
    $$
    frac{1}{2i}int_{gamma_epsilon} e^{-z},frac{mathrm{d}z}{z}tag{2}
    $$
    where $gamma_epsilon$ comes in along the line $(1+i)x$, makes a quarter circle clockwise along $|z|=epsilon$, goes out along the line $(1-i)x$ and then back a quarter circle counter-clockwise along $|z|=1/epsilon$. There are no poles inside this path, so the integral in $(2)$ is $0$.



    The part along $|z|=1/epsilon$ dies away exponentially as $epsilonto0$. The two parts along the lines sum to our integral, $(1)$, and the part along $|z|=epsilon$ tends to $frac14$ of the integral of $frac{1}{2iz}$ clockwise around the origin; that is, $-pi/4$. Since the sum of these parts is $0$, the limit in $(1)$ must be $pi/4$. That is,
    $$
    int_0^inftyfrac{e^{-x}sin(x)}{x}mathrm{d}x=frac{pi}{4}tag{3}
    $$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jun 18 '12 at 10:45









    robjohn♦robjohn

    266k27306630




    266k27306630












    • $begingroup$
      @robjohn that's a nice solution! But I think that you should also justify that principal value of the integral equals integral itself.
      $endgroup$
      – qoqosz
      Jun 18 '12 at 11:08










    • $begingroup$
      @qoqosz: I am not using a principal value since $frac{sin(x)}{x}$ is a nice bounded function. I use the $epsilon$ and $1/epsilon$ to connect with the contour integral. Perhaps I am missing your concern.
      $endgroup$
      – robjohn♦
      Jun 18 '12 at 11:17










    • $begingroup$
      @robjohn $frac{x}{1+x^2}$ is also bounded but when integrating over $mathbb{R}$ it exists only in P.V. sense. Well, maybe it's not a 'nice' function ;).
      $endgroup$
      – qoqosz
      Jun 18 '12 at 11:21










    • $begingroup$
      @qoqosz: $e^{-x}frac{sin(x)}{x}$ is a nice bounded function that dies away exponentially. It converges absolutely over $[0,infty)$. In fact, all of the integrals above converge absolutely. What am I missing?
      $endgroup$
      – robjohn♦
      Jun 18 '12 at 11:31






    • 1




      $begingroup$
      @qoqosz: Ah, I just saw the improper-integral tag. In the Riemann integral sense, it is improper because the domain of integration is not bounded, but it is not improper in the Lebesgue sense.
      $endgroup$
      – robjohn♦
      Jun 18 '12 at 11:36




















    • $begingroup$
      @robjohn that's a nice solution! But I think that you should also justify that principal value of the integral equals integral itself.
      $endgroup$
      – qoqosz
      Jun 18 '12 at 11:08










    • $begingroup$
      @qoqosz: I am not using a principal value since $frac{sin(x)}{x}$ is a nice bounded function. I use the $epsilon$ and $1/epsilon$ to connect with the contour integral. Perhaps I am missing your concern.
      $endgroup$
      – robjohn♦
      Jun 18 '12 at 11:17










    • $begingroup$
      @robjohn $frac{x}{1+x^2}$ is also bounded but when integrating over $mathbb{R}$ it exists only in P.V. sense. Well, maybe it's not a 'nice' function ;).
      $endgroup$
      – qoqosz
      Jun 18 '12 at 11:21










    • $begingroup$
      @qoqosz: $e^{-x}frac{sin(x)}{x}$ is a nice bounded function that dies away exponentially. It converges absolutely over $[0,infty)$. In fact, all of the integrals above converge absolutely. What am I missing?
      $endgroup$
      – robjohn♦
      Jun 18 '12 at 11:31






    • 1




      $begingroup$
      @qoqosz: Ah, I just saw the improper-integral tag. In the Riemann integral sense, it is improper because the domain of integration is not bounded, but it is not improper in the Lebesgue sense.
      $endgroup$
      – robjohn♦
      Jun 18 '12 at 11:36


















    $begingroup$
    @robjohn that's a nice solution! But I think that you should also justify that principal value of the integral equals integral itself.
    $endgroup$
    – qoqosz
    Jun 18 '12 at 11:08




    $begingroup$
    @robjohn that's a nice solution! But I think that you should also justify that principal value of the integral equals integral itself.
    $endgroup$
    – qoqosz
    Jun 18 '12 at 11:08












