calculus - Evaluate $int_0^infty 2 u sin u, exp(-su^2)mathrm du$

Evaluate $\int_0^\infty 2u \sin u \,e^{-su^2} \mathrm du$.

This integral emerged while finding Laplace Transform of $\sin(\sqrt u)$

As a start, I used:

$$\begin{align*} I &= 2\int_{u=0}^{\infty} \int_{v=0}^{u} \cos v~\mathrm dv\mathrm du\\ &= \frac{1}{s} \int_{0}^\infty \cos v \, e^{-sv^2 }\mathrm dv \end{align*}$$

Again here I am stuck.

Answer

(We assume $\operatorname{Re}s >0$)
$$I= \int_0^\infty 2u \sin u e^{-s u^2} \,du = \operatorname{Im} \int_{-\infty}^\infty u e^{i u - s u^2}\,du = \operatorname{Im} \frac{d}{i d \alpha} \int_{-\infty}^\infty e^{i \alpha u - s u^2}\,du \Big|_{\alpha=1}\,.$$

Now, we have

$$\int_{-\infty}^\infty e^{i \alpha u - s u^2} \,du = e^{-\alpha^2/4s} \int_{-\infty}^\infty e^{-s(u-i\alpha/2 s)^2}\,du= \sqrt{\frac{\pi}{s}}e^{-\alpha^2/4s}\,.$$

As a result, we obtain
$$I = \frac{\sqrt{\pi}}{2 s^{3/2}} e^{-1/4 s}\,.$$