Could anyone show me how to perform this integral $$\int_{-\infty}^{\infty}\frac{e^{-x^2}}{z-x}dx, z \in \mathbb{C}$$ Thanks. The following is what I have tried.
Set $z = iz^\prime$, also make the substitution $v = iz^\prime - x$, then
$$-e^{z^{\prime 2}}\int_{-\infty}^{\infty}\frac{e^{-v^2+2ivz^\prime}}{v}dv.$$
Then we can set $f(z) = \int_{-\infty}^{\infty}\frac{e^{-v^2+2ivz}}{v}dv$, so
$$f^\prime(z) = 2i\int_{-\infty}^{\infty}{e^{-v^2+2ivz}}dv
=2ie^{-z^2}.$$
Also we have $f(0) = 0$. Then
$$f(z) = \int_0^z f^\prime(z)dz=i\int_{-z}^z e^{-t^2}dt=i\mathrm{erf}(z).$$
So the final result should be like
$$\int_{-\infty}^{\infty}\frac{e^{-x^2}}{z-x}dx = -ie^{-z^2}\mathrm{erf}(-iz).$$
But the answer is $e^{z^2}(1-\mathrm{erf}(z))$. It seems the answer is wrong, since when $z=0$, the value should be 0 because the integrand is an odd function. But then I tried the wollfram alpha, The result is
The result is in accordance with the answer. I know that there are some steps which are not rigid, could someone give me the full derivation of this problem? Thanks.
Assuming $z\not\in\mathbb{R}$, the given integral equals
$$ \int_{0}^{+\infty}\frac{2z}{z^2-x^2}e^{-x^2}\,dx \stackrel{x\mapsto\sqrt{u}}{=}z\int_{0}^{+\infty}\frac{e^{-u}}{\sqrt{u}(z^2-u)}\,du $$ that just depends on the Laplace transform of $\frac{1}{\sqrt{u}(z^2-u)}$, that can be computed from the Laplace transform of $\frac{2}{z}\text{arctanh}\left(\frac{\sqrt{u}}{z}\right)$ or simply from the Laplace transform of $\text{arctanh}(\sqrt{u})$.