i am trying to solve the next integral: \begin{align} & \int_{-\infty}^{\infty} \frac{xe^{-a^2 x^2}}{\sqrt{(b-x)^2+c}}\,dx \\ & {} \quad\end{align} with $c>0$ and $a,b,c\in\mathbb{R}$ but i cannot quite do it. We have two branches ($b+i\sqrt{c}$ and $b-i\sqrt{c}$)
my idea was to do it by residues in the positive semicircle and the circle around $b+i\sqrt{c}$ of radius $r \to 0$. However the function $e^{-z^2}=e^{-R^2e^{i2\theta}}$ is unbounded (as $R\to \infty$) for $\theta\in (\pi/4,3\pi/4)$
A friend recommended me to do it by Euler substitution but i'm not sure how to do it so i did \begin{align} \sqrt{x^2-2bx+(c+b^2)}=x+t \end{align} \begin{align} x=\frac{c+b^2-t^2}{2(b+t)}; dx=\frac{-t^2-2bt-c-b^2}{2(b+t)^2} \end{align} obtaining after some manipulation \begin{align} \frac{-(t+b)e^{-a^2(\frac{c+b^2-t^2}{2(b+t)})^2}}{a^2(t^2+2bt+c+b^2)}+\frac{1}{a^2}\int_{}^{} (\frac{2c}{(t^2+2bt+c+b^2)^2}-\frac{1}{(t^2+2bt+c+b^2)})e^{-a^2(\frac{c+b^2-t^2}{2(b+t)})^2} \end{align}
how should i continue? the only informations i have about the result is that i should obtain something with the error function (erf). i'm also ok with approximations