I am trying to evaluate the following integral: \begin{equation} I=\int_{-\infty}^{\infty}\exp\left \{-\frac{(u-1)^2}{2\sigma^2}\right\}\frac{1}{(u-x)^2+y^2}\mathrm{d}u \end{equation} where $x,y\in\mathbb{R}$.

Trying on mathematica tells me that the integral does not converge. Moreover I cannot use the residue theorem due to the $\exp\{-u^2\}$ term. However when plotting the function for certain $(x,y)$ values the function seems to have a closed area so I would expect the integral to converge given some restraints on $x$ and $y$. How does one tackle such integrals?

In a broader context, under what conditions does the following integral converge? \begin{equation} I_\text{gen}=\int_{-\infty}^{\infty}\exp\left \{-\frac{(x-\mu)^2}{2\sigma^2}\right\}\frac{1}{ax^2+bx+c}\mathrm{d}x \end{equation}


1 Answer 1


$\DeclareMathOperator{\erfc}{erfc}$ The integral $$\begin{align}\int_{-\infty}^{\infty}\frac{\mathrm{e}^{-t^2}\mathrm{d}t}{\left\lvert t- z\right\rvert^2}&=\frac{\pi}{\Im z}\Re\left(\mathrm{e}^{-z^2}\erfc(-\mathrm{i}z)\right) &&(\Im z \neq 0)\end{align}$$ is standard (DLMF 7.19.4).

Shift and rescale your integrals so that they look like the form on the left.

  • $\begingroup$ Thanks, I did not know this identity! By rescaling, is the change of variables $u=\sqrt{2}\sigma t +1$ enough? It does take care of the exponential but then I get $|t\sqrt{2}\sigma+1-z|$ in the denominator. Sorry for asking something that is probably very simple, I just don't see it! $\endgroup$
    – Matt
    Commented Aug 24, 2019 at 16:39
  • 1
    $\begingroup$ pull out the factor of $\sqrt{2}\sigma$, then shift and rescale $z$ $\endgroup$
    – K B Dave
    Commented Aug 24, 2019 at 16:52

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