Calculating improper integral $int limits_{0}^{infty}frac{mathrm{e}^{-x}}{sqrt{x}},mathrm{d}x$












1












$begingroup$



I want to calculate the improper integral $displaystyle int
limits_{0}^{infty}dfrac{mathrm{e}^{-x}}{sqrt{x}},mathrm{d}x$
$DeclareMathOperatorerf{erf}$




Therefore
begin{align}
I(b)&=limlimits_{bto0}left(displaystyle int limits_{b}^{infty}dfrac{mathrm{e}^{-x}}{sqrt{x}},mathrm{d}xright) qquad forall binmathbb{R}:0<b<infty\
&=limlimits_{bto0}left(sqrt{pi} erf(sqrt{b}) right)=sqrt{pi}erf(sqrt{0})=sqrt{pi}
end{align}



This looks way to easy. Is this correct or am I missing something? Do you know a better way while using the following equation from our lectures?: $$displaystyle intlimits_0^infty e^{-x^2},mathrm{d}x=frac{1}{2}sqrt{pi}$$






integration definite-integrals improper-integrals







share|cite|improve this question











$endgroup$












  • $begingroup$
    I corrected that, thank you!
    $endgroup$
    – Doesbaddel
    Jan 24 at 11:36
















1












$begingroup$



I want to calculate the improper integral $displaystyle int
limits_{0}^{infty}dfrac{mathrm{e}^{-x}}{sqrt{x}},mathrm{d}x$
$DeclareMathOperatorerf{erf}$




Therefore
begin{align}
I(b)&=limlimits_{bto0}left(displaystyle int limits_{b}^{infty}dfrac{mathrm{e}^{-x}}{sqrt{x}},mathrm{d}xright) qquad forall binmathbb{R}:0<b<infty\
&=limlimits_{bto0}left(sqrt{pi} erf(sqrt{b}) right)=sqrt{pi}erf(sqrt{0})=sqrt{pi}
end{align}



This looks way to easy. Is this correct or am I missing something? Do you know a better way while using the following equation from our lectures?: $$displaystyle intlimits_0^infty e^{-x^2},mathrm{d}x=frac{1}{2}sqrt{pi}$$






integration definite-integrals improper-integrals







share|cite|improve this question











$endgroup$












  • $begingroup$
    I corrected that, thank you!
    $endgroup$
    – Doesbaddel
    Jan 24 at 11:36














1












1








1





$begingroup$



I want to calculate the improper integral $displaystyle int
limits_{0}^{infty}dfrac{mathrm{e}^{-x}}{sqrt{x}},mathrm{d}x$
$DeclareMathOperatorerf{erf}$




Therefore
begin{align}
I(b)&=limlimits_{bto0}left(displaystyle int limits_{b}^{infty}dfrac{mathrm{e}^{-x}}{sqrt{x}},mathrm{d}xright) qquad forall binmathbb{R}:0<b<infty\
&=limlimits_{bto0}left(sqrt{pi} erf(sqrt{b}) right)=sqrt{pi}erf(sqrt{0})=sqrt{pi}
end{align}



This looks way to easy. Is this correct or am I missing something? Do you know a better way while using the following equation from our lectures?: $$displaystyle intlimits_0^infty e^{-x^2},mathrm{d}x=frac{1}{2}sqrt{pi}$$






integration definite-integrals improper-integrals







share|cite|improve this question











$endgroup$





I want to calculate the improper integral $displaystyle int
limits_{0}^{infty}dfrac{mathrm{e}^{-x}}{sqrt{x}},mathrm{d}x$
$DeclareMathOperatorerf{erf}$




Therefore
begin{align}
I(b)&=limlimits_{bto0}left(displaystyle int limits_{b}^{infty}dfrac{mathrm{e}^{-x}}{sqrt{x}},mathrm{d}xright) qquad forall binmathbb{R}:0<b<infty\
&=limlimits_{bto0}left(sqrt{pi} erf(sqrt{b}) right)=sqrt{pi}erf(sqrt{0})=sqrt{pi}
end{align}



This looks way to easy. Is this correct or am I missing something? Do you know a better way while using the following equation from our lectures?: $$displaystyle intlimits_0^infty e^{-x^2},mathrm{d}x=frac{1}{2}sqrt{pi}$$






integration definite-integrals improper-integrals




integration definite-integrals improper-integrals






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share|cite|improve this question













