<s>[INST] 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}
 [/INST] 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
$${\cal L}(\sin\sqrt{t})=\dfrac{1}{p^{3/2}}\sum_{n=0}^{\infty}(-1)^n\dfrac{\Gamma(n+3/2)}{\sqrt{\pi}^{-1}2^{2n+1}\Gamma(n+1)\Gamma(n+3/2)p^n}$$
$$=\dfrac{\sqrt{\pi}}{p^{3/2}}\sum_{n=0}^{\infty}\dfrac{\left(\frac{-1}{4p}\right)^{n}}{n!}=\color{blue}{\dfrac{\sqrt{\pi}}{p^{3/2}}e^{\frac{-1}{4p}}}$$
</s>