Integrate $int_{0}^{infty}e^{-pt}sinleft(sqrt{t}right)mathrm dt$












2












$begingroup$


I need the following Laplace transform to solve the Differential Equation



$$int_{0}^{infty}e^{-pt}sinsqrt{t}, dt, quad text{where} p>0$$



I tried Integration by parts after substituting $t=x^2$, but didn't work.




begin{align}
int_{0}^{infty}e^{-pt}sinsqrt{t}dt &overset{t=x^2}= int_{0}^{infty}e^{-px^2}2xsin xdx = text{I}\
& = sin x frac{e^{-px^2}}{-p} + frac{1}{p}int_{0}^{infty}e^{-px^2}cos xdx \
& = sin x frac{e^{-px^2}}{-p} + frac{1}{p}left(-e^{-px^2}sin x - int_{0}^{infty}e^{-px^2}(-2px)(-sin x)dxright) \
& = -sin x frac{e^{-px^2}}{p} + frac{1}{p}left(-sin xe^{-px^2} - pint_{0}^{infty}e^{-px^2}(2x)(sin x)dxright) \
& = -sin x frac{e^{-px^2}}{p} + frac{1}{p}left(-sin xe^{-px^2} - ptext{I}right) \
end{align}
begin{align}
& text{I} = -sin x frac{e^{-px^2}}{p} + frac{1}{p}left(-sin xe^{-px^2} - ptext{I}right) \
& 2text{I} = -2sin x frac{e^{-px^2}}{p}Big|_0^infty \
& 2text{I} = 0
end{align}






integration definite-integrals laplace-transform







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  • 1




    $begingroup$
    Have you tried using Moivre's relationship? $sin(theta)=frac{ e^{itheta}-e^{-itheta}}{2i}$, It could help
    $endgroup$
    – Ramiro Scorolli
    Dec 31 '18 at 12:30








  • 1




    $begingroup$
    Substitute $sqrt{t}=y$, to get $$ int_{0}^{infty}e^{-pt}sinsqrt{t}~dt=2int_{0}^{infty}e^{-py^2}ysin y~dy=2mathrm{Im}int_{0}^{infty}e^{-py^2+mathrm{i}y}y~dy , $$ where Im stands for imaginary part.
    $endgroup$
    – Pierpaolo Vivo
    Dec 31 '18 at 12:32










  • $begingroup$
    Why did you delete your answer? It was a great method, and if you're not familiar with contour integration, it's fine to just leave that link to justify your calculations.
    $endgroup$
    – Zachary
    Dec 31 '18 at 15:04










  • $begingroup$
    Are you sure that is fine? I really don't have knowledge for that.
    $endgroup$
    – Zacky
    Dec 31 '18 at 15:05






  • 1




    $begingroup$
    Well I'm not a moderator, but I would assume that posting such an answer, which in this case is mathematically correct and relevant, would be acceptable - even if you leave a link to justify a certain computation. The answer may help somebody who may not see it in the comments.
    $endgroup$
    – Zachary
    Dec 31 '18 at 15:13
















2












$begingroup$


I need the following Laplace transform to solve the Differential Equation



$$int_{0}^{infty}e^{-pt}sinsqrt{t}, dt, quad text{where} p>0$$



I tried Integration by parts after substituting $t=x^2$, but didn't work.




begin{align}
int_{0}^{infty}e^{-pt}sinsqrt{t}dt &overset{t=x^2}= int_{0}^{infty}e^{-px^2}2xsin xdx = text{I}\
& = sin x frac{e^{-px^2}}{-p} + frac{1}{p}int_{0}^{infty}e^{-px^2}cos xdx \
& = sin x frac{e^{-px^2}}{-p} + frac{1}{p}left(-e^{-px^2}sin x - int_{0}^{infty}e^{-px^2}(-2px)(-sin x)dxright) \
& = -sin x frac{e^{-px^2}}{p} + frac{1}{p}left(-sin xe^{-px^2} - pint_{0}^{infty}e^{-px^2}(2x)(sin x)dxright) \
& = -sin x frac{e^{-px^2}}{p} + frac{1}{p}left(-sin xe^{-px^2} - ptext{I}right) \
end{align}
begin{align}
& text{I} = -sin x frac{e^{-px^2}}{p} + frac{1}{p}left(-sin xe^{-px^2} - ptext{I}right) \
& 2text{I} = -2sin x frac{e^{-px^2}}{p}Big|_0^infty \
& 2text{I} = 0
end{align}






integration definite-integrals laplace-transform







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$endgroup$








  • 1




    $begingroup$
    Have you tried using Moivre's relationship? $sin(theta)=frac{ e^{itheta}-e^{-itheta}}{2i}$, It could help
    $endgroup$
    – Ramiro Scorolli
    Dec 31 '18 at 12:30








  • 1




    $begingroup$
    Substitute $sqrt{t}=y$, to get $$ int_{0}^{infty}e^{-pt}sinsqrt{t}~dt=2int_{0}^{infty}e^{-py^2}ysin y~dy=2mathrm{Im}int_{0}^{infty}e^{-py^2+mathrm{i}y}y~dy , $$ where Im stands for imaginary part.
    $endgroup$
    – Pierpaolo Vivo
    Dec 31 '18 at 12:32