    $begingroup$
    @qoqosz: I am not using a principal value since $frac{sin(x)}{x}$ is a nice bounded function. I use the $epsilon$ and $1/epsilon$ to connect with the contour integral. Perhaps I am missing your concern.
    $endgroup$
    – robjohn♦
    Jun 18 '12 at 11:17




    $begingroup$
    @qoqosz: I am not using a principal value since $frac{sin(x)}{x}$ is a nice bounded function. I use the $epsilon$ and $1/epsilon$ to connect with the contour integral. Perhaps I am missing your concern.
    $endgroup$
    – robjohn♦
    Jun 18 '12 at 11:17












    $begingroup$
    @robjohn $frac{x}{1+x^2}$ is also bounded but when integrating over $mathbb{R}$ it exists only in P.V. sense. Well, maybe it's not a 'nice' function ;).
    $endgroup$
    – qoqosz
    Jun 18 '12 at 11:21




    $begingroup$
    @robjohn $frac{x}{1+x^2}$ is also bounded but when integrating over $mathbb{R}$ it exists only in P.V. sense. Well, maybe it's not a 'nice' function ;).
    $endgroup$
    – qoqosz
    Jun 18 '12 at 11:21












    $begingroup$
    @qoqosz: $e^{-x}frac{sin(x)}{x}$ is a nice bounded function that dies away exponentially. It converges absolutely over $[0,infty)$. In fact, all of the integrals above converge absolutely. What am I missing?
    $endgroup$
    – robjohn♦
    Jun 18 '12 at 11:31




    $begingroup$
    @qoqosz: $e^{-x}frac{sin(x)}{x}$ is a nice bounded function that dies away exponentially. It converges absolutely over $[0,infty)$. In fact, all of the integrals above converge absolutely. What am I missing?
    $endgroup$
    – robjohn♦
    Jun 18 '12 at 11:31




    1




    1




    $begingroup$
    @qoqosz: Ah, I just saw the improper-integral tag. In the Riemann integral sense, it is improper because the domain of integration is not bounded, but it is not improper in the Lebesgue sense.
    $endgroup$
    – robjohn♦
    Jun 18 '12 at 11:36






    $begingroup$
    @qoqosz: Ah, I just saw the improper-integral tag. In the Riemann integral sense, it is improper because the domain of integration is not bounded, but it is not improper in the Lebesgue sense.
    $endgroup$
    – robjohn♦
    Jun 18 '12 at 11:36













    9












    $begingroup$

    If you know a bit about Fourier theory. You could Parseval's theorem
    $$int !dx ,f(x) g(x)^* = int !dxi,hat f(xi) hat g(xi)^* $$
    with $f(x) = sin(x)/x$, $g(x) = Theta(x) e^{-x}$ and $hat{f}$, $hat{g}$ their Fourier transforms and $Theta(x)$ the Heaviside step function.



    Hint: $hat{f}(xi) = tfrac12sqrt{frac{pi}{2}} [Theta(1-xi) + Theta(1+xi)] =sqrt{frac{pi}{2}} mathop{rm rect}(xi) $.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      @Anon: Sorry for not defining: $Theta(x)$ is the Heaviside step function.
      $endgroup$
      – Fabian
      Jun 18 '12 at 10:24


















    9












    $begingroup$

    If you know a bit about Fourier theory. You could Parseval's theorem
    $$int !dx ,f(x) g(x)^* = int !dxi,hat f(xi) hat g(xi)^* $$
    with $f(x) = sin(x)/x$, $g(x) = Theta(x) e^{-x}$ and $hat{f}$, $hat{g}$ their Fourier transforms and $Theta(x)$ the Heaviside step function.



    Hint: $hat{f}(xi) = tfrac12sqrt{frac{pi}{2}} [Theta(1-xi) + Theta(1+xi)] =sqrt{frac{pi}{2}} mathop{rm rect}(xi) $.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      @Anon: Sorry for not defining: $Theta(x)$ is the Heaviside step function.
      $endgroup$
      – Fabian
      Jun 18 '12 at 10:24
















    9












    9








    9





    $begingroup$

    If you know a bit about Fourier theory. You could Parseval's theorem
    $$int !dx ,f(x) g(x)^* = int !dxi,hat f(xi) hat g(xi)^* $$
    with $f(x) = sin(x)/x$, $g(x) = Theta(x) e^{-x}$ and $hat{f}$, $hat{g}$ their Fourier transforms and $Theta(x)$ the Heaviside step function.



