Well, RV, I will go ahead and post my workings to that one. All of the little intricacies were cumbersome. For me anyway. But, I got it assembled.

What I have here is for a = 1.

Write \(\displaystyle \int_{C}\frac{e^{iz}}{b^{2n}+z^{2n}}dz\), where C is a typical semicircle in the UHP.

The poles are at \(\displaystyle z=be^{\frac{(2k+1)\pi i}{2n}}\) and are equally spaced at \(\displaystyle \frac{\pi}{n}\) intervals.

The poles that contribute are in the UHP. So, I am not going to bother writing out the others because they don't contribute anything anyway.

\(\displaystyle \sum_{k=0}^{n-1}\frac{1}{2n\left(be^{\frac{(2k+1)}{2n}\pi i}\right)^{2n-1}\left(z-be^{\frac{(2k+1)\pi i}{2n}}\right)}\)

The poles in the contour then contribute, due to the sum of the residues, in total (I hope this is correct. It appears to be. One itty-bitty negative or whatever and it's blown out of the water

):

\(\displaystyle -\sum_{k=0}^{n-1}\frac{\pi i}{nb^{2n-1}}e^{-b\sin(\frac{2k+1}{2n}\pi}\left[\cos\left(\frac{2k+1}{2n}\pi +b\cos(\frac{2k+1}{2n}\pi)\right)+i\cdot \sin\left(\frac{2k+1}{2n}\pi +b\cos(\frac{2k+1}{2n}\pi)\right)\right]\).

The outside arc goes to 0 as R goes to infinity because of the \(\displaystyle e^{-R\sin(t)}\) thing. So, I will not write it up.

Assembling:

\(\displaystyle \int_{-\infty}^{0}\frac{e^{ix}}{b^{2n}+x^{2n}}dx+\int_{0}^{\infty}\frac{e^{ix}}{b^{2n}+x^{2n}}dx+\int_{0}^{\pi}\frac{e^{iRe^{it}}iRe^{it}}{b^{2n}+R^{2n}e^{2nit}}dt=\)

\(\displaystyle \sum_{k=0}^{n-1}\frac{\pi i}{nb^{2n-1}}e^{-b\sin(\frac{2k+1}{2n}\pi}\left[\cos\left(\frac{2k+1}{2n}\pi +b\cos(\frac{2k+1}{2n}\pi)\right)+i\cdot \sin\left(\frac{2k+1}{2n}\pi +b\cos(\frac{2k+1}{2n}\pi)\right)\right]\)

Now, on the left, let \(\displaystyle x\to -x\) in the first integral and add to the second one. Thus, giving:

\(\displaystyle \int_{0}^{\infty}\frac{\cos(x)}{b^{2n}+x^{2n}}dx=\sum_{k=0}^{n-1}\frac{\pi}{2nb^{2n-1}}e^{-b\sin(\frac{2k+1}{2n}\pi)}\sin\left(\frac{2k+1}{2n}\pi +b\cos(\frac{2k+1}{2n}\pi)\right)\)

There certainly could be a typo in all of that anyway.

For example, if n = 4, then the result is:

\(\displaystyle \frac{\pi}{4}e^{-\sqrt{\sqrt{2}+2}\sin(\frac{\sqrt{2-\sqrt{2}}}{2}+\frac{3\pi}{8})/2}+\frac{\pi}{4}e^{-\sqrt{2-\sqrt{2}}\sin(\frac{\sqrt{2+\sqrt{2}}}{2}+\frac{\pi}{8})/2}\)

which, I am sure, can be written in various forms, and appears to check numerically.

BTW, here is another general one to try with or without contours. Whatever you wish.\(\displaystyle \int_{0}^{\infty}\frac{\cos(x)}{(b^{2}+x^{2})^{n+1}}dx\)

Maybe a little tougher?. Anyway, the next time an integral like this pops up on SE or wherever, one can just show the general case and maybe show off a little bit

:):).