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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}

$$I=int_0^infty sinleft(sqrt t right)e^{-pt}dtoverset{sqrt t=x}=2int_0^infty xsin x e^{-px^2}dx=int_0^infty sin xleft(-frac1pe^{-px^2}right)'dx$$
$$overset{IBP}=underbrace{-frac1psin xe^{-px^2}bigg|_0^infty}_{=0}+frac1pint_0^infty cos x,e^{-px^2}dx=frac1{2p}int_{-infty}^infty cos x,e^{-px^2}dx$$
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.

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
$${cal L}(sinsqrt{t})=dfrac{1}{p^{3/2}}sum_{n=0}^{infty}(-1)^ndfrac{Gamma(n+3/2)}{sqrt{pi}^{-1}2^{2n+1}Gamma(n+1)Gamma(n+3/2)p^n}$$
$$=dfrac{sqrt{pi}}{p^{3/2}}sum_{n=0}^{infty}dfrac{left(frac{-1}{4p}right)^{n}}{n!}=color{blue}{dfrac{sqrt{pi}}{p^{3/2}}e^{frac{-1}{4p}}}$$

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