  • $begingroup$
    Why did you delete your answer? It was a great method, and if you're not familiar with contour integration, it's fine to just leave that link to justify your calculations.
    $endgroup$
    – Zachary
    Dec 31 '18 at 15:04










  • $begingroup$
    Are you sure that is fine? I really don't have knowledge for that.
    $endgroup$
    – Zacky
    Dec 31 '18 at 15:05






  • 1




    $begingroup$
    Well I'm not a moderator, but I would assume that posting such an answer, which in this case is mathematically correct and relevant, would be acceptable - even if you leave a link to justify a certain computation. The answer may help somebody who may not see it in the comments.
    $endgroup$
    – Zachary
    Dec 31 '18 at 15:13














2












2








2





$begingroup$


I need the following Laplace transform to solve the Differential Equation



$$int_{0}^{infty}e^{-pt}sinsqrt{t}, dt, quad text{where} p>0$$



I tried Integration by parts after substituting $t=x^2$, but didn't work.




begin{align}
int_{0}^{infty}e^{-pt}sinsqrt{t}dt &overset{t=x^2}= int_{0}^{infty}e^{-px^2}2xsin xdx = text{I}\
& = sin x frac{e^{-px^2}}{-p} + frac{1}{p}int_{0}^{infty}e^{-px^2}cos xdx \
& = sin x frac{e^{-px^2}}{-p} + frac{1}{p}left(-e^{-px^2}sin x - int_{0}^{infty}e^{-px^2}(-2px)(-sin x)dxright) \
& = -sin x frac{e^{-px^2}}{p} + frac{1}{p}left(-sin xe^{-px^2} - pint_{0}^{infty}e^{-px^2}(2x)(sin x)dxright) \
& = -sin x frac{e^{-px^2}}{p} + frac{1}{p}left(-sin xe^{-px^2} - ptext{I}right) \
end{align}
begin{align}
& text{I} = -sin x frac{e^{-px^2}}{p} + frac{1}{p}left(-sin xe^{-px^2} - ptext{I}right) \
& 2text{I} = -2sin x frac{e^{-px^2}}{p}Big|_0^infty \
& 2text{I} = 0
end{align}






integration definite-integrals laplace-transform







share|cite|improve this question











$endgroup$




I need the following Laplace transform to solve the Differential Equation



$$int_{0}^{infty}e^{-pt}sinsqrt{t}, dt, quad text{where} p>0$$



I tried Integration by parts after substituting $t=x^2$, but didn't work.




begin{align}
int_{0}^{infty}e^{-pt}sinsqrt{t}dt &overset{t=x^2}= int_{0}^{infty}e^{-px^2}2xsin xdx = text{I}\
& = sin x frac{e^{-px^2}}{-p} + frac{1}{p}int_{0}^{infty}e^{-px^2}cos xdx \
& = sin x frac{e^{-px^2}}{-p} + frac{1}{p}left(-e^{-px^2}sin x - int_{0}^{infty}e^{-px^2}(-2px)(-sin x)dxright) \
& = -sin x frac{e^{-px^2}}{p} + frac{1}{p}left(-sin xe^{-px^2} - pint_{0}^{infty}e^{-px^2}(2x)(sin x)dxright) \
& = -sin x frac{e^{-px^2}}{p} + frac{1}{p}left(-sin xe^{-px^2} - ptext{I}right) \
end{align}
begin{align}
& text{I} = -sin x frac{e^{-px^2}}{p} + frac{1}{p}left(-sin xe^{-px^2} - ptext{I}right) \
& 2text{I} = -2sin x frac{e^{-px^2}}{p}Big|_0^infty \
& 2text{I} = 0
end{align}






integration definite-integrals laplace-transform




integration definite-integrals laplace-transform






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edited Dec 31 '18 at 16:36







Ajay Choudhary

















asked Dec 31 '18 at 12:23









Ajay ChoudharyAjay Choudhary

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888








  • 1




    $begingroup$
    Have you tried using Moivre's relationship? $sin(theta)=frac{ e^{itheta}-e^{-itheta}}{2i}$, It could help
    $endgroup$
    – Ramiro Scorolli
    Dec 31 '18 at 12:30








  • 1




    $begingroup$
    Substitute $sqrt{t}=y$, to get $$ int_{0}^{infty}e^{-pt}sinsqrt{t}~dt=2int_{0}^{infty}e^{-py^2}ysin y~dy=2mathrm{Im}int_{0}^{infty}e^{-py^2+mathrm{i}y}y~dy , $$ where Im stands for imaginary part.
    $endgroup$
    – Pierpaolo Vivo
    Dec 31 '18 at 12:32










  • $begingroup$
    Why did you delete your answer? It was a great method, and if you're not familiar with contour integration, it's fine to just leave that link to justify your calculations.
    $endgroup$
    – Zachary
    Dec 31 '18 at 15:04










  • $begingroup$
    Are you sure that is fine? I really don't have knowledge for that.
    $endgroup$
    – Zacky
    Dec 31 '18 at 15:05






  • 1




    $begingroup$
    Well I'm not a moderator, but I would assume that posting such an answer, which in this case is mathematically correct and relevant, would be acceptable - even if you leave a link to justify a certain computation. The answer may help somebody who may not see it in the comments.
    $endgroup$
    – Zachary
    Dec 31 '18 at 15:13














