Prove that $$\int_\infty^{-\infty} \frac {cos x} {e^x + e^-x} dx = \frac {\pi} {e^{\pi/2} + e^{-\pi/2}}$$.
I have a question about the proof shown below: why does the third integral (the union of left and right side contour) vanish?
I think I understand intuition wise, i.e. $i\pi$ becomes insignificant as $R$'s in both $R+i\pi$ and $R$ get infinitely large. However, as practice I tried to calculate the third integral myself to show it equal to 0 and couldn't do it.
I am wondering if someone could show me how to calculate the third integral.
The third integral is indeed zero but not because of taking $R$ to be infinitely large. It is zero no matter what $R$ we use. I integrate it by taking $z=R+\theta i$ in the upwards path in positive side and taking $z=-R+\theta i$ in the negative downward path. I put a negative in front of the downward integral which make $\theta$ in both integral going from $0$ to $\pi$ . Making two integral into one you will find that the numerator cancel to $0$. Be careful that the denominator is not the same. I guess you have made some careless mistake in your calculation maybe.