    Hint: $hat{f}(xi) = tfrac12sqrt{frac{pi}{2}} [Theta(1-xi) + Theta(1+xi)] =sqrt{frac{pi}{2}} mathop{rm rect}(xi) $.






    share|cite|improve this answer









    $endgroup$



    If you know a bit about Fourier theory. You could Parseval's theorem
    $$int !dx ,f(x) g(x)^* = int !dxi,hat f(xi) hat g(xi)^* $$
    with $f(x) = sin(x)/x$, $g(x) = Theta(x) e^{-x}$ and $hat{f}$, $hat{g}$ their Fourier transforms and $Theta(x)$ the Heaviside step function.



    Hint: $hat{f}(xi) = tfrac12sqrt{frac{pi}{2}} [Theta(1-xi) + Theta(1+xi)] =sqrt{frac{pi}{2}} mathop{rm rect}(xi) $.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jun 18 '12 at 10:20









    FabianFabian

    19.7k3674




    19.7k3674












    • $begingroup$
      @Anon: Sorry for not defining: $Theta(x)$ is the Heaviside step function.
      $endgroup$
      – Fabian
      Jun 18 '12 at 10:24




















    • $begingroup$
      @Anon: Sorry for not defining: $Theta(x)$ is the Heaviside step function.
      $endgroup$
      – Fabian
      Jun 18 '12 at 10:24


















    $begingroup$
    @Anon: Sorry for not defining: $Theta(x)$ is the Heaviside step function.
    $endgroup$
    – Fabian
    Jun 18 '12 at 10:24






    $begingroup$
    @Anon: Sorry for not defining: $Theta(x)$ is the Heaviside step function.
    $endgroup$
    – Fabian
    Jun 18 '12 at 10:24













    5












    $begingroup$

    Let



    $$f(z) = frac{e^{-z+iz}}{z}$$



    and let $C$ be the contour that travels along $0$ to $R$, makes a quarter of a circle around to $iR$ and back to $0$, properly indented around $0$ with a quarter circle of radius $delta$ to avoid the pole.



    Contour



    As $R to infty$, the integral over rounded part of the contour tends to $0$ and the part around $0$ tends to $-ifrac{pi}{2}$ (N.B. this is $-ifrac{pi}{2}$ of the residue at $z=0$) as $delta to 0$. Then by Cauchy's theorem:



    $$
    0=oint_C f(z),dz =\
    int_0^infty frac{e^{-z+iz}}{z},dz -int_0^{infty} frac{e^{-iz-z}}{z},dz - ifrac{pi}{2}
    $$



    And upon taking imaginary parts and solving:



    $$
    frac{pi}{4}=int_0^infty frac{e^{-z}sin(z)}{z},dx
    $$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      hehe (+1). Glad to have the opportunity to learn more on complex analysis. :-)
      $endgroup$
      – user 1357113
      Jul 4 '13 at 16:06










    • $begingroup$
      @Chris'swisesister I thought you may be interested in a slightly different contour for variety
      $endgroup$
      – Argon
      Jul 4 '13 at 16:06










    • $begingroup$
      sure. I'm always interested in various ways. Thanks! :-)
      $endgroup$
      – user 1357113
      Jul 4 '13 at 16:07










    • $begingroup$
      Could you also add the contour (a picture) in case you have one ready?
      $endgroup$
      – user 1357113
      Jul 4 '13 at 16:09












    • $begingroup$
      @Chris'swisesister I could try and draw one, I will update this soon.
      $endgroup$
      – Argon
      Jul 4 '13 at 16:11
















    5












    $begingroup$

    Let



    $$f(z) = frac{e^{-z+iz}}{z}$$



    and let $C$ be the contour that travels along $0$ to $R$, makes a quarter of a circle around to $iR$ and back to $0$, properly indented around $0$ with a quarter circle of radius $delta$ to avoid the pole.