  • 1




    $begingroup$
    Have you tried using Moivre's relationship? $sin(theta)=frac{ e^{itheta}-e^{-itheta}}{2i}$, It could help
    $endgroup$
    – Ramiro Scorolli
    Dec 31 '18 at 12:30








  • 1




    $begingroup$
    Substitute $sqrt{t}=y$, to get $$ int_{0}^{infty}e^{-pt}sinsqrt{t}~dt=2int_{0}^{infty}e^{-py^2}ysin y~dy=2mathrm{Im}int_{0}^{infty}e^{-py^2+mathrm{i}y}y~dy , $$ where Im stands for imaginary part.
    $endgroup$
    – Pierpaolo Vivo
    Dec 31 '18 at 12:32










  • $begingroup$
    Why did you delete your answer? It was a great method, and if you're not familiar with contour integration, it's fine to just leave that link to justify your calculations.
    $endgroup$
    – Zachary
    Dec 31 '18 at 15:04










  • $begingroup$
    Are you sure that is fine? I really don't have knowledge for that.
    $endgroup$
    – Zacky
    Dec 31 '18 at 15:05






  • 1




    $begingroup$
    Well I'm not a moderator, but I would assume that posting such an answer, which in this case is mathematically correct and relevant, would be acceptable - even if you leave a link to justify a certain computation. The answer may help somebody who may not see it in the comments.
    $endgroup$
    – Zachary
    Dec 31 '18 at 15:13








1




1




$begingroup$
Have you tried using Moivre's relationship? $sin(theta)=frac{ e^{itheta}-e^{-itheta}}{2i}$, It could help
$endgroup$
– Ramiro Scorolli
Dec 31 '18 at 12:30






$begingroup$
Have you tried using Moivre's relationship? $sin(theta)=frac{ e^{itheta}-e^{-itheta}}{2i}$, It could help
$endgroup$
– Ramiro Scorolli
Dec 31 '18 at 12:30






1




1




$begingroup$
Substitute $sqrt{t}=y$, to get $$ int_{0}^{infty}e^{-pt}sinsqrt{t}~dt=2int_{0}^{infty}e^{-py^2}ysin y~dy=2mathrm{Im}int_{0}^{infty}e^{-py^2+mathrm{i}y}y~dy , $$ where Im stands for imaginary part.
$endgroup$
– Pierpaolo Vivo
Dec 31 '18 at 12:32




$begingroup$
Substitute $sqrt{t}=y$, to get $$ int_{0}^{infty}e^{-pt}sinsqrt{t}~dt=2int_{0}^{infty}e^{-py^2}ysin y~dy=2mathrm{Im}int_{0}^{infty}e^{-py^2+mathrm{i}y}y~dy , $$ where Im stands for imaginary part.
$endgroup$
– Pierpaolo Vivo
Dec 31 '18 at 12:32












$begingroup$
Why did you delete your answer? It was a great method, and if you're not familiar with contour integration, it's fine to just leave that link to justify your calculations.
$endgroup$
– Zachary
Dec 31 '18 at 15:04




$begingroup$
Why did you delete your answer? It was a great method, and if you're not familiar with contour integration, it's fine to just leave that link to justify your calculations.
$endgroup$
– Zachary
Dec 31 '18 at 15:04












$begingroup$
Are you sure that is fine? I really don't have knowledge for that.
$endgroup$
– Zacky
Dec 31 '18 at 15:05




$begingroup$
Are you sure that is fine? I really don't have knowledge for that.
$endgroup$
– Zacky
Dec 31 '18 at 15:05




1




1




$begingroup$
Well I'm not a moderator, but I would assume that posting such an answer, which in this case is mathematically correct and relevant, would be acceptable - even if you leave a link to justify a certain computation. The answer may help somebody who may not see it in the comments.
$endgroup$
– Zachary
Dec 31 '18 at 15:13




$begingroup$
Well I'm not a moderator, but I would assume that posting such an answer, which in this case is mathematically correct and relevant, would be acceptable - even if you leave a link to justify a certain computation. The answer may help somebody who may not see it in the comments.
$endgroup$
– Zachary
Dec 31 '18 at 15:13










3 Answers
3






active

oldest

votes


















2












$begingroup$

$$I=int_0^infty sinleft(sqrt t right)e^{-pt}dtoverset{sqrt t=x}=2int_0^infty xsin x e^{-px^2}dx=int_0^infty sin xleft(-frac1pe^{-px^2}right)'dx$$
$$overset{IBP}=underbrace{-frac1psin xe^{-px^2}bigg|_0^infty}_{=0}+frac1pint_0^infty cos x,e^{-px^2}dx=frac1{2p}int_{-infty}^infty cos x,e^{-px^2}dx$$
We can also make use of the fact that $cos x$ is the real part of $e^{ix}=cos x+isin x$.
$$I=frac1{2p}Re left(int_{-infty}^infty e^{ix}e^{-px^2}dxright)=frac1{2p}Re left(int_{-infty}^infty e^{large-(px^2-ix)+frac{1}{4p}-frac{1}{4p}}dxright)$$
$$=frac1{2p}Re left(int_{-infty}^infty e^{-largeleft(sqrt{p}x-frac{i}{2sqrt p}right)^2 -frac{1}{4p}}dxright)=frac{e^{-frac{1}{4p}}}{2p}Re left(int_{-infty}^infty e^{-largeleft(sqrt{p}x-frac{i}{2sqrt p}right)^2}dxright)$$
Substituting $,displaystyle{sqrt{p}x-frac{i}{2sqrt p}=tRightarrow dx=frac{dt}{sqrt p}}$ gives:
$$I=frac{e^{-frac{1}{4p}}}{2p} frac{1}{sqrt p}Releft(int_{-infty-largefrac{i}{2sqrt p}}^{infty-largefrac{i}{2sqrt p}} e^{-t^2}dtright)=frac{e^{-frac{1}{4p}}}{2p} frac{1}{sqrt p}Releft(int_{-infty}^infty e^{-t^2}dtright)=frac{e^{-frac{1}{4p}}}{2p} sqrt{frac{pi}{p}}$$
For the last line see here and here.






