$int_{0}^{infty} frac{e^{-x} sin(x)}{x} dx$ Evaluate Integral
$begingroup$
Compute the following integral:
$$int_{0}^{infty} frac{e^{-x} sin(x)}{x} dx$$
Any hint, suggestion is welcome.
calculus integration limits improper-integrals
edited Jan 12 at 6:49
Abcd
3,02821235
asked Jun 18 '12 at 10:02
user 1357113user 1357113
22.4k877226
$endgroup$
add a comment |
$begingroup$
Compute the following integral:
$$int_{0}^{infty} frac{e^{-x} sin(x)}{x} dx$$
Any hint, suggestion is welcome.
calculus integration limits improper-integrals
edited Jan 12 at 6:49
Abcd
3,02821235
asked Jun 18 '12 at 10:02
user 1357113user 1357113
22.4k877226
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
Compute the following integral:
$$int_{0}^{infty} frac{e^{-x} sin(x)}{x} dx$$
Any hint, suggestion is welcome.
calculus integration limits improper-integrals
edited Jan 12 at 6:49
Abcd
3,02821235
asked Jun 18 '12 at 10:02
user 1357113user 1357113
22.4k877226
$endgroup$
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
edited Jan 12 at 6:49
Abcd
3,02821235
asked Jun 18 '12 at 10:02
user 1357113user 1357113
22.4k877226
edited Jan 12 at 6:49
Abcd
3,02821235
asked Jun 18 '12 at 10:02
user 1357113user 1357113
22.4k877226
edited Jan 12 at 6:49
Abcd
3,02821235
edited Jan 12 at 6:49
Abcd
3,02821235
edited Jan 12 at 6:49
Abcd
3,02821235
3,02821235
asked Jun 18 '12 at 10:02
user 1357113user 1357113
22.4k877226
asked Jun 18 '12 at 10:02
user 1357113user 1357113
22.4k877226
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
add a comment |
$begingroup$
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
add a comment |
8 Answers
8
active
oldest
votes
$begingroup$
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
$$
int_{0}^{infty} frac{e^{-x} sin(x)}{x},dx=F(1)=fracpi4.
$$
answered Jun 18 '12 at 11:07
Julián AguirreJulián Aguirre
68.2k24094
$endgroup$
$begingroup$
@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
add a comment |
$begingroup$
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,
$$int_0^infty e^{-x} frac{sin(x)}{x} dx = frac{pi}{4}$$
answered Jun 18 '12 at 10:41
TenaliRamanTenaliRaman
3,1641223
$endgroup$
$begingroup$
thanks for your delicate solution!
$endgroup$
– user 1357113
Jun 18 '12 at 10:57
add a comment |
$begingroup$
Another approach:
$$begin{eqnarray*}
int_{0}^{infty} dx, frac{e^{-x} sin(x)}{x}
&=& int_{0}^{infty}dx, frac{e^{-x}}{x} sum_{k=0}^infty frac{(-1)^k x^{2k+1}}{(2k+1)!} \
&=& sum_{k=0}^infty frac{(-1)^k}{(2k+1)!}
int_{0}^{infty}dx, x^{2k} e^{-x} \
&=& sum_{k=0}^infty frac{(-1)^k}{(2k+1)!}(2k)! \
&=& sum_{k=0}^infty frac{(-1)^k}{2k+1}
hspace{5ex} textrm{(Leibniz series for $pi$)}\
&=& frac{pi}{4}.
end{eqnarray*}$$
answered Jun 19 '12 at 0:13
user26872user26872
14.9k22773
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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,
$$
int_0^inftyfrac{e^{-x}sin(x)}{x}mathrm{d}x=frac{pi}{4}tag{3}
$$
answered Jun 18 '12 at 10:45
robjohn♦robjohn
266k27306630
$endgroup$
$begingroup$
@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
add a comment |
$begingroup$
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) $.
answered Jun 18 '12 at 10:20
FabianFabian
19.7k3674
$endgroup$
$begingroup$
@Anon: Sorry for not defining: $Theta(x)$ is the Heaviside step function.