    Contour



    As $R to infty$, the integral over rounded part of the contour tends to $0$ and the part around $0$ tends to $-ifrac{pi}{2}$ (N.B. this is $-ifrac{pi}{2}$ of the residue at $z=0$) as $delta to 0$. Then by Cauchy's theorem:



    $$
    0=oint_C f(z),dz =\
    int_0^infty frac{e^{-z+iz}}{z},dz -int_0^{infty} frac{e^{-iz-z}}{z},dz - ifrac{pi}{2}
    $$



    And upon taking imaginary parts and solving:



    $$
    frac{pi}{4}=int_0^infty frac{e^{-z}sin(z)}{z},dx
    $$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      hehe (+1). Glad to have the opportunity to learn more on complex analysis. :-)
      $endgroup$
      – user 1357113
      Jul 4 '13 at 16:06










    • $begingroup$
      @Chris'swisesister I thought you may be interested in a slightly different contour for variety
      $endgroup$
      – Argon
      Jul 4 '13 at 16:06










    • $begingroup$
      sure. I'm always interested in various ways. Thanks! :-)
      $endgroup$
      – user 1357113
      Jul 4 '13 at 16:07










    • $begingroup$
      Could you also add the contour (a picture) in case you have one ready?
      $endgroup$
      – user 1357113
      Jul 4 '13 at 16:09












    • $begingroup$
      @Chris'swisesister I could try and draw one, I will update this soon.
      $endgroup$
      – Argon
      Jul 4 '13 at 16:11














    5












    5








    5





    $begingroup$

    Let



    $$f(z) = frac{e^{-z+iz}}{z}$$



    and let $C$ be the contour that travels along $0$ to $R$, makes a quarter of a circle around to $iR$ and back to $0$, properly indented around $0$ with a quarter circle of radius $delta$ to avoid the pole.



    Contour



    As $R to infty$, the integral over rounded part of the contour tends to $0$ and the part around $0$ tends to $-ifrac{pi}{2}$ (N.B. this is $-ifrac{pi}{2}$ of the residue at $z=0$) as $delta to 0$. Then by Cauchy's theorem:



    $$
    0=oint_C f(z),dz =\
    int_0^infty frac{e^{-z+iz}}{z},dz -int_0^{infty} frac{e^{-iz-z}}{z},dz - ifrac{pi}{2}
    $$



    And upon taking imaginary parts and solving:



    $$
    frac{pi}{4}=int_0^infty frac{e^{-z}sin(z)}{z},dx
    $$






    share|cite|improve this answer











    $endgroup$



    Let



    $$f(z) = frac{e^{-z+iz}}{z}$$



    and let $C$ be the contour that travels along $0$ to $R$, makes a quarter of a circle around to $iR$ and back to $0$, properly indented around $0$ with a quarter circle of radius $delta$ to avoid the pole.



    Contour



    As $R to infty$, the integral over rounded part of the contour tends to $0$ and the part around $0$ tends to $-ifrac{pi}{2}$ (N.B. this is $-ifrac{pi}{2}$ of the residue at $z=0$) as $delta to 0$. Then by Cauchy's theorem:



    $$
    0=oint_C f(z),dz =\
    int_0^infty frac{e^{-z+iz}}{z},dz -int_0^{infty} frac{e^{-iz-z}}{z},dz - ifrac{pi}{2}
    $$



    And upon taking imaginary parts and solving:



    $$
    frac{pi}{4}=int_0^infty frac{e^{-z}sin(z)}{z},dx
    $$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jul 4 '13 at 16:40

























    answered Jul 4 '13 at 15:57









    ArgonArgon

    16.3k673122




    16.3k673122












    • $begingroup$
      hehe (+1). Glad to have the opportunity to learn more on complex analysis. :-)
      $endgroup$
      – user 1357113
      Jul 4 '13 at 16:06










    • $begingroup$
      @Chris'swisesister I thought you may be interested in a slightly different contour for variety
      $endgroup$
      – Argon
      Jul 4 '13 at 16:06










    • $begingroup$
      sure. I'm always interested in various ways. Thanks! :-)
      $endgroup$
      – user 1357113
      Jul 4 '13 at 16:07










    • $begingroup$
      Could you also add the contour (a picture) in case you have one ready?
      $endgroup$
      – user 1357113
      Jul 4 '13 at 16:09












    • $begingroup$
      @Chris'swisesister I could try and draw one, I will update this soon.
      $endgroup$
      – Argon
      Jul 4 '13 at 16:11


