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  • 1




    $begingroup$
    Careful there. After your substitution $sqrt{p}x-i/2sqrt{p}=t$, your new bounds are $infty - i/2sqrt{p}$, and $-infty - i/2sqrt{p}$. You'd have to show that the extra imaginary part does not matter in the calculation of the integral.
    $endgroup$
    – Zachary
    Dec 31 '18 at 14:57








  • 1




    $begingroup$
    Yup. Are you familiar with Contour integration? Use Cauchy's theorem, as in this post: math.stackexchange.com/questions/648043/….
    $endgroup$
    – Zachary
    Dec 31 '18 at 14:59












  • $begingroup$
    @Zachary thanks alot! Over time I alway struggled with contour integration and I tend to rely on real analysis as much as I can. I have also done the same mistake in the past: math.stackexchange.com/q/2861708/515527
    $endgroup$
    – Zacky
    Dec 31 '18 at 15:14





















2












$begingroup$

Hint:
Use the expansion of $sin$
$$sinsqrt{t}=sum_{n=0}^{infty}(-1)^ndfrac{t^{n+1/2}}{Gamma(2n+2)}$$
Edit:
$${cal L}(sinsqrt{t})=sum_{n=0}^{infty}(-1)^ndfrac{Gamma(n+3/2)}{Gamma(2n+2)p^{n+3/2}}$$
with Legendre Duplication Formula we have
$${cal L}(sinsqrt{t})=dfrac{1}{p^{3/2}}sum_{n=0}^{infty}(-1)^ndfrac{Gamma(n+3/2)}{sqrt{pi}^{-1}2^{2n+1}Gamma(n+1)Gamma(n+3/2)p^n}$$
$$=dfrac{sqrt{pi}}{p^{3/2}}sum_{n=0}^{infty}dfrac{left(frac{-1}{4p}right)^{n}}{n!}=color{blue}{dfrac{sqrt{pi}}{p^{3/2}}e^{frac{-1}{4p}}}$$






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  • $begingroup$
    I know it can be solved by using the expansion. But I was looking for some other methods to do the same. Can you see if I applied by parts correctly?
    $endgroup$
    – Ajay Choudhary
    Dec 31 '18 at 13:30










  • $begingroup$
    How exactly would you use that?
    $endgroup$
    – Zachary
    Dec 31 '18 at 13:31










  • $begingroup$
    @AjayChoudhary you should say that in your question.
    $endgroup$
    – Nosrati
    Dec 31 '18 at 13:39



















0












$begingroup$

Hint: Use integration by parts two times. The result should be
$$frac{sqrt{pi}e^{-1/(4p)}}{2p^{2/3}}$$






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$endgroup$













  • $begingroup$
    I've added my attempt of Integration by parts, don't know where I'm going wrong. Have a look.
    $endgroup$
    – Ajay Choudhary
    Dec 31 '18 at 13:04











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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

$$I=int_0^infty sinleft(sqrt t right)e^{-pt}dtoverset{sqrt t=x}=2int_0^infty xsin x e^{-px^2}dx=int_0^infty sin xleft(-frac1pe^{-px^2}right)'dx$$
$$overset{IBP}=underbrace{-frac1psin xe^{-px^2}bigg|_0^infty}_{=0}+frac1pint_0^infty cos x,e^{-px^2}dx=frac1{2p}int_{-infty}^infty cos x,e^{-px^2}dx$$
We can also make use of the fact that $cos x$ is the real part of $e^{ix}=cos x+isin x$.
$$I=frac1{2p}Re left(int_{-infty}^infty e^{ix}e^{-px^2}dxright)=frac1{2p}Re left(int_{-infty}^infty e^{large-(px^2-ix)+frac{1}{4p}-frac{1}{4p}}dxright)$$
$$=frac1{2p}Re left(int_{-infty}^infty e^{-largeleft(sqrt{p}x-frac{i}{2sqrt p}right)^2 -frac{1}{4p}}dxright)=frac{e^{-frac{1}{4p}}}{2p}Re left(int_{-infty}^infty e^{-largeleft(sqrt{p}x-frac{i}{2sqrt p}right)^2}dxright)$$
Substituting $,displaystyle{sqrt{p}x-frac{i}{2sqrt p}=tRightarrow dx=frac{dt}{sqrt p}}$ gives:
$$I=frac{e^{-frac{1}{4p}}}{2p} frac{1}{sqrt p}Releft(int_{-infty-largefrac{i}{2sqrt p}}^{infty-largefrac{i}{2sqrt p}} e^{-t^2}dtright)=frac{e^{-frac{1}{4p}}}{2p} frac{1}{sqrt p}Releft(int_{-infty}^infty e^{-t^2}dtright)=frac{e^{-frac{1}{4p}}}{2p} sqrt{frac{pi}{p}}$$
For the last line see here and here.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Careful there. After your substitution $sqrt{p}x-i/2sqrt{p}=t$, your new bounds are $infty - i/2sqrt{p}$, and $-infty - i/2sqrt{p}$. You'd have to show that the extra imaginary part does not matter in the calculation of the integral.
    $endgroup$
    – Zachary
    Dec 31 '18 at 14:57








