Laplace transform of $\sin(\sqrt{t})$ I'm tired and completely blank on how to find a solution of Laplace transformation of $\sin(\sqrt{t})$.Please specify the method used and if possible something other than using series method.
thank you
 A: You want to evaluate the integral
$$\underbrace{\int_0^{\infty} dt \, e^{-s t} \sin{\sqrt{t}}}_{t=u^2} = 2 \int_0^{\infty} du \, u \, e^{-s u^2} \sin{u} $$
The integral may be evaluated via Cauchy's theorem by considering the contour integral
$$\oint_C dz \, z \, e^{-s z^2} $$
where $C$ is the rectangle having vertices $\pm R \pm i/(2 s)$, and the integral traverses $C$ in a positive sense.  Because the integral is zero, we may then show that, when we take the limit $R \to \infty$, we get
$$ 2 \int_0^{\infty} du \, u \, e^{-s u^2} \sin{u} = \frac{1}{s} e^{1/(4 s)}\: \Re{\left [ \int_{-\infty}^{\infty} dx \, e^{-s \left ( x-\frac{i}{2s}\right )^2}\right ]} $$
which may be evaluated via Cauchy's theorem, again, this time on the integral
$$\oint_{C'} dz \, e^{-s z^2} = 0$$
where $C'$ is the rectangle having vertices $\pm R$ and $\pm R-i/(2 s)$.  By taking the limit as $R\to\infty$, we find that
$$\Re{\left [ \int_{-\infty}^{\infty} dx \, e^{-s \left ( x-\frac{i}{2 s}\right )^2}\right ]} = \int_{-\infty}^{\infty} dx \, e^{-s x^2} = \sqrt{\frac{\pi}{s}}$$
Therefore,
$$\int_0^{\infty} dt \, e^{-s t} \sin{\sqrt{t}} = \frac12 \sqrt{\pi} \, s^{-3/2} \, e^{-1/(4 s)}$$
