In this video, the following nice integral expression is derived:
$$\int_0^{\infty}\frac{\cos(x)}{(x^2+1)} \,\mathrm{d}x = \frac{\pi}{2\,\mathrm{e}}$$
Wondered about the more general form:
$$f(s) = \int_0^{\infty}\frac{\cos(x)}{(x^2+1)^s} \,\mathrm{d}x$$
and managed to derive:
$$g(s) = f(s)\,\Gamma(s)\,2^s = {}_2F_0 \left([s, 1-s],[], -\frac12\right)\cdot \frac{\pi}{\mathrm{e}} \qquad s \in \mathbb{C}$$
which has the functional equation:
$$g(s) = g(1-s)$$
All roots of $g(s)$ seem to be complex and to reside on the critical line with $\Re(s) = \frac12$, e.g.:
0.5000000000 + 2.962548535*I
0.5000000000 + 4.534490718*I
0.5000000000 + 5.879867200*I
0.5000000000 + 7.107583837*I
0.5000000000 + 8.258936409*I
0.5000000000 + 9.355093826*I
...
Since these roots look quite regularly spaced, could the conjecture that they all reside on the critical be in reach of a proof?
Too long for comments.
$$\Re(s)>0\qquad\implies \qquad g(s)=\sqrt{2 \pi } \,\,K_{\frac{1}{2}-s}(1)$$ which is easier to analyse than the hypergeometric function.
$$g\left(\frac{1}{2}+i y\right)=\sqrt{2 \pi } \,\, K_{-i y}(1)$$ is a real function.
An estimate of the $n^{\text{th}}$ root is given by $$y_n \sim \frac{31 }{13}n^{8/11}+\frac{18}{31}$$ Trying for the tenth root, Newton iterates are $$\left( \begin{array}{cc} k & y_{(k)} \\ 0 & 13.306620250 \\ 1 & 13.378420422 \\ 2 & 13.385796461 \\ 3 & 13.385882874 \\ 4 & 13.385882886 \\ \end{array} \right)$$
Edit
In comments, @Gary mentioned this very recent paper which exactly adresses this problem. It also improves (equation $2.17$) the asymptotics given here $${\large y_n=\frac 12 e^{1+W \left(\frac{ (4 n-1)\pi}{2 e}\right)}}\left(1+O\left(\frac{\log (n)}{n^2}\right)\right)$$ which is already extremely good even for small values of $n$ (for example $y_1=2.98934$ instead of $ 2.96255$; $y_{10}=13.3896$ instead of $13.3859$).