Is it possible to evaluate this integral to some reasonable formula?
$$\int \frac{1-\text{sech}(\pi x)}{x} \, dx$$
I used integration by parts and got this result:
$$\int \frac{1-\text{sech}(\pi x)}{x} \, dx=-\frac{2}{\pi} \int \frac{\tan ^{-1}\left(\tanh \left(\frac{\pi x}{2}\right)\right)}{x^2} \, dx+\log (x)-\frac{2 \tan ^{-1}\left(\tanh \left(\frac{\pi x}{2}\right)\right)}{\pi x}$$
Hoping for simplifying somehow $\tan ^{-1}\left(\tanh \left(\frac{\pi x}{2}\right)\right)$ but without any success.
Best Answer
One method is to use the series expansion for $\text{sech}(x)$ in the form $$ \text{sech}(x) = \sum_{n=0}^{\infty} \frac{E_{2n} \, x^{2n}}{(2n)!} \hspace{10mm} |x| < \frac{\pi}{2}.$$ This leads to \begin{align} I &= \int \frac{1 - \text{sech}(\pi x)}{x} \, dx \\ &= - \sum_{n=1}^{\infty} \frac{E_{2n} \, \pi^{2n}}{(2n)!} \, \int x^{2n-1} \, dx \\ &= - \sum_{n=1}^{\infty} \frac{E_{2n} \, (\pi \, x)^{2n}}{(2n) \, (2n)!} + c_{0} \end{align} for $|x| < \frac{1}{2}$.