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Profile · Posted 2020-07-23, 06:27 PM

I got this idea from an integral on Wikipedia:

$\reverse \int_{-\infty}^{\infty} \frac{e^{itx}}{x^2+1}\,\mathrm{d}x = \pi e^{-|t|}$

This integral is quite difficult to solve with standard techniques from elementary calculus. Instead, the usual approach is to rewrite it as $\reverse \int_{-\infty}^{\infty} \frac{e^{itx}}{(x-i)(x+i)}\,\mathrm{d}x$ , then evaluate it as a contour integral by using the following contour that surrounds the pole $\reverse x=i$ :

The point of this approach is that it leaves the other pole $\reverse x=-i$ outside the nice semicircle contour no matter the value of a, which lets you convert the problem into a "simple" calculation of limits (e.g. by using the residue theorem). Compared to integration, the calculation of limits is a relatively algorithmic process that usually works without any "art".

So I figured, why not add more poles to the lower half-plane? After all, this shouldn't affect the contour, and you should still be able to use the residue theorem on a single pole. In other words, you should, at least in principle, still be able to find exact solutions to such integrals by calculating a finite number of limits, even if they might be somewhat... messy.

Well, as far as I can tell, they are. I played around a bit with this one:

$\reverse I_n = \int_{-\infty}^{\infty} \frac{e^{itx}}{(x^2+1)\prod_{k=1}^{n}(x-k+i)(x+k+i)}\,\mathrm{d}x$

Obviously $\reverse I_n$ has a single pole $\reverse x=i$ in the upper half-plane, and its remaining $\reverse 2n+1$ poles in the lower half-plane.

Apparently I failed to evaluate any such integrals (except $\reverse I_0$ ) manually, but Mathematica returned some... interesting results for n=1 and n=2 after half an hour or so:

Have you seen any other particularly hideous integrals?