  • 1




    $begingroup$
    Yup. Are you familiar with Contour integration? Use Cauchy's theorem, as in this post: math.stackexchange.com/questions/648043/….
    $endgroup$
    – Zachary
    Dec 31 '18 at 14:59












  • $begingroup$
    @Zachary thanks alot! Over time I alway struggled with contour integration and I tend to rely on real analysis as much as I can. I have also done the same mistake in the past: math.stackexchange.com/q/2861708/515527
    $endgroup$
    – Zacky
    Dec 31 '18 at 15:14


















2












$begingroup$

$$I=int_0^infty sinleft(sqrt t right)e^{-pt}dtoverset{sqrt t=x}=2int_0^infty xsin x e^{-px^2}dx=int_0^infty sin xleft(-frac1pe^{-px^2}right)'dx$$
$$overset{IBP}=underbrace{-frac1psin xe^{-px^2}bigg|_0^infty}_{=0}+frac1pint_0^infty cos x,e^{-px^2}dx=frac1{2p}int_{-infty}^infty cos x,e^{-px^2}dx$$
We can also make use of the fact that $cos x$ is the real part of $e^{ix}=cos x+isin x$.
$$I=frac1{2p}Re left(int_{-infty}^infty e^{ix}e^{-px^2}dxright)=frac1{2p}Re left(int_{-infty}^infty e^{large-(px^2-ix)+frac{1}{4p}-frac{1}{4p}}dxright)$$
$$=frac1{2p}Re left(int_{-infty}^infty e^{-largeleft(sqrt{p}x-frac{i}{2sqrt p}right)^2 -frac{1}{4p}}dxright)=frac{e^{-frac{1}{4p}}}{2p}Re left(int_{-infty}^infty e^{-largeleft(sqrt{p}x-frac{i}{2sqrt p}right)^2}dxright)$$
Substituting $,displaystyle{sqrt{p}x-frac{i}{2sqrt p}=tRightarrow dx=frac{dt}{sqrt p}}$ gives:
$$I=frac{e^{-frac{1}{4p}}}{2p} frac{1}{sqrt p}Releft(int_{-infty-largefrac{i}{2sqrt p}}^{infty-largefrac{i}{2sqrt p}} e^{-t^2}dtright)=frac{e^{-frac{1}{4p}}}{2p} frac{1}{sqrt p}Releft(int_{-infty}^infty e^{-t^2}dtright)=frac{e^{-frac{1}{4p}}}{2p} sqrt{frac{pi}{p}}$$
For the last line see here and here.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Careful there. After your substitution $sqrt{p}x-i/2sqrt{p}=t$, your new bounds are $infty - i/2sqrt{p}$, and $-infty - i/2sqrt{p}$. You'd have to show that the extra imaginary part does not matter in the calculation of the integral.
    $endgroup$
    – Zachary
    Dec 31 '18 at 14:57








  • 1




    $begingroup$
    Yup. Are you familiar with Contour integration? Use Cauchy's theorem, as in this post: math.stackexchange.com/questions/648043/….
    $endgroup$
    – Zachary
    Dec 31 '18 at 14:59












  • $begingroup$
    @Zachary thanks alot! Over time I alway struggled with contour integration and I tend to rely on real analysis as much as I can. I have also done the same mistake in the past: math.stackexchange.com/q/2861708/515527
    $endgroup$
    – Zacky
    Dec 31 '18 at 15:14
















2












2








2





$begingroup$

$$I=int_0^infty sinleft(sqrt t right)e^{-pt}dtoverset{sqrt t=x}=2int_0^infty xsin x e^{-px^2}dx=int_0^infty sin xleft(-frac1pe^{-px^2}right)'dx$$
$$overset{IBP}=underbrace{-frac1psin xe^{-px^2}bigg|_0^infty}_{=0}+frac1pint_0^infty cos x,e^{-px^2}dx=frac1{2p}int_{-infty}^infty cos x,e^{-px^2}dx$$
We can also make use of the fact that $cos x$ is the real part of $e^{ix}=cos x+isin x$.
$$I=frac1{2p}Re left(int_{-infty}^infty e^{ix}e^{-px^2}dxright)=frac1{2p}Re left(int_{-infty}^infty e^{large-(px^2-ix)+frac{1}{4p}-frac{1}{4p}}dxright)$$
$$=frac1{2p}Re left(int_{-infty}^infty e^{-largeleft(sqrt{p}x-frac{i}{2sqrt p}right)^2 -frac{1}{4p}}dxright)=frac{e^{-frac{1}{4p}}}{2p}Re left(int_{-infty}^infty e^{-largeleft(sqrt{p}x-frac{i}{2sqrt p}right)^2}dxright)$$
Substituting $,displaystyle{sqrt{p}x-frac{i}{2sqrt p}=tRightarrow dx=frac{dt}{sqrt p}}$ gives:
$$I=frac{e^{-frac{1}{4p}}}{2p} frac{1}{sqrt p}Releft(int_{-infty-largefrac{i}{2sqrt p}}^{infty-largefrac{i}{2sqrt p}} e^{-t^2}dtright)=frac{e^{-frac{1}{4p}}}{2p} frac{1}{sqrt p}Releft(int_{-infty}^infty e^{-t^2}dtright)=frac{e^{-frac{1}{4p}}}{2p} sqrt{frac{pi}{p}}$$
For the last line see here and here.






