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







share|cite|improve this question











$endgroup$








  • 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












$begingroup$


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











$endgroup$








  • 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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 31 '18 at 16:36







Ajay Choudhary

















asked Dec 31 '18 at 12:23









Ajay ChoudharyAjay Choudhary

888




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.






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












$begingroup$

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



















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











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
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3057658%2fintegrate-int-0-inftye-pt-sin-left-sqrtt-right-mathrm-dt%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























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


















draft saved

draft discarded




















































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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3057658%2fintegrate-int-0-inftye-pt-sin-left-sqrtt-right-mathrm-dt%23new-answer', 'question_page');
}
);

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







-

Popular posts from this blog

Bressuire

Image
.mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid
Read more

Cabo Verde

Image
  Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º)
Read more

Gyllenstierna

Image
.mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid
Read more