Residue of $p.v.\int_{-\infty}^{\infty}\frac{e^{2x}}{\cosh(\pi x)}dx=\text{sec}1$

Consider the integral

$$\oint_C dz \frac{e^{2 z}}{\cosh{\pi z}}$$

where $C$ is the above-described rectangle. On the one hand, this integral is equal to the integral about the individual legs of the contour, viz.:

$$\int_{-p}^p dx \frac{e^{2 x}}{\cosh{\pi x}} + i \int_0^1 dy \frac{e^{2 (p+i y)}}{\cos{\pi (p+i y)}} + \int_{p}^{-p} dx \frac{e^{2 (x+i)}}{\cosh{\pi (x+i)}} + i \int_1^0 dy \frac{e^{2 (-p+i y)}}{\cos{\pi (-p+i y)}}$$

Take the limit as $p \rightarrow \infty$. It should be clear that the 2nd and 4th integrals - those over the vertical legs of $C$ - will vanish in this limit. That leaves the 1st and 3rd integrals over the horizontal sections; these may be combined to produce

$$\left(1+e^{i 2}\right) \int_{-\infty}^{\infty} dx \frac{e^{2 x}}{\cosh{\pi x}}$$

This equals, by the residue theorem, the residue at the only pole within $C$, namely, $z=i/2$:

$$i 2 \pi \lim_{z->i/2} \frac{e^{2 z}}{\cosh{\pi z}} = i 2 \pi\frac{e^{i}}{\pi \sinh{(i \pi/2)}} = 2 e^{i}$$

Therefore

$$\int_{\infty}^{\infty} dx \frac{e^{2 x}}{\cosh{\pi x}} = \frac{2 e^{i}}{1+e^{i 2}}= \sec{1}$$

You should note that no $PV$ was needed here.