share|cite|improve this answer











$endgroup$



$$I=int_0^infty sinleft(sqrt t right)e^{-pt}dtoverset{sqrt t=x}=2int_0^infty xsin x e^{-px^2}dx=int_0^infty sin xleft(-frac1pe^{-px^2}right)'dx$$
$$overset{IBP}=underbrace{-frac1psin xe^{-px^2}bigg|_0^infty}_{=0}+frac1pint_0^infty cos x,e^{-px^2}dx=frac1{2p}int_{-infty}^infty cos x,e^{-px^2}dx$$
We can also make use of the fact that $cos x$ is the real part of $e^{ix}=cos x+isin x$.
$$I=frac1{2p}Re left(int_{-infty}^infty e^{ix}e^{-px^2}dxright)=frac1{2p}Re left(int_{-infty}^infty e^{large-(px^2-ix)+frac{1}{4p}-frac{1}{4p}}dxright)$$
$$=frac1{2p}Re left(int_{-infty}^infty e^{-largeleft(sqrt{p}x-frac{i}{2sqrt p}right)^2 -frac{1}{4p}}dxright)=frac{e^{-frac{1}{4p}}}{2p}Re left(int_{-infty}^infty e^{-largeleft(sqrt{p}x-frac{i}{2sqrt p}right)^2}dxright)$$
Substituting $,displaystyle{sqrt{p}x-frac{i}{2sqrt p}=tRightarrow dx=frac{dt}{sqrt p}}$ gives:
$$I=frac{e^{-frac{1}{4p}}}{2p} frac{1}{sqrt p}Releft(int_{-infty-largefrac{i}{2sqrt p}}^{infty-largefrac{i}{2sqrt p}} e^{-t^2}dtright)=frac{e^{-frac{1}{4p}}}{2p} frac{1}{sqrt p}Releft(int_{-infty}^infty e^{-t^2}dtright)=frac{e^{-frac{1}{4p}}}{2p} sqrt{frac{pi}{p}}$$
For the last line see here and here.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 31 '18 at 20:50









Ajay Choudhary

888




888










answered Dec 31 '18 at 14:48









ZackyZacky

7,3101961




7,3101961








  • 1




    $begingroup$
    Careful there. After your substitution $sqrt{p}x-i/2sqrt{p}=t$, your new bounds are $infty - i/2sqrt{p}$, and $-infty - i/2sqrt{p}$. You'd have to show that the extra imaginary part does not matter in the calculation of the integral.
    $endgroup$
    – Zachary
    Dec 31 '18 at 14:57








  • 1




    $begingroup$
    Yup. Are you familiar with Contour integration? Use Cauchy's theorem, as in this post: math.stackexchange.com/questions/648043/….
    $endgroup$
    – Zachary
    Dec 31 '18 at 14:59












  • $begingroup$
    @Zachary thanks alot! Over time I alway struggled with contour integration and I tend to rely on real analysis as much as I can. I have also done the same mistake in the past: math.stackexchange.com/q/2861708/515527
    $endgroup$
    – Zacky
    Dec 31 '18 at 15:14
















  • 1




    $begingroup$
    Careful there. After your substitution $sqrt{p}x-i/2sqrt{p}=t$, your new bounds are $infty - i/2sqrt{p}$, and $-infty - i/2sqrt{p}$. You'd have to show that the extra imaginary part does not matter in the calculation of the integral.
    $endgroup$
    – Zachary
    Dec 31 '18 at 14:57








  • 1




    $begingroup$
    Yup. Are you familiar with Contour integration? Use Cauchy's theorem, as in this post: math.stackexchange.com/questions/648043/….
    $endgroup$
    – Zachary
    Dec 31 '18 at 14:59












  • $begingroup$
    @Zachary thanks alot! Over time I alway struggled with contour integration and I tend to rely on real analysis as much as I can. I have also done the same mistake in the past: math.stackexchange.com/q/2861708/515527
    $endgroup$
    – Zacky
    Dec 31 '18 at 15:14










1




1




$begingroup$
Careful there. After your substitution $sqrt{p}x-i/2sqrt{p}=t$, your new bounds are $infty - i/2sqrt{p}$, and $-infty - i/2sqrt{p}$. You'd have to show that the extra imaginary part does not matter in the calculation of the integral.
$endgroup$
– Zachary
Dec 31 '18 at 14:57






