Working with a Fourier transform problem, I've encountered the following integral:
$$ \int_{-\infty}^{\infty}\frac{\exp\left(-a^2x^2+ibx\right)}{x^2+c^2}dx $$
where $a$, $b$, and $c$ are real coefficients. Mathematica claims that this has no closed-form solution, but I suspect (hope?) that's not the case. Unfortunately, my background in complex analysis is limited to some panic-studying I did to make it through a field theory course a decade ago. I've spent the weekend dusting that off, but am still stumped...
I know there's a pole at $z=ci$. If I take (to me) the obvious extension of the integral to the complex plane by simply replacing $x$ with $z$, I can even calculate its residue as:
$$ \frac{\exp\left(a^2c^2-bc\right)}{2ci} $$
My hope was to integrate over a semicircle in the positive half-plane, and use this residue to get the integral along the real axis. However, with this extension, the integral over the semicircle of radius R doesn't seem to be zero (or possibly even converge) with R going to infinity -- or at least, Jordan's lemma doesn't give me any reason to believe so, since the $-a^2z^2$ part becomes unfriendly on the imaginary axis.
I vaguely recall there being a strategy for dealing with this sort of problem. I specifically recall the strategy not being to use $zz^*$, since explicit dependence on the complex conjugate was a no-go. For the life of me, though, I can't recall what the strategy was.
If anybody can (a) refresh my memory on this; (b) give a solution method; or (c) explain why no closed form solution exists, that would be much appreciated. Thanks in advance for any help you can offer!
The gaussian term makes complex analysis iffy. This integral may be done much more easily using the convolution theorem.
To begin, note that
$$\int_{-\infty}^{\infty} dx \, e^{-a x^2} e^{i k x} = \sqrt{\frac{\pi}{a}} e^{-k^2/(4 a)}$$
$$\int_{-\infty}^{\infty} dx \frac{e^{i k x}}{x^2+c^2} = \frac{\pi}{c} e^{-c |k|}$$
The integral sought is the convolution of these transforms. That is,
$$\int_{-\infty}^{\infty} dx \frac{e^{-a x^2}}{x^2+c^2} e^{i b x} = \frac{1}{2 \pi} \sqrt{\frac{\pi}{a}} \frac{\pi}{c} \int_{-\infty}^{\infty} dk \,e^{-k^2/(4 a)} \, e^{-c |b-k|} $$
This integral is very doable, despite what Mathematica says. For the time being, I will spare you the details, which involve splitting the integral about $k=b$ and rescaling to get a nice form. The result is that the sought-after integral is
$$\frac{\pi}{c} e^{a c^2} \left [e^{-b c} \left (1+\text{erf}\left (\frac{b-2 a c}{2 \sqrt{a}} \right ) \right ) + e^{b c} \left (1-\text{erf}\left (\frac{b+2 a c}{2 \sqrt{a}} \right ) \right )\right ] $$
where erf is the error function.
NOTE
My $a$ is the OP's $a^2$ - this was not intentional, but to get an answer the OP wants, sub $a^2$ for $a$ in the above result.