$endgroup$
– Fabian
Jun 18 '12 at 10:24
add a comment |
$begingroup$
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.
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:
$$
frac{pi}{4}=int_0^infty frac{e^{-z}sin(z)}{z},dx
$$
answered Jul 4 '13 at 15:57
ArgonArgon
16.3k673122
$endgroup$
$begingroup$
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
|
show 1 more comment
$begingroup$
$$int_0^{infty} frac{e^{-x}sin x}{x},dx=int_0^{infty}int_0^{infty} e^{-x}sin x ,e^{-xy},dy,dx=int_0^{infty} int_0^{infty} e^{-x(1+y)}sin x,dx,dy$$
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
$$int_0^{infty} int_0^{infty} e^{-x(1+y)}sin x,dx,dy=int_0^{infty} frac{dy}{1+(1+y)^2}=left(arctan(1+y)right|_0^{infty}=boxed{dfrac{pi}{4}}$$
answered May 8 '14 at 15:32
Pranav AroraPranav Arora
8,5472562
$endgroup$
$begingroup$
Neat, but isn't that more or less the same as Julián Aguirre's answer?
$endgroup$
– IAmNoOne
May 9 '14 at 0:12
add a comment |
$begingroup$
$newcommand{+}{^{dagger}}%
newcommand{angles}[1]{leftlangle #1 rightrangle}%
newcommand{braces}[1]{leftlbrace #1 rightrbrace}%
newcommand{bracks}[1]{leftlbrack #1 rightrbrack}%
newcommand{ceil}[1]{,leftlceil #1 rightrceil,}%
newcommand{dd}{{rm d}}%
newcommand{down}{downarrow}%
newcommand{ds}[1]{displaystyle{#1}}%
newcommand{equalby}[1]{{#1 atop {= atop vphantom{huge A}}}}%
newcommand{expo}[1]{,{rm e}^{#1},}%
newcommand{fermi}{,{rm f}}%
newcommand{floor}[1]{,leftlfloor #1 rightrfloor,}%
newcommand{half}{{1 over 2}}%
newcommand{ic}{{rm i}}%
newcommand{iff}{Longleftrightarrow}
newcommand{imp}{Longrightarrow}%
newcommand{isdiv}{,left.rightvert,}%
newcommand{ket}[1]{leftvert #1rightrangle}%
newcommand{ol}[1]{overline{#1}}%
newcommand{pars}[1]{left( #1 right)}%
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{pp}{{cal P}}%
newcommand{root}[2]{,sqrt[#1]{,#2,},}%
newcommand{sech}{,{rm sech}}%
newcommand{sgn}{,{rm sgn}}%
newcommand{totald}[3]{frac{{rm d}^{#1} #2}{{rm d} #3^{#1}}}
newcommand{ul}[1]{underline{#1}}%
newcommand{verts}[1]{leftvert, #1 ,rightvert}$
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}
answered Jan 28 '14 at 12:11
Felix MarinFelix Marin
67.5k7107141
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f159836%2fint-0-infty-frace-x-sinxx-dx-evaluate-integral%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
8 Answers
8
active
oldest
votes
8 Answers
8
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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
$$
int_{0}^{infty} frac{e^{-x} sin(x)}{x},dx=F(1)=fracpi4.
$$
answered Jun 18 '12 at 11:07
Julián AguirreJulián Aguirre
68.2k24094
$endgroup$
$begingroup$
@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
add a comment |
$begingroup$
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
$$
int_{0}^{infty} frac{e^{-x} sin(x)}{x},dx=F(1)=fracpi4.
$$
answered Jun 18 '12 at 11:07
Julián AguirreJulián Aguirre
68.2k24094
$endgroup$
$begingroup$
@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
add a comment |
$begingroup$
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
$$
int_{0}^{infty} frac{e^{-x} sin(x)}{x},dx=F(1)=fracpi4.