    • $begingroup$
      hehe (+1). Glad to have the opportunity to learn more on complex analysis. :-)
      $endgroup$
      – user 1357113
      Jul 4 '13 at 16:06










    • $begingroup$
      @Chris'swisesister I thought you may be interested in a slightly different contour for variety
      $endgroup$
      – Argon
      Jul 4 '13 at 16:06










    • $begingroup$
      sure. I'm always interested in various ways. Thanks! :-)
      $endgroup$
      – user 1357113
      Jul 4 '13 at 16:07










    • $begingroup$
      Could you also add the contour (a picture) in case you have one ready?
      $endgroup$
      – user 1357113
      Jul 4 '13 at 16:09












    • $begingroup$
      @Chris'swisesister I could try and draw one, I will update this soon.
      $endgroup$
      – Argon
      Jul 4 '13 at 16:11
















    $begingroup$
    hehe (+1). Glad to have the opportunity to learn more on complex analysis. :-)
    $endgroup$
    – user 1357113
    Jul 4 '13 at 16:06




    $begingroup$
    hehe (+1). Glad to have the opportunity to learn more on complex analysis. :-)
    $endgroup$
    – user 1357113
    Jul 4 '13 at 16:06












    $begingroup$
    @Chris'swisesister I thought you may be interested in a slightly different contour for variety
    $endgroup$
    – Argon
    Jul 4 '13 at 16:06




    $begingroup$
    @Chris'swisesister I thought you may be interested in a slightly different contour for variety
    $endgroup$
    – Argon
    Jul 4 '13 at 16:06












    $begingroup$
    sure. I'm always interested in various ways. Thanks! :-)
    $endgroup$
    – user 1357113
    Jul 4 '13 at 16:07




    $begingroup$
    sure. I'm always interested in various ways. Thanks! :-)
    $endgroup$
    – user 1357113
    Jul 4 '13 at 16:07












    $begingroup$
    Could you also add the contour (a picture) in case you have one ready?
    $endgroup$
    – user 1357113
    Jul 4 '13 at 16:09






    $begingroup$
    Could you also add the contour (a picture) in case you have one ready?
    $endgroup$
    – user 1357113
    Jul 4 '13 at 16:09














    $begingroup$
    @Chris'swisesister I could try and draw one, I will update this soon.
    $endgroup$
    – Argon
    Jul 4 '13 at 16:11




    $begingroup$
    @Chris'swisesister I could try and draw one, I will update this soon.
    $endgroup$
    – Argon
    Jul 4 '13 at 16:11











    3












    $begingroup$

    $$int_0^{infty} frac{e^{-x}sin x}{x},dx=int_0^{infty}int_0^{infty} e^{-x}sin x ,e^{-xy},dy,dx=int_0^{infty} int_0^{infty} e^{-x(1+y)}sin x,dx,dy$$
    From integration by parts or otherwise, one can show that:
    $$int_0^{infty}e^{-x(1+y)}sin x,dx=frac{1}{1+(1+y)^2}$$
    Hence
    $$int_0^{infty} int_0^{infty} e^{-x(1+y)}sin x,dx,dy=int_0^{infty} frac{dy}{1+(1+y)^2}=left(arctan(1+y)right|_0^{infty}=boxed{dfrac{pi}{4}}$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Neat, but isn't that more or less the same as Julián Aguirre's answer?
      $endgroup$
      – IAmNoOne
      May 9 '14 at 0:12
















    3












    $begingroup$

    $$int_0^{infty} frac{e^{-x}sin x}{x},dx=int_0^{infty}int_0^{infty} e^{-x}sin x ,e^{-xy},dy,dx=int_0^{infty} int_0^{infty} e^{-x(1+y)}sin x,dx,dy$$
    From integration by parts or otherwise, one can show that:
    $$int_0^{infty}e^{-x(1+y)}sin x,dx=frac{1}{1+(1+y)^2}$$
    Hence
    $$int_0^{infty} int_0^{infty} e^{-x(1+y)}sin x,dx,dy=int_0^{infty} frac{dy}{1+(1+y)^2}=left(arctan(1+y)right|_0^{infty}=boxed{dfrac{pi}{4}}$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Neat, but isn't that more or less the same as Julián Aguirre's answer?
      $endgroup$
      – IAmNoOne
      May 9 '14 at 0:12














