# Integrate $\int_{0}^{\infty}e^{-pt}\sin\left(\sqrt{t}\right)\mathrm dt$

I need the following Laplace transform to solve the Differential Equation

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

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

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

• Have you tried using Moivre's relationship? $\sin(\theta)=\frac{ e^{i\theta}-e^{-i\theta}}{2i}$, It could help – RScrlli Dec 31 '18 at 12:30
• Substitute $\sqrt{t}=y$, to get $$\int_{0}^{\infty}e^{-pt}\sin\sqrt{t}~dt=2\int_{0}^{\infty}e^{-py^2}y\sin y~dy=2\mathrm{Im}\int_{0}^{\infty}e^{-py^2+\mathrm{i}y}y~dy\ ,$$ where Im stands for imaginary part. – Pierpaolo Vivo Dec 31 '18 at 12:32
• 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. – Zachary Dec 31 '18 at 15:04
• Are you sure that is fine? I really don't have knowledge for that. – LeBlanc Dec 31 '18 at 15:05
• 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. – Zachary Dec 31 '18 at 15:13

$$I=\int_0^\infty \sin\left(\sqrt t \right)e^{-pt}dt\overset{\sqrt t=x}=2\int_0^\infty x\sin x e^{-px^2}dx=\int_0^\infty \sin x\left(-\frac1pe^{-px^2}\right)'dx$$ $$\overset{IBP}=\underbrace{-\frac1p\sin xe^{-px^2}\bigg|_0^\infty}_{=0}+\frac1p\int_0^\infty \cos x\,e^{-px^2}dx=\frac1{2p}\int_{-\infty}^\infty \cos x\,e^{-px^2}dx$$ We can also make use of the fact that $$\cos x$$ is the real part of $$e^{ix}=\cos x+i\sin x$$. $$I=\frac1{2p}\Re \left(\int_{-\infty}^\infty e^{ix}e^{-px^2}dx\right)=\frac1{2p}\Re \left(\int_{-\infty}^\infty e^{\large-(px^2-ix)+\frac{1}{4p}-\frac{1}{4p}}dx\right)$$ $$=\frac1{2p}\Re \left(\int_{-\infty}^\infty e^{-\large\left(\sqrt{p}x-\frac{i}{2\sqrt p}\right)^2 -\frac{1}{4p}}dx\right)=\frac{e^{-\frac{1}{4p}}}{2p}\Re \left(\int_{-\infty}^\infty e^{-\large\left(\sqrt{p}x-\frac{i}{2\sqrt p}\right)^2}dx\right)$$ Substituting $$\,\displaystyle{\sqrt{p}x-\frac{i}{2\sqrt p}=t\Rightarrow dx=\frac{dt}{\sqrt p}}$$ gives: $$I=\frac{e^{-\frac{1}{4p}}}{2p} \frac{1}{\sqrt p}\Re\left(\int_{-\infty-\large\frac{i}{2\sqrt p}}^{\infty-\large\frac{i}{2\sqrt p}} e^{-t^2}dt\right)=\frac{e^{-\frac{1}{4p}}}{2p} \frac{1}{\sqrt p}\Re\left(\int_{-\infty}^\infty e^{-t^2}dt\right)=\frac{e^{-\frac{1}{4p}}}{2p} \sqrt{\frac{\pi}{p}}$$ For the last line see here and here.

• Careful there. After your substitution $\sqrt{p}x-i/2\sqrt{p}=t$, your new bounds are $\infty - i/2\sqrt{p}$, and $-\infty - i/2\sqrt{p}$. You'd have to show that the extra imaginary part does not matter in the calculation of the integral. – Zachary Dec 31 '18 at 14:57
• Yup. Are you familiar with Contour integration? Use Cauchy's theorem, as in this post: math.stackexchange.com/questions/648043/…. – Zachary Dec 31 '18 at 14:59
• @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 – LeBlanc Dec 31 '18 at 15:14

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

• 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? – Ajay Choudhary Dec 31 '18 at 13:30
• How exactly would you use that? – Zachary Dec 31 '18 at 13:31
• @AjayChoudhary you should say that in your question. – Nosrati Dec 31 '18 at 13:39

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

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