$$
answered Jun 18 '12 at 11:07
Julián AguirreJulián Aguirre
68.2k24094
$endgroup$
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
$$
int_{0}^{infty} frac{e^{-x} sin(x)}{x},dx=F(1)=fracpi4.
$$
answered Jun 18 '12 at 11:07
Julián AguirreJulián Aguirre
68.2k24094
answered Jun 18 '12 at 11:07
Julián AguirreJulián Aguirre
68.2k24094
answered Jun 18 '12 at 11:07
Julián AguirreJulián Aguirre
68.2k24094
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
add a comment |
$begingroup$
@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
add a comment |
$begingroup$
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,
$$int_0^infty e^{-x} frac{sin(x)}{x} dx = frac{pi}{4}$$
answered Jun 18 '12 at 10:41
TenaliRamanTenaliRaman
3,1641223
$endgroup$
$begingroup$
thanks for your delicate solution!
$endgroup$
– user 1357113
Jun 18 '12 at 10:57
add a comment |
$begingroup$
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,
$$int_0^infty e^{-x} frac{sin(x)}{x} dx = frac{pi}{4}$$
answered Jun 18 '12 at 10:41
TenaliRamanTenaliRaman
3,1641223
$endgroup$
$begingroup$
thanks for your delicate solution!
$endgroup$
– user 1357113
Jun 18 '12 at 10:57
add a comment |
$begingroup$
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,
$$int_0^infty e^{-x} frac{sin(x)}{x} dx = frac{pi}{4}$$
answered Jun 18 '12 at 10:41
TenaliRamanTenaliRaman
3,1641223
$endgroup$
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,
$$int_0^infty e^{-x} frac{sin(x)}{x} dx = frac{pi}{4}$$
answered Jun 18 '12 at 10:41
TenaliRamanTenaliRaman
3,1641223
edited Jun 18 '12 at 10:55
answered Jun 18 '12 at 10:41
TenaliRamanTenaliRaman
3,1641223
answered Jun 18 '12 at 10:41
TenaliRamanTenaliRaman
3,1641223
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
Another approach:
$$begin{eqnarray*}
int_{0}^{infty} dx, frac{e^{-x} sin(x)}{x}
&=& int_{0}^{infty}dx, frac{e^{-x}}{x} sum_{k=0}^infty frac{(-1)^k x^{2k+1}}{(2k+1)!} \
&=& sum_{k=0}^infty frac{(-1)^k}{(2k+1)!}
int_{0}^{infty}dx, x^{2k} e^{-x} \
&=& sum_{k=0}^infty frac{(-1)^k}{(2k+1)!}(2k)! \
&=& sum_{k=0}^infty frac{(-1)^k}{2k+1}
hspace{5ex} textrm{(Leibniz series for $pi$)}\
&=& frac{pi}{4}.
end{eqnarray*}$$
answered Jun 19 '12 at 0:13
user26872user26872
14.9k22773
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
Another approach:
$$begin{eqnarray*}
int_{0}^{infty} dx, frac{e^{-x} sin(x)}{x}
&=& int_{0}^{infty}dx, frac{e^{-x}}{x} sum_{k=0}^infty frac{(-1)^k x^{2k+1}}{(2k+1)!} \
&=& sum_{k=0}^infty frac{(-1)^k}{(2k+1)!}
int_{0}^{infty}dx, x^{2k} e^{-x} \
&=& sum_{k=0}^infty frac{(-1)^k}{(2k+1)!}(2k)! \
&=& sum_{k=0}^infty frac{(-1)^k}{2k+1}
hspace{5ex} textrm{(Leibniz series for $pi$)}\
&=& frac{pi}{4}.