    3












    3








    3





    $begingroup$

    $$int_0^{infty} frac{e^{-x}sin x}{x},dx=int_0^{infty}int_0^{infty} e^{-x}sin x ,e^{-xy},dy,dx=int_0^{infty} int_0^{infty} e^{-x(1+y)}sin x,dx,dy$$
    From integration by parts or otherwise, one can show that:
    $$int_0^{infty}e^{-x(1+y)}sin x,dx=frac{1}{1+(1+y)^2}$$
    Hence
    $$int_0^{infty} int_0^{infty} e^{-x(1+y)}sin x,dx,dy=int_0^{infty} frac{dy}{1+(1+y)^2}=left(arctan(1+y)right|_0^{infty}=boxed{dfrac{pi}{4}}$$






    share|cite|improve this answer









    $endgroup$



    $$int_0^{infty} frac{e^{-x}sin x}{x},dx=int_0^{infty}int_0^{infty} e^{-x}sin x ,e^{-xy},dy,dx=int_0^{infty} int_0^{infty} e^{-x(1+y)}sin x,dx,dy$$
    From integration by parts or otherwise, one can show that:
    $$int_0^{infty}e^{-x(1+y)}sin x,dx=frac{1}{1+(1+y)^2}$$
    Hence
    $$int_0^{infty} int_0^{infty} e^{-x(1+y)}sin x,dx,dy=int_0^{infty} frac{dy}{1+(1+y)^2}=left(arctan(1+y)right|_0^{infty}=boxed{dfrac{pi}{4}}$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered May 8 '14 at 15:32









    Pranav AroraPranav Arora

    8,5472562




    8,5472562












    • $begingroup$
      Neat, but isn't that more or less the same as Julián Aguirre's answer?
      $endgroup$
      – IAmNoOne
      May 9 '14 at 0:12


















    • $begingroup$
      Neat, but isn't that more or less the same as Julián Aguirre's answer?
      $endgroup$
      – IAmNoOne
      May 9 '14 at 0:12
















    $begingroup$
    Neat, but isn't that more or less the same as Julián Aguirre's answer?
    $endgroup$
    – IAmNoOne
    May 9 '14 at 0:12




    $begingroup$
    Neat, but isn't that more or less the same as Julián Aguirre's answer?
    $endgroup$
    – IAmNoOne
    May 9 '14 at 0:12











    2












    $begingroup$

    $newcommand{+}{^{dagger}}%
    newcommand{angles}[1]{leftlangle #1 rightrangle}%
    newcommand{braces}[1]{leftlbrace #1 rightrbrace}%
    newcommand{bracks}[1]{leftlbrack #1 rightrbrack}%
    newcommand{ceil}[1]{,leftlceil #1 rightrceil,}%
    newcommand{dd}{{rm d}}%
    newcommand{down}{downarrow}%
    newcommand{ds}[1]{displaystyle{#1}}%
    newcommand{equalby}[1]{{#1 atop {= atop vphantom{huge A}}}}%
    newcommand{expo}[1]{,{rm e}^{#1},}%
    newcommand{fermi}{,{rm f}}%
    newcommand{floor}[1]{,leftlfloor #1 rightrfloor,}%
    newcommand{half}{{1 over 2}}%
    newcommand{ic}{{rm i}}%
    newcommand{iff}{Longleftrightarrow}
    newcommand{imp}{Longrightarrow}%
    newcommand{isdiv}{,left.rightvert,}%
    newcommand{ket}[1]{leftvert #1rightrangle}%
    newcommand{ol}[1]{overline{#1}}%
    newcommand{pars}[1]{left( #1 right)}%
    newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
    newcommand{pp}{{cal P}}%
    newcommand{root}[2]{,sqrt[#1]{,#2,},}%
    newcommand{sech}{,{rm sech}}%
    newcommand{sgn}{,{rm sgn}}%
    newcommand{totald}[3]{frac{{rm d}^{#1} #2}{{rm d} #3^{#1}}}
    newcommand{ul}[1]{underline{#1}}%
    newcommand{verts}[1]{leftvert, #1 ,rightvert}$
    begin{align}
    &color{#00f}{largeint_{0}^{infty}{expo{-x}sinpars{x} over x},dd x}
    =int_{0}^{infty}expo{-x}pars{halfint_{-1}^{1}expo{ic kx},dd k}
    =halfint_{-1}^{1}int_{0}^{infty}expo{pars{ic k - 1}x},dd k
    \[3mm]&=halfint_{-1}^{1}{1 over 1 - ic k},dd k
    =int_{0}^{1}{dd k over 1 + k^{2}} = arctanpars{1}
    =color{#00f}{Large{pi over 4}}
    end{align}