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edited Jan 24 at 12:23









Martin Sleziak

44.8k10119273




44.8k10119273










asked Jan 24 at 11:26









DoesbaddelDoesbaddel

34211




34211












  • $begingroup$
    I corrected that, thank you!
    $endgroup$
    – Doesbaddel
    Jan 24 at 11:36


















  • $begingroup$
    I corrected that, thank you!
    $endgroup$
    – Doesbaddel
    Jan 24 at 11:36
















$begingroup$
I corrected that, thank you!
$endgroup$
– Doesbaddel
Jan 24 at 11:36




$begingroup$
I corrected that, thank you!
$endgroup$
– Doesbaddel
Jan 24 at 11:36










2 Answers
2






active

oldest

votes


















5












$begingroup$

Hint:



Just substitute $x= u^2$. So, you get
$$int limits_{0}^{infty}dfrac{mathrm{e}^{-x}}{sqrt{x}},mathrm{d}x =2int_0^{infty}e^{-u^2}du$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Ok, so $int limits_{0}^{infty}dfrac{mathrm{e}^{-x}}{sqrt{x}},mathrm{d}x =2int_0^{infty}e^{-u^2}du =2cdot frac{1}{2}sqrt{pi}=sqrt{pi}$ does the job?
    $endgroup$
    – Doesbaddel
    Jan 24 at 11:44












  • $begingroup$
    Yes, it is the famous Gaussian integral.
    $endgroup$
    – Larry
    Jan 24 at 11:46










  • $begingroup$
    @Doesbaddel : So ist es. :-)
    $endgroup$
    – trancelocation
    Jan 24 at 11:46










  • $begingroup$
    You've switched from Gamma function to Gaussian integral. Why do you need it, if Gaussian integral is calculated through Gamma function
    $endgroup$
    – Yauhen Mardan
    Jan 24 at 11:47






  • 1




    $begingroup$
    @YauhenMardan Because Doesbaddel asked exactly for something like this. Just read the post :-)
    $endgroup$
    – trancelocation
    Jan 24 at 11:49





















3












$begingroup$

It's Gamma function: $int_0^infty x^{1/2-1}e^{-x}dx=Г(1/2)=sqrt{pi}$






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    In particular $Gamma^2(1/2)=operatorname{B}(1/2,,1/2)=2int_0^{pi/2}dx=pi$, so we don't need to already know the Gaussian integral; in fact, this argument is one way to compute the Gaussian integral.
    $endgroup$
    – J.G.
    Jan 24 at 12:32











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









5












$begingroup$

Hint:



Just substitute $x= u^2$. So, you get
$$int limits_{0}^{infty}dfrac{mathrm{e}^{-x}}{sqrt{x}},mathrm{d}x =2int_0^{infty}e^{-u^2}du$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Ok, so $int limits_{0}^{infty}dfrac{mathrm{e}^{-x}}{sqrt{x}},mathrm{d}x =2int_0^{infty}e^{-u^2}du =2cdot frac{1}{2}sqrt{pi}=sqrt{pi}$ does the job?
    $endgroup$
    – Doesbaddel
    Jan 24 at 11:44












  • $begingroup$
    Yes, it is the famous Gaussian integral.
    $endgroup$
    – Larry
    Jan 24 at 11:46










  • $begingroup$
    @Doesbaddel : So ist es. :-)
    $endgroup$
    – trancelocation
    Jan 24 at 11:46










  • $begingroup$
    You've switched from Gamma function to Gaussian integral. Why do you need it, if Gaussian integral is calculated through Gamma function
    $endgroup$
    – Yauhen Mardan
    Jan 24 at 11:47






  • 1




    $begingroup$
    @YauhenMardan Because Doesbaddel asked exactly for something like this. Just read the post :-)
    $endgroup$
    – trancelocation
    Jan 24 at 11:49


















5












$begingroup$

Hint:



Just substitute $x= u^2$. So, you get
$$int limits_{0}^{infty}dfrac{mathrm{e}^{-x}}{sqrt{x}},mathrm{d}x =2int_0^{infty}e^{-u^2}du$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Ok, so $int limits_{0}^{infty}dfrac{mathrm{e}^{-x}}{sqrt{x}},mathrm{d}x =2int_0^{infty}e^{-u^2}du =2cdot frac{1}{2}sqrt{pi}=sqrt{pi}$ does the job?
    $endgroup$
    – Doesbaddel
    Jan 24 at 11:44












  • $begingroup$
    Yes, it is the famous Gaussian integral.
    $endgroup$
    – Larry
    Jan 24 at 11:46










  • $begingroup$
    @Doesbaddel : So ist es. :-)
    $endgroup$
    – trancelocation
    Jan 24 at 11:46










  • $begingroup$
    You've switched from Gamma function to Gaussian integral. Why do you need it, if Gaussian integral is calculated through Gamma function
    $endgroup$
    – Yauhen Mardan
    Jan 24 at 11:47






  • 1




    $begingroup$
    @YauhenMardan Because Doesbaddel asked exactly for something like this. Just read the post :-)
    $endgroup$
    – trancelocation
    Jan 24 at 11:49
















5












5








5





$begingroup$

Hint:



Just substitute $x= u^2$. So, you get
$$int limits_{0}^{infty}dfrac{mathrm{e}^{-x}}{sqrt{x}},mathrm{d}x =2int_0^{infty}e^{-u^2}du$$






share|cite|improve this answer









$endgroup$



Hint:



Just substitute $x= u^2$. So, you get
$$int limits_{0}^{infty}dfrac{mathrm{e}^{-x}}{sqrt{x}},mathrm{d}x =2int_0^{infty}e^{-u^2}du$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 24 at 11:31









trancelocationtrancelocation

12.6k1826




12.6k1826












  • $begingroup$
    Ok, so $int limits_{0}^{infty}dfrac{mathrm{e}^{-x}}{sqrt{x}},mathrm{d}x =2int_0^{infty}e^{-u^2}du =2cdot frac{1}{2}sqrt{pi}=sqrt{pi}$ does the job?
    $endgroup$
    – Doesbaddel
    Jan 24 at 11:44












  • $begingroup$
    Yes, it is the famous Gaussian integral.
    $endgroup$
    – Larry
    Jan 24 at 11:46










  • $begingroup$
    @Doesbaddel : So ist es. :-)
    $endgroup$
    – trancelocation
    Jan 24 at 11:46










  • $begingroup$
    You've switched from Gamma function to Gaussian integral. Why do you need it, if Gaussian integral is calculated through Gamma function
    $endgroup$
    – Yauhen Mardan
    Jan 24 at 11:47






  • 1




    $begingroup$
    @YauhenMardan Because Doesbaddel asked exactly for something like this. Just read the post :-)
    $endgroup$
    – trancelocation
    Jan 24 at 11:49




















  • $begingroup$
    Ok, so $int limits_{0}^{infty}dfrac{mathrm{e}^{-x}}{sqrt{x}},mathrm{d}x =2int_0^{infty}e^{-u^2}du =2cdot frac{1}{2}sqrt{pi}=sqrt{pi}$ does the job?
    $endgroup$
    – Doesbaddel
    Jan 24 at 11:44












  • $begingroup$
    Yes, it is the famous Gaussian integral.
    $endgroup$
    – Larry
    Jan 24 at 11:46










  • $begingroup$
    @Doesbaddel : So ist es. :-)
    $endgroup$
    – trancelocation
    Jan 24 at 11:46










  • $begingroup$
    You've switched from Gamma function to Gaussian integral. Why do you need it, if Gaussian integral is calculated through Gamma function
    $endgroup$
    – Yauhen Mardan
    Jan 24 at 11:47






  • 1




    $begingroup$
    @YauhenMardan Because Doesbaddel asked exactly for something like this. Just read the post :-)
    $endgroup$
    – trancelocation
    Jan 24 at 11:49


















$begingroup$
Ok, so $int limits_{0}^{infty}dfrac{mathrm{e}^{-x}}{sqrt{x}},mathrm{d}x =2int_0^{infty}e^{-u^2}du =2cdot frac{1}{2}sqrt{pi}=sqrt{pi}$ does the job?
$endgroup$
– Doesbaddel
Jan 24 at 11:44






