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}
Best Answer
$$I=\int_0^\infty \sin\left(\sqrt t \right)e^{-pt}dt\overset{\sqrt t=x}=2\int_0^\infty x\sin x e^{-px^2}dx=\int_0^\infty \sin x\left(-\frac1pe^{-px^2}\right)'dx$$ $$\overset{IBP}=\underbrace{-\frac1p\sin xe^{-px^2}\bigg|_0^\infty}_{=0}+\frac1p\int_0^\infty \cos x\,e^{-px^2}dx=\frac1{2p}\int_{-\infty}^\infty \cos x\,e^{-px^2}dx$$ 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.