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      $newcommand{+}{^{dagger}}%
      newcommand{angles}[1]{leftlangle #1 rightrangle}%
      newcommand{braces}[1]{leftlbrace #1 rightrbrace}%
      newcommand{bracks}[1]{leftlbrack #1 rightrbrack}%
      newcommand{ceil}[1]{,leftlceil #1 rightrceil,}%
      newcommand{dd}{{rm d}}%
      newcommand{down}{downarrow}%
      newcommand{ds}[1]{displaystyle{#1}}%
      newcommand{equalby}[1]{{#1 atop {= atop vphantom{huge A}}}}%
      newcommand{expo}[1]{,{rm e}^{#1},}%
      newcommand{fermi}{,{rm f}}%
      newcommand{floor}[1]{,leftlfloor #1 rightrfloor,}%
      newcommand{half}{{1 over 2}}%
      newcommand{ic}{{rm i}}%
      newcommand{iff}{Longleftrightarrow}
      newcommand{imp}{Longrightarrow}%
      newcommand{isdiv}{,left.rightvert,}%
      newcommand{ket}[1]{leftvert #1rightrangle}%
      newcommand{ol}[1]{overline{#1}}%
      newcommand{pars}[1]{left( #1 right)}%
      newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
      newcommand{pp}{{cal P}}%
      newcommand{root}[2]{,sqrt[#1]{,#2,},}%
      newcommand{sech}{,{rm sech}}%
      newcommand{sgn}{,{rm sgn}}%
      newcommand{totald}[3]{frac{{rm d}^{#1} #2}{{rm d} #3^{#1}}}
      newcommand{ul}[1]{underline{#1}}%
      newcommand{verts}[1]{leftvert, #1 ,rightvert}$
      begin{align}
      &color{#00f}{largeint_{0}^{infty}{expo{-x}sinpars{x} over x},dd x}
      =int_{0}^{infty}expo{-x}pars{halfint_{-1}^{1}expo{ic kx},dd k}
      =halfint_{-1}^{1}int_{0}^{infty}expo{pars{ic k - 1}x},dd k
      \[3mm]&=halfint_{-1}^{1}{1 over 1 - ic k},dd k
      =int_{0}^{1}{dd k over 1 + k^{2}} = arctanpars{1}
      =color{#00f}{Large{pi over 4}}
      end{align}






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        $newcommand{+}{^{dagger}}%
        newcommand{angles}[1]{leftlangle #1 rightrangle}%
        newcommand{braces}[1]{leftlbrace #1 rightrbrace}%
        newcommand{bracks}[1]{leftlbrack #1 rightrbrack}%
        newcommand{ceil}[1]{,leftlceil #1 rightrceil,}%
        newcommand{dd}{{rm d}}%
        newcommand{down}{downarrow}%
        newcommand{ds}[1]{displaystyle{#1}}%
        newcommand{equalby}[1]{{#1 atop {= atop vphantom{huge A}}}}%
        newcommand{expo}[1]{,{rm e}^{#1},}%
        newcommand{fermi}{,{rm f}}%
        newcommand{floor}[1]{,leftlfloor #1 rightrfloor,}%
        newcommand{half}{{1 over 2}}%
        newcommand{ic}{{rm i}}%
        newcommand{iff}{Longleftrightarrow}
        newcommand{imp}{Longrightarrow}%
        newcommand{isdiv}{,left.rightvert,}%
        newcommand{ket}[1]{leftvert #1rightrangle}%
        newcommand{ol}[1]{overline{#1}}%
        newcommand{pars}[1]{left( #1 right)}%
        newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
        newcommand{pp}{{cal P}}%
        newcommand{root}[2]{,sqrt[#1]{,#2,},}%
        newcommand{sech}{,{rm sech}}%
        newcommand{sgn}{,{rm sgn}}%
        newcommand{totald}[3]{frac{{rm d}^{#1} #2}{{rm d} #3^{#1}}}
        newcommand{ul}[1]{underline{#1}}%
        newcommand{verts}[1]{leftvert, #1 ,rightvert}$
        begin{align}
        &color{#00f}{largeint_{0}^{infty}{expo{-x}sinpars{x} over x},dd x}
        =int_{0}^{infty}expo{-x}pars{halfint_{-1}^{1}expo{ic kx},dd k}
        =halfint_{-1}^{1}int_{0}^{infty}expo{pars{ic k - 1}x},dd k
        \[3mm]&=halfint_{-1}^{1}{1 over 1 - ic k},dd k
        =int_{0}^{1}{dd k over 1 + k^{2}} = arctanpars{1}
        =color{#00f}{Large{pi over 4}}
        end{align}






