Contour integral in complex

Question

I have a problem with this integral : $\int_{R} ^{R+i \pi} \frac {1}{\cosh x}dx$, So...$\int_{R} ^{R+i \pi} \frac {1}{\cosh x}dx$ $=$ $\int_{0}^{\pi} \frac {2}{e^{R+ti}+e^{-R-ti}}i \space dt$ $->$ $h=e^{R+ti}$(new variable) $->$ $\int_{e^R}^{e^{R+\pi i}} \frac {2}{h^2+1}dh$ $=2(\arctan (e^{R+\pi i})-\arctan(e^R))$=$lim_{R->\infty}\space 2(\arctan (-e^{R})-\arctan(e^R))=-2\pi$, but the result should be 0, what am I missing ?

Answer 1

The issue with your work is that, in order for the fundamental theorem of calculus (for complex functions) to work here, you have to assume $2\tan^{-1}(h)$ is a primitive for $\frac{2}{1+h^2}$. However, $\tan^{-1}$ requires a branch cut in order to be defined almost everywhere on $\Bbb{C}$. The most common branch cut happens on the imaginary axis where $-1 \geq \Im(z)$ and where $1 \leq \Im(z)$. As it so happens, your change of variables causes the integral to be taken along the top half of the circle of radius $e^R\geq 1$ centered at the origin, and so it passes through the branch cut, and so the value of $\tan^{-1}$ "jumps" by a value of $2\pi$ in the middle of the curve, causing the discrepency you got.

Be sure to be really, really careful about poles and branch cuts when taking contour integrals.