I wanted an interesting one, so I started with
$\lim\limits_{N\to\infty}\int\limits_{-\infty}^{\infty} \operatorname{sech}\left(x-\sum\limits_{n=1}^{N}\frac{1}{x+n^{2}}\right)dx$
The idea was that, by Glasser's master theorem, this direction should be easy to evaluate, as it should just become
$\lim\limits_{N\to\infty}\int\limits_{-\infty}^{\infty} \operatorname{sech}(x)dx=\lim\limits_{N\to\infty}\pi=\pi$
Then, the other direction will be
$\int\limits_{-\infty}^{\infty}\lim\limits_{N\to\infty}\operatorname{sech}\left(x-\sum\limits_{n=1}^{N}\frac{1}{x+n^{2}}\right)dx=\int\limits_{-\infty}^{\infty}\operatorname{sech}\left(x-\sum\limits_{n=1}^{\infty}\frac{1}{x+n^{2}}\right)dx$
Using the result from complex analysis that states
$\sum\limits_{k=1}^{\infty}\frac{1}{k^{2}+a^{2}}=-\frac{1}{2a^2}+\frac{\pi}{2a}\coth(\pi a)$
the integral should transform into
$\int\limits_{-\infty}^{0}\operatorname{sech}(x+\frac{1}{2x}+\frac{\pi}{2\sqrt{-x}}\cot(\pi\sqrt{-x}))dx+\int\limits_{0}^{\infty}\operatorname{sech}(x+\frac{1}{2x}-\frac{\pi}{2\sqrt{x}}\coth(\pi\sqrt{x}))dx$
A substitution $-x\mapsto x$ on the first makes it
$\int\limits_{0}^{\infty} \operatorname{sech}\left(x + \frac{1}{2x} - \frac{\pi}{2\sqrt{x}} \cot{\left(\pi\sqrt{x}\right)}\right) + \operatorname{sech}\left(x + \frac{1}{2x} - \frac{\pi}{2\sqrt{x}} \coth{\left(\pi\sqrt{x}\right)}\right)dx$
Now I need to either evaluate it exactly, or show it's not equal to $\pi$. I initially thought Feynman's technique might help. If I consider
$I(p)=\int\limits_{0}^{\infty} \operatorname{sech}\left(x + \frac{1}{2x} - \frac{\pi}{2\sqrt{px}} \cot{\left(\pi\sqrt{px}\right)}\right) + \operatorname{sech}\left(x + \frac{1}{2x} - \frac{\pi}{2\sqrt{px}} \coth{\left(\pi\sqrt{px}\right)}\right)dx$
then I want $I(1)$. The idea was, the terms with $p$ should both vanish almost everywhere as $p\to\infty$, so I would end up with
$\lim\limits_{p\to\infty} I(p)=\int\limits_{0}^{\infty}2\operatorname{sech}(x+\frac{1}{2x})dx$
I would then apply Glasser's master theorem again to get $\lim\limits_{p\to\infty}I(p)=\pi$, and use sign analysis on $I'(p)$ to show it wouldn't take this value for a finite $p$. The issue is that Glasser's master theorem would only apply if I was subtracting the $\frac{1}{2x}$, not adding it, so $\lim\limits_{p\to\infty}I(p)$ may be different from $\pi$. So, I don't know what technique would allow this to progress further, and how we'd find the closed form, if there somehow is one.