        share|cite|improve this answer











        $endgroup$



        $newcommand{+}{^{dagger}}%
        newcommand{angles}[1]{leftlangle #1 rightrangle}%
        newcommand{braces}[1]{leftlbrace #1 rightrbrace}%
        newcommand{bracks}[1]{leftlbrack #1 rightrbrack}%
        newcommand{ceil}[1]{,leftlceil #1 rightrceil,}%
        newcommand{dd}{{rm d}}%
        newcommand{down}{downarrow}%
        newcommand{ds}[1]{displaystyle{#1}}%
        newcommand{equalby}[1]{{#1 atop {= atop vphantom{huge A}}}}%
        newcommand{expo}[1]{,{rm e}^{#1},}%
        newcommand{fermi}{,{rm f}}%
        newcommand{floor}[1]{,leftlfloor #1 rightrfloor,}%
        newcommand{half}{{1 over 2}}%
        newcommand{ic}{{rm i}}%
        newcommand{iff}{Longleftrightarrow}
        newcommand{imp}{Longrightarrow}%
        newcommand{isdiv}{,left.rightvert,}%
        newcommand{ket}[1]{leftvert #1rightrangle}%
        newcommand{ol}[1]{overline{#1}}%
        newcommand{pars}[1]{left( #1 right)}%
        newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
        newcommand{pp}{{cal P}}%
        newcommand{root}[2]{,sqrt[#1]{,#2,},}%
        newcommand{sech}{,{rm sech}}%
        newcommand{sgn}{,{rm sgn}}%
        newcommand{totald}[3]{frac{{rm d}^{#1} #2}{{rm d} #3^{#1}}}
        newcommand{ul}[1]{underline{#1}}%
        newcommand{verts}[1]{leftvert, #1 ,rightvert}$
        begin{align}
        &color{#00f}{largeint_{0}^{infty}{expo{-x}sinpars{x} over x},dd x}
        =int_{0}^{infty}expo{-x}pars{halfint_{-1}^{1}expo{ic kx},dd k}
        =halfint_{-1}^{1}int_{0}^{infty}expo{pars{ic k - 1}x},dd k
        \[3mm]&=halfint_{-1}^{1}{1 over 1 - ic k},dd k
        =int_{0}^{1}{dd k over 1 + k^{2}} = arctanpars{1}
        =color{#00f}{Large{pi over 4}}
        end{align}







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited May 8 '14 at 5:09

























        answered Jan 28 '14 at 12:11









        Felix MarinFelix Marin

        67.5k7107141




        67.5k7107141






























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            Mario Kart Wii

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            Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro...
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            The Binding of Isaac: Rebirth/Afterbirth

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            Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! The Binding of Isaac: Rebirth/Afterbirth From The Cutting Room Floor < The Binding of Isaac: Rebirth Jump to: navigation, search This is a sub-page of The Binding of Isaac: Rebirth . Afterbirth is the first expansion pack for Rebirth and features a variety of improvements. Contents 1 Unused Graphics 1.1 Old Man Keeper 1.2 Skinless Hush 1.3 Miscellaneous Unused Hush Sprites 1.4 Generosity achievement 1.5 Afterbirth+ Room Editor 1.6 Bearded Ultra Greed 1.7 Early Ultra Greed 1.8 Unused Brownie Expression 1.9 Frail's Spinning Head 1.10 Mom's Dead Hand Alt 2 Unused Audio 2.1 Unused Dead Womb Layer 2.2 Unused Hush Battle Outro 2.3 The Stain Grunt Unused Graphics Old Man Keeper Image file: gfxuimain me...
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            What does “Dominus providebit” mean?

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            5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1 ...
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