end{eqnarray*}$$
answered Jun 19 '12 at 0:13
user26872user26872
14.9k22773
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
Another approach:
$$begin{eqnarray*}
int_{0}^{infty} dx, frac{e^{-x} sin(x)}{x}
&=& int_{0}^{infty}dx, frac{e^{-x}}{x} sum_{k=0}^infty frac{(-1)^k x^{2k+1}}{(2k+1)!} \
&=& sum_{k=0}^infty frac{(-1)^k}{(2k+1)!}
int_{0}^{infty}dx, x^{2k} e^{-x} \
&=& sum_{k=0}^infty frac{(-1)^k}{(2k+1)!}(2k)! \
&=& sum_{k=0}^infty frac{(-1)^k}{2k+1}
hspace{5ex} textrm{(Leibniz series for $pi$)}\
&=& frac{pi}{4}.
end{eqnarray*}$$
answered Jun 19 '12 at 0:13
user26872user26872
14.9k22773
$endgroup$
Another approach:
$$begin{eqnarray*}
int_{0}^{infty} dx, frac{e^{-x} sin(x)}{x}
&=& int_{0}^{infty}dx, frac{e^{-x}}{x} sum_{k=0}^infty frac{(-1)^k x^{2k+1}}{(2k+1)!} \
&=& sum_{k=0}^infty frac{(-1)^k}{(2k+1)!}
int_{0}^{infty}dx, x^{2k} e^{-x} \
&=& sum_{k=0}^infty frac{(-1)^k}{(2k+1)!}(2k)! \
&=& sum_{k=0}^infty frac{(-1)^k}{2k+1}
hspace{5ex} textrm{(Leibniz series for $pi$)}\
&=& frac{pi}{4}.
end{eqnarray*}$$
answered Jun 19 '12 at 0:13
user26872user26872
14.9k22773
answered Jun 19 '12 at 0:13
user26872user26872
14.9k22773
answered Jun 19 '12 at 0:13
user26872user26872
14.9k22773
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
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,
$$
int_0^inftyfrac{e^{-x}sin(x)}{x}mathrm{d}x=frac{pi}{4}tag{3}
$$
answered Jun 18 '12 at 10:45
robjohn♦robjohn
266k27306630
$endgroup$
$begingroup$
@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
add a comment |
$begingroup$
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,
$$
int_0^inftyfrac{e^{-x}sin(x)}{x}mathrm{d}x=frac{pi}{4}tag{3}
$$
answered Jun 18 '12 at 10:45
robjohn♦robjohn
266k27306630
$endgroup$
$begingroup$
@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
add a comment |
$begingroup$
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,
$$
int_0^inftyfrac{e^{-x}sin(x)}{x}mathrm{d}x=frac{pi}{4}tag{3}
$$
answered Jun 18 '12 at 10:45
robjohn♦robjohn
266k27306630
$endgroup$
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,
$$
int_0^inftyfrac{e^{-x}sin(x)}{x}mathrm{d}x=frac{pi}{4}tag{3}
$$
answered Jun 18 '12 at 10:45
robjohn♦robjohn
266k27306630
answered Jun 18 '12 at 10:45
robjohn♦robjohn
266k27306630
answered Jun 18 '12 at 10:45
robjohn♦robjohn
266k27306630
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
add a comment |
$begingroup$
@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
add a comment |
$begingroup$
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) $.
answered Jun 18 '12 at 10:20
FabianFabian
19.7k3674
$endgroup$
$begingroup$
@Anon: Sorry for not defining: $Theta(x)$ is the Heaviside step function.
$endgroup$
– Fabian
Jun 18 '12 at 10:24
add a comment |
$begingroup$
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) $.
answered Jun 18 '12 at 10:20
FabianFabian
19.7k3674
$endgroup$
$begingroup$
@Anon: Sorry for not defining: $Theta(x)$ is the Heaviside step function.
$endgroup$
– Fabian
Jun 18 '12 at 10:24
add a comment |
$begingroup$
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) $.
answered Jun 18 '12 at 10:20
FabianFabian
19.7k3674
$endgroup$
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) $.
answered Jun 18 '12 at 10:20
FabianFabian
19.7k3674
answered Jun 18 '12 at 10:20
FabianFabian
19.7k3674
answered Jun 18 '12 at 10:20
FabianFabian
19.7k3674
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
add a comment |
$begingroup$
@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
add a comment |
$begingroup$
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.
