How to evaluate this gaussian integral Is there a way to solve this integral?
$$\int_{-\infty}^{\infty} e^{-(s^2+2\sqrt{t}s)}\sin(y+2\sqrt{t}s)\,ds$$
I usually apply this formula for integrals of this kind
$$\int_{-\infty}^{\infty} e^{-(as^2+bs+c)}\,ds=\sqrt{\frac{\pi}{a}}e^{\frac{b^2}{4a}-c}$$
Thank you in advance
 A: If you're OK with working with functions of a complex variable, the formula you gave still applies if $a,b,c\in\mathbb C$ provided that $\mathrm{Re}[a] > 0$. It gives
\begin{multline}
\int_{-\infty}^{\infty} e^{-(s^2+2\sqrt{t}s)}\sin(y+2\sqrt{t}s)\,ds = \mathrm{Im}\left[\int_{-\infty}^{\infty} e^{-(s^2+2\sqrt{t}s)}e^{i\left(y+2\sqrt{t}s\right)}ds\right]
\\= \mathrm{Im}\left[\int_{-\infty}^{\infty} e^{-[s^2+2(1-i)\sqrt{t}s-iy]}ds\right] = \mathrm{Im}\left[\sqrt{\pi}\exp\left(-2it +i y\right)\right] = \sqrt{\pi}\sin(y-2t).
\end{multline}
If you're not OK with working with complex numbers, there actually are formulas for these:
\begin{eqnarray}
\int_{-\infty}^\infty e^{-(as^2 + bs + c)}\sin(ks + m)ds &=& \sqrt{\frac{\pi}{a}}e^{\frac{b^2-k^2}{4a} - c}\sin\left(m-\frac{b k}{2a}\right)\\
\int_{-\infty}^\infty e^{-(as^2 + bs + c)}\cos(ks + m)ds &=& \sqrt{\frac{\pi}{a}}e^{\frac{b^2-k^2}{4a} - c}\cos\left(m-\frac{b k}{2a}\right)
\end{eqnarray}
These formulas follow directly from the complex form of the integral I used before, and if you plug in the values, you'll get $\sqrt{\pi}\sin(y-2t)$ again. 
If you want to know how to get these formulas without using complex numbers, well, that's a more involved post.