$begingroup$
Ok, so $int limits_{0}^{infty}dfrac{mathrm{e}^{-x}}{sqrt{x}},mathrm{d}x =2int_0^{infty}e^{-u^2}du =2cdot frac{1}{2}sqrt{pi}=sqrt{pi}$ does the job?
$endgroup$
– Doesbaddel
Jan 24 at 11:44














$begingroup$
Yes, it is the famous Gaussian integral.
$endgroup$
– Larry
Jan 24 at 11:46




$begingroup$
Yes, it is the famous Gaussian integral.
$endgroup$
– Larry
Jan 24 at 11:46












$begingroup$
@Doesbaddel : So ist es. :-)
$endgroup$
– trancelocation
Jan 24 at 11:46




$begingroup$
@Doesbaddel : So ist es. :-)
$endgroup$
– trancelocation
Jan 24 at 11:46












$begingroup$
You've switched from Gamma function to Gaussian integral. Why do you need it, if Gaussian integral is calculated through Gamma function
$endgroup$
– Yauhen Mardan
Jan 24 at 11:47




$begingroup$
You've switched from Gamma function to Gaussian integral. Why do you need it, if Gaussian integral is calculated through Gamma function
$endgroup$
– Yauhen Mardan
Jan 24 at 11:47




1




1




$begingroup$
@YauhenMardan Because Doesbaddel asked exactly for something like this. Just read the post :-)
$endgroup$
– trancelocation
Jan 24 at 11:49






$begingroup$
@YauhenMardan Because Doesbaddel asked exactly for something like this. Just read the post :-)
$endgroup$
– trancelocation
Jan 24 at 11:49













3












$begingroup$

It's Gamma function: $int_0^infty x^{1/2-1}e^{-x}dx=Г(1/2)=sqrt{pi}$






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    In particular $Gamma^2(1/2)=operatorname{B}(1/2,,1/2)=2int_0^{pi/2}dx=pi$, so we don't need to already know the Gaussian integral; in fact, this argument is one way to compute the Gaussian integral.
    $endgroup$
    – J.G.
    Jan 24 at 12:32
















3












$begingroup$

It's Gamma function: $int_0^infty x^{1/2-1}e^{-x}dx=Г(1/2)=sqrt{pi}$






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    In particular $Gamma^2(1/2)=operatorname{B}(1/2,,1/2)=2int_0^{pi/2}dx=pi$, so we don't need to already know the Gaussian integral; in fact, this argument is one way to compute the Gaussian integral.
    $endgroup$
    – J.G.
    Jan 24 at 12:32














3












3








3





$begingroup$

It's Gamma function: $int_0^infty x^{1/2-1}e^{-x}dx=Г(1/2)=sqrt{pi}$






share|cite|improve this answer









$endgroup$



It's Gamma function: $int_0^infty x^{1/2-1}e^{-x}dx=Г(1/2)=sqrt{pi}$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 24 at 11:42









Yauhen MardanYauhen Mardan

936




936








  • 1




    $begingroup$
    In particular $Gamma^2(1/2)=operatorname{B}(1/2,,1/2)=2int_0^{pi/2}dx=pi$, so we don't need to already know the Gaussian integral; in fact, this argument is one way to compute the Gaussian integral.
    $endgroup$
    – J.G.
    Jan 24 at 12:32














  • 1




    $begingroup$
    In particular $Gamma^2(1/2)=operatorname{B}(1/2,,1/2)=2int_0^{pi/2}dx=pi$, so we don't need to already know the Gaussian integral; in fact, this argument is one way to compute the Gaussian integral.
    $endgroup$
    – J.G.
    Jan 24 at 12:32








1




1




$begingroup$
In particular $Gamma^2(1/2)=operatorname{B}(1/2,,1/2)=2int_0^{pi/2}dx=pi$, so we don't need to already know the Gaussian integral; in fact, this argument is one way to compute the Gaussian integral.
$endgroup$
– J.G.
Jan 24 at 12:32




$begingroup$
In particular $Gamma^2(1/2)=operatorname{B}(1/2,,1/2)=2int_0^{pi/2}dx=pi$, so we don't need to already know the Gaussian integral; in fact, this argument is one way to compute the Gaussian integral.
$endgroup$
– J.G.
Jan 24 at 12:32


















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