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:
$$
frac{pi}{4}=int_0^infty frac{e^{-z}sin(z)}{z},dx
$$
answered Jul 4 '13 at 15:57
ArgonArgon
16.3k673122
$endgroup$
$begingroup$
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
|
show 1 more comment
$begingroup$
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.
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:
$$
frac{pi}{4}=int_0^infty frac{e^{-z}sin(z)}{z},dx
$$
answered Jul 4 '13 at 15:57
ArgonArgon
16.3k673122
$endgroup$
$begingroup$
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
|
show 1 more comment
$begingroup$
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.
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:
$$
frac{pi}{4}=int_0^infty frac{e^{-z}sin(z)}{z},dx
$$
answered Jul 4 '13 at 15:57
ArgonArgon
16.3k673122
$endgroup$
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.
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:
$$
frac{pi}{4}=int_0^infty frac{e^{-z}sin(z)}{z},dx
$$
answered Jul 4 '13 at 15:57
ArgonArgon
16.3k673122
edited Jul 4 '13 at 16:40
answered Jul 4 '13 at 15:57
ArgonArgon
16.3k673122
answered Jul 4 '13 at 15:57
ArgonArgon
16.3k673122
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
|
show 1 more comment
$begingroup$
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
|
show 1 more comment
$begingroup$
$$int_0^{infty} frac{e^{-x}sin x}{x},dx=int_0^{infty}int_0^{infty} e^{-x}sin x ,e^{-xy},dy,dx=int_0^{infty} int_0^{infty} e^{-x(1+y)}sin x,dx,dy$$
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
$$int_0^{infty} int_0^{infty} e^{-x(1+y)}sin x,dx,dy=int_0^{infty} frac{dy}{1+(1+y)^2}=left(arctan(1+y)right|_0^{infty}=boxed{dfrac{pi}{4}}$$
answered May 8 '14 at 15:32
Pranav AroraPranav Arora
8,5472562
$endgroup$
$begingroup$
Neat, but isn't that more or less the same as Julián Aguirre's answer?
$endgroup$
– IAmNoOne
May 9 '14 at 0:12
add a comment |
$begingroup$
$$int_0^{infty} frac{e^{-x}sin x}{x},dx=int_0^{infty}int_0^{infty} e^{-x}sin x ,e^{-xy},dy,dx=int_0^{infty} int_0^{infty} e^{-x(1+y)}sin x,dx,dy$$
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
$$int_0^{infty} int_0^{infty} e^{-x(1+y)}sin x,dx,dy=int_0^{infty} frac{dy}{1+(1+y)^2}=left(arctan(1+y)right|_0^{infty}=boxed{dfrac{pi}{4}}$$
answered May 8 '14 at 15:32
Pranav AroraPranav Arora
8,5472562
$endgroup$
$begingroup$
Neat, but isn't that more or less the same as Julián Aguirre's answer?