$begingroup$
Careful there. After your substitution $sqrt{p}x-i/2sqrt{p}=t$, your new bounds are $infty - i/2sqrt{p}$, and $-infty - i/2sqrt{p}$. You'd have to show that the extra imaginary part does not matter in the calculation of the integral.
$endgroup$
– Zachary
Dec 31 '18 at 14:57






1




1




$begingroup$
Yup. Are you familiar with Contour integration? Use Cauchy's theorem, as in this post: math.stackexchange.com/questions/648043/….
$endgroup$
– Zachary
Dec 31 '18 at 14:59






$begingroup$
Yup. Are you familiar with Contour integration? Use Cauchy's theorem, as in this post: math.stackexchange.com/questions/648043/….
$endgroup$
– Zachary
Dec 31 '18 at 14:59














$begingroup$
@Zachary thanks alot! Over time I alway struggled with contour integration and I tend to rely on real analysis as much as I can. I have also done the same mistake in the past: math.stackexchange.com/q/2861708/515527
$endgroup$
– Zacky
Dec 31 '18 at 15:14






$begingroup$
@Zachary thanks alot! Over time I alway struggled with contour integration and I tend to rely on real analysis as much as I can. I have also done the same mistake in the past: math.stackexchange.com/q/2861708/515527
$endgroup$
– Zacky
Dec 31 '18 at 15:14













2












$begingroup$

Hint:
Use the expansion of $sin$
$$sinsqrt{t}=sum_{n=0}^{infty}(-1)^ndfrac{t^{n+1/2}}{Gamma(2n+2)}$$
Edit:
$${cal L}(sinsqrt{t})=sum_{n=0}^{infty}(-1)^ndfrac{Gamma(n+3/2)}{Gamma(2n+2)p^{n+3/2}}$$
with Legendre Duplication Formula we have
$${cal L}(sinsqrt{t})=dfrac{1}{p^{3/2}}sum_{n=0}^{infty}(-1)^ndfrac{Gamma(n+3/2)}{sqrt{pi}^{-1}2^{2n+1}Gamma(n+1)Gamma(n+3/2)p^n}$$
$$=dfrac{sqrt{pi}}{p^{3/2}}sum_{n=0}^{infty}dfrac{left(frac{-1}{4p}right)^{n}}{n!}=color{blue}{dfrac{sqrt{pi}}{p^{3/2}}e^{frac{-1}{4p}}}$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I know it can be solved by using the expansion. But I was looking for some other methods to do the same. Can you see if I applied by parts correctly?
    $endgroup$
    – Ajay Choudhary
    Dec 31 '18 at 13:30










  • $begingroup$
    How exactly would you use that?
    $endgroup$
    – Zachary
    Dec 31 '18 at 13:31










  • $begingroup$
    @AjayChoudhary you should say that in your question.
    $endgroup$
    – Nosrati
    Dec 31 '18 at 13:39
















2












$begingroup$

Hint:
Use the expansion of $sin$
$$sinsqrt{t}=sum_{n=0}^{infty}(-1)^ndfrac{t^{n+1/2}}{Gamma(2n+2)}$$
Edit:
$${cal L}(sinsqrt{t})=sum_{n=0}^{infty}(-1)^ndfrac{Gamma(n+3/2)}{Gamma(2n+2)p^{n+3/2}}$$
with Legendre Duplication Formula we have
$${cal L}(sinsqrt{t})=dfrac{1}{p^{3/2}}sum_{n=0}^{infty}(-1)^ndfrac{Gamma(n+3/2)}{sqrt{pi}^{-1}2^{2n+1}Gamma(n+1)Gamma(n+3/2)p^n}$$
$$=dfrac{sqrt{pi}}{p^{3/2}}sum_{n=0}^{infty}dfrac{left(frac{-1}{4p}right)^{n}}{n!}=color{blue}{dfrac{sqrt{pi}}{p^{3/2}}e^{frac{-1}{4p}}}$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I know it can be solved by using the expansion. But I was looking for some other methods to do the same. Can you see if I applied by parts correctly?
    $endgroup$
    – Ajay Choudhary
    Dec 31 '18 at 13:30










  • $begingroup$
    How exactly would you use that?
    $endgroup$
    – Zachary
    Dec 31 '18 at 13:31










  • $begingroup$
    @AjayChoudhary you should say that in your question.
    $endgroup$
    – Nosrati
    Dec 31 '18 at 13:39














2












2








2





$begingroup$

Hint:
Use the expansion of $sin$
$$sinsqrt{t}=sum_{n=0}^{infty}(-1)^ndfrac{t^{n+1/2}}{Gamma(2n+2)}$$
Edit:
$${cal L}(sinsqrt{t})=sum_{n=0}^{infty}(-1)^ndfrac{Gamma(n+3/2)}{Gamma(2n+2)p^{n+3/2}}$$
with Legendre Duplication Formula we have
$${cal L}(sinsqrt{t})=dfrac{1}{p^{3/2}}sum_{n=0}^{infty}(-1)^ndfrac{Gamma(n+3/2)}{sqrt{pi}^{-1}2^{2n+1}Gamma(n+1)Gamma(n+3/2)p^n}$$
$$=dfrac{sqrt{pi}}{p^{3/2}}sum_{n=0}^{infty}dfrac{left(frac{-1}{4p}right)^{n}}{n!}=color{blue}{dfrac{sqrt{pi}}{p^{3/2}}e^{frac{-1}{4p}}}$$