$endgroup$
– IAmNoOne
May 9 '14 at 0:12
add a comment |
$begingroup$
$$int_0^{infty} frac{e^{-x}sin x}{x},dx=int_0^{infty}int_0^{infty} e^{-x}sin x ,e^{-xy},dy,dx=int_0^{infty} int_0^{infty} e^{-x(1+y)}sin x,dx,dy$$
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
$$int_0^{infty} int_0^{infty} e^{-x(1+y)}sin x,dx,dy=int_0^{infty} frac{dy}{1+(1+y)^2}=left(arctan(1+y)right|_0^{infty}=boxed{dfrac{pi}{4}}$$
answered May 8 '14 at 15:32
Pranav AroraPranav Arora
8,5472562
$endgroup$
$$int_0^{infty} frac{e^{-x}sin x}{x},dx=int_0^{infty}int_0^{infty} e^{-x}sin x ,e^{-xy},dy,dx=int_0^{infty} int_0^{infty} e^{-x(1+y)}sin x,dx,dy$$
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
$$int_0^{infty} int_0^{infty} e^{-x(1+y)}sin x,dx,dy=int_0^{infty} frac{dy}{1+(1+y)^2}=left(arctan(1+y)right|_0^{infty}=boxed{dfrac{pi}{4}}$$
answered May 8 '14 at 15:32
Pranav AroraPranav Arora
8,5472562
answered May 8 '14 at 15:32
Pranav AroraPranav Arora
8,5472562
answered May 8 '14 at 15:32
Pranav AroraPranav Arora
8,5472562
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
$newcommand{+}{^{dagger}}%
newcommand{angles}[1]{leftlangle #1 rightrangle}%
newcommand{braces}[1]{leftlbrace #1 rightrbrace}%
newcommand{bracks}[1]{leftlbrack #1 rightrbrack}%
newcommand{ceil}[1]{,leftlceil #1 rightrceil,}%
newcommand{dd}{{rm d}}%
newcommand{down}{downarrow}%
newcommand{ds}[1]{displaystyle{#1}}%
newcommand{equalby}[1]{{#1 atop {= atop vphantom{huge A}}}}%
newcommand{expo}[1]{,{rm e}^{#1},}%
newcommand{fermi}{,{rm f}}%
newcommand{floor}[1]{,leftlfloor #1 rightrfloor,}%
newcommand{half}{{1 over 2}}%
newcommand{ic}{{rm i}}%
newcommand{iff}{Longleftrightarrow}
newcommand{imp}{Longrightarrow}%
newcommand{isdiv}{,left.rightvert,}%
newcommand{ket}[1]{leftvert #1rightrangle}%
newcommand{ol}[1]{overline{#1}}%
newcommand{pars}[1]{left( #1 right)}%
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{pp}{{cal P}}%
newcommand{root}[2]{,sqrt[#1]{,#2,},}%
newcommand{sech}{,{rm sech}}%
newcommand{sgn}{,{rm sgn}}%
newcommand{totald}[3]{frac{{rm d}^{#1} #2}{{rm d} #3^{#1}}}
newcommand{ul}[1]{underline{#1}}%
newcommand{verts}[1]{leftvert, #1 ,rightvert}$
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}
answered Jan 28 '14 at 12:11
Felix MarinFelix Marin
67.5k7107141
$endgroup$
add a comment |
$begingroup$
$newcommand{+}{^{dagger}}%
newcommand{angles}[1]{leftlangle #1 rightrangle}%
newcommand{braces}[1]{leftlbrace #1 rightrbrace}%
newcommand{bracks}[1]{leftlbrack #1 rightrbrack}%
newcommand{ceil}[1]{,leftlceil #1 rightrceil,}%
newcommand{dd}{{rm d}}%
newcommand{down}{downarrow}%
newcommand{ds}[1]{displaystyle{#1}}%
newcommand{equalby}[1]{{#1 atop {= atop vphantom{huge A}}}}%
newcommand{expo}[1]{,{rm e}^{#1},}%
newcommand{fermi}{,{rm f}}%
newcommand{floor}[1]{,leftlfloor #1 rightrfloor,}%
newcommand{half}{{1 over 2}}%
newcommand{ic}{{rm i}}%
newcommand{iff}{Longleftrightarrow}
newcommand{imp}{Longrightarrow}%
newcommand{isdiv}{,left.rightvert,}%
newcommand{ket}[1]{leftvert #1rightrangle}%
newcommand{ol}[1]{overline{#1}}%
newcommand{pars}[1]{left( #1 right)}%
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{pp}{{cal P}}%
newcommand{root}[2]{,sqrt[#1]{,#2,},}%