share|cite|improve this answer











$endgroup$



Hint:
Use the expansion of $sin$
$$sinsqrt{t}=sum_{n=0}^{infty}(-1)^ndfrac{t^{n+1/2}}{Gamma(2n+2)}$$
Edit:
$${cal L}(sinsqrt{t})=sum_{n=0}^{infty}(-1)^ndfrac{Gamma(n+3/2)}{Gamma(2n+2)p^{n+3/2}}$$
with Legendre Duplication Formula we have
$${cal L}(sinsqrt{t})=dfrac{1}{p^{3/2}}sum_{n=0}^{infty}(-1)^ndfrac{Gamma(n+3/2)}{sqrt{pi}^{-1}2^{2n+1}Gamma(n+1)Gamma(n+3/2)p^n}$$
$$=dfrac{sqrt{pi}}{p^{3/2}}sum_{n=0}^{infty}dfrac{left(frac{-1}{4p}right)^{n}}{n!}=color{blue}{dfrac{sqrt{pi}}{p^{3/2}}e^{frac{-1}{4p}}}$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 31 '18 at 13:38

























answered Dec 31 '18 at 13:18









NosratiNosrati

26.6k62354




26.6k62354












  • $begingroup$
    I know it can be solved by using the expansion. But I was looking for some other methods to do the same. Can you see if I applied by parts correctly?
    $endgroup$
    – Ajay Choudhary
    Dec 31 '18 at 13:30










  • $begingroup$
    How exactly would you use that?
    $endgroup$
    – Zachary
    Dec 31 '18 at 13:31










  • $begingroup$
    @AjayChoudhary you should say that in your question.
    $endgroup$
    – Nosrati
    Dec 31 '18 at 13:39


















  • $begingroup$
    I know it can be solved by using the expansion. But I was looking for some other methods to do the same. Can you see if I applied by parts correctly?
    $endgroup$
    – Ajay Choudhary
    Dec 31 '18 at 13:30










  • $begingroup$
    How exactly would you use that?
    $endgroup$
    – Zachary
    Dec 31 '18 at 13:31










  • $begingroup$
    @AjayChoudhary you should say that in your question.
    $endgroup$
    – Nosrati
    Dec 31 '18 at 13:39
















$begingroup$
I know it can be solved by using the expansion. But I was looking for some other methods to do the same. Can you see if I applied by parts correctly?
$endgroup$
– Ajay Choudhary
Dec 31 '18 at 13:30




$begingroup$
I know it can be solved by using the expansion. But I was looking for some other methods to do the same. Can you see if I applied by parts correctly?
$endgroup$
– Ajay Choudhary
Dec 31 '18 at 13:30












$begingroup$
How exactly would you use that?
$endgroup$
– Zachary
Dec 31 '18 at 13:31




$begingroup$
How exactly would you use that?
$endgroup$
– Zachary
Dec 31 '18 at 13:31












$begingroup$
@AjayChoudhary you should say that in your question.
$endgroup$
– Nosrati
Dec 31 '18 at 13:39




$begingroup$
@AjayChoudhary you should say that in your question.
$endgroup$
– Nosrati
Dec 31 '18 at 13:39











0












$begingroup$

Hint: Use integration by parts two times. The result should be
$$frac{sqrt{pi}e^{-1/(4p)}}{2p^{2/3}}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I've added my attempt of Integration by parts, don't know where I'm going wrong. Have a look.
    $endgroup$
    – Ajay Choudhary
    Dec 31 '18 at 13:04
















0












$begingroup$

Hint: Use integration by parts two times. The result should be
$$frac{sqrt{pi}e^{-1/(4p)}}{2p^{2/3}}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I've added my attempt of Integration by parts, don't know where I'm going wrong. Have a look.
    $endgroup$
    – Ajay Choudhary
    Dec 31 '18 at 13:04














0












0








0





$begingroup$

Hint: Use integration by parts two times. The result should be
$$frac{sqrt{pi}e^{-1/(4p)}}{2p^{2/3}}$$






share|cite|improve this answer









$endgroup$



Hint: Use integration by parts two times. The result should be
$$frac{sqrt{pi}e^{-1/(4p)}}{2p^{2/3}}$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 31 '18 at 12:33









Dr. Sonnhard GraubnerDr. Sonnhard Graubner

76.7k42866




76.7k42866












  • $begingroup$
    I've added my attempt of Integration by parts, don't know where I'm going wrong. Have a look.
    $endgroup$
    – Ajay Choudhary
    Dec 31 '18 at 13:04


















  • $begingroup$
    I've added my attempt of Integration by parts, don't know where I'm going wrong. Have a look.
    $endgroup$
    – Ajay Choudhary
    Dec 31 '18 at 13:04
















$begingroup$
I've added my attempt of Integration by parts, don't know where I'm going wrong. Have a look.
$endgroup$
– Ajay Choudhary
Dec 31 '18 at 13:04




$begingroup$
I've added my attempt of Integration by parts, don't know where I'm going wrong. Have a look.
$endgroup$
– Ajay Choudhary
Dec 31 '18 at 13:04


















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