newcommand{sech}{,{rm sech}}%
newcommand{sgn}{,{rm sgn}}%
newcommand{totald}[3]{frac{{rm d}^{#1} #2}{{rm d} #3^{#1}}}
newcommand{ul}[1]{underline{#1}}%
newcommand{verts}[1]{leftvert, #1 ,rightvert}$
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}
answered Jan 28 '14 at 12:11
Felix MarinFelix Marin
67.5k7107141
$endgroup$
add a comment |
$begingroup$
$newcommand{+}{^{dagger}}%
newcommand{angles}[1]{leftlangle #1 rightrangle}%
newcommand{braces}[1]{leftlbrace #1 rightrbrace}%
newcommand{bracks}[1]{leftlbrack #1 rightrbrack}%
newcommand{ceil}[1]{,leftlceil #1 rightrceil,}%
newcommand{dd}{{rm d}}%
newcommand{down}{downarrow}%
newcommand{ds}[1]{displaystyle{#1}}%
newcommand{equalby}[1]{{#1 atop {= atop vphantom{huge A}}}}%
newcommand{expo}[1]{,{rm e}^{#1},}%
newcommand{fermi}{,{rm f}}%
newcommand{floor}[1]{,leftlfloor #1 rightrfloor,}%
newcommand{half}{{1 over 2}}%
newcommand{ic}{{rm i}}%
newcommand{iff}{Longleftrightarrow}
newcommand{imp}{Longrightarrow}%
newcommand{isdiv}{,left.rightvert,}%
newcommand{ket}[1]{leftvert #1rightrangle}%
newcommand{ol}[1]{overline{#1}}%
newcommand{pars}[1]{left( #1 right)}%
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{pp}{{cal P}}%
newcommand{root}[2]{,sqrt[#1]{,#2,},}%
newcommand{sech}{,{rm sech}}%
newcommand{sgn}{,{rm sgn}}%
newcommand{totald}[3]{frac{{rm d}^{#1} #2}{{rm d} #3^{#1}}}
newcommand{ul}[1]{underline{#1}}%
newcommand{verts}[1]{leftvert, #1 ,rightvert}$
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}
answered Jan 28 '14 at 12:11
Felix MarinFelix Marin
67.5k7107141
$endgroup$
$newcommand{+}{^{dagger}}%
newcommand{angles}[1]{leftlangle #1 rightrangle}%
newcommand{braces}[1]{leftlbrace #1 rightrbrace}%
newcommand{bracks}[1]{leftlbrack #1 rightrbrack}%
newcommand{ceil}[1]{,leftlceil #1 rightrceil,}%
newcommand{dd}{{rm d}}%
newcommand{down}{downarrow}%
newcommand{ds}[1]{displaystyle{#1}}%
newcommand{equalby}[1]{{#1 atop {= atop vphantom{huge A}}}}%
newcommand{expo}[1]{,{rm e}^{#1},}%
newcommand{fermi}{,{rm f}}%
newcommand{floor}[1]{,leftlfloor #1 rightrfloor,}%
newcommand{half}{{1 over 2}}%
newcommand{ic}{{rm i}}%
newcommand{iff}{Longleftrightarrow}
newcommand{imp}{Longrightarrow}%
newcommand{isdiv}{,left.rightvert,}%
newcommand{ket}[1]{leftvert #1rightrangle}%
newcommand{ol}[1]{overline{#1}}%
newcommand{pars}[1]{left( #1 right)}%
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{pp}{{cal P}}%
newcommand{root}[2]{,sqrt[#1]{,#2,},}%
newcommand{sech}{,{rm sech}}%
newcommand{sgn}{,{rm sgn}}%
newcommand{totald}[3]{frac{{rm d}^{#1} #2}{{rm d} #3^{#1}}}
newcommand{ul}[1]{underline{#1}}%
newcommand{verts}[1]{leftvert, #1 ,rightvert}$
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}
answered Jan 28 '14 at 12:11
Felix MarinFelix Marin
67.5k7107141
edited May 8 '14 at 5:09
answered Jan 28 '14 at 12:11
Felix MarinFelix Marin
67.5k7107141
answered Jan 28 '14 at 12:11
Felix MarinFelix Marin
67.5k7107141
answered Jan 28 '14 at 12:11
Felix MarinFelix Marin
67.5k7107141
67.5k7107141
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f159836%2fint-0-infty-frace-x-sinxx-dx-evaluate-integral%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
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