A general integral

Let us define $X(n) \; = \; \int_0^\infty \frac{e^{\pi x}}{(e^{2\pi x} + 1)(n^2 + 4x^2)}\,dx \hspace{2cm} n > 0$ noting that $X(n) \; = \; \int_0^\infty \frac{1}{(e^{\pi x} + e^{-\pi x})(n^2 + 4x^2)}\,dx \; = \; \frac12\int_0^\infty \frac{\mathrm{sech}\,(\pi x)}{n^2 + 4x^2}\,dx \; = \; \tfrac14\int_\mathbb{R} \frac{\mathrm{sech}\,(\pi x)}{n^2 + 4x^2}\,dx$ If, for any positive integer $N$ , $C_N$ is the square with vertices $N+Ni$ , $-N+Ni$ , $-N-Ni$ , $N-Ni$ , then it is standard that $|\cosh(\pi z)| \ge 1$ for all $z \in C_N$ . Thus, if $R_N$ is the positively oriented rectangle with vertices $N$ , $N+Ni$ , $-N+Ni$ $-N$ , then we deduce that $\lim_{N \to \infty} \int_{R_N} \frac{\mathrm{sech}\,(\pi z)}{n^2 + 4z^2}\,dz \; = \; \int_{\mathbb{R}} \frac{\mathrm{sech}\,(\pi x)}{n^2 + 4x^2}\,dx \; = \; 4X(n)$ Suppose for the moment that $n > 0$ is not an odd positive integer. Then $\frac{\mathrm{sech}\,(\pi z)}{n^2 + 4z^2}$ is meromorphic in the upper half plane, with simple poles at $\tfrac12ni$ and $(m - \tfrac12)i$ for all $m \in \mathbb{N}$ . Moreover $\begin{aligned} \mathrm{Res}_{z = \frac12ni} \frac{\mathrm{sech}\,(\pi z)}{n^2 + 4z^2} & = \; \frac{\mathrm{sech}\,(\frac12n\pi i)}{4 \times in} \; = \; \frac{1}{4in}\sec(\tfrac12n\pi) \\ \mathrm{Res}_{z = (m-\frac12)i} \frac{\mathrm{sech}\,(\pi z)}{n^2 + 4z^2} & = \; \frac{1}{\pi \sinh (\pi(m-\frac12)i)\big[n^2 - (2m-1)^2\big]} \; = \; \frac{(-1)^{m-1}}{\pi i(n+2m-1)(n-2m+1)} \end{aligned}$ and since the residue calculus tells us that $\begin{aligned} \int_{R_N} \frac{\mathrm{sech}\,(\pi z)}{n^2 + 4z^2}\,dz & = \; 2\pi i \left(\mathrm{Res}_{z = \frac12ni} + \sum_{m=1}^N \mathrm{Res}_{z = (m-\frac12)i}\right) \frac{\mathrm{sech}\,(\pi z)}{n^2 + 4z^2} \\ & = \; \frac{\pi}{2n}\sec(\tfrac12 \pi n) + 2\sum_{m=1}^N \frac{(-1)^{m-1}}{(n+2m-1)(n-2m+1)} \end{aligned}$ we deduce that $X(n) \; = \; \frac{\pi}{8n}\sec(\tfrac12n\pi) + \frac12\sum_{m=1}^\infty \frac{(-1)^{m-1}}{(n+2m-1)(n-2m+1)}$ Now $\begin{aligned} 8n\sum_{m=1}^\infty \frac{(-1)^{m-1}}{(n-2m+1)(n+2m-1)} & = \; 4\sum_{m=1}^\infty \left(\frac{(-1)^{m-1}}{n+2m-1} + \frac{(-1)^{m-1}}{n-2m+1}\right) \\ & = \; 4\sum_{m=1}^\infty \left(\frac{1}{n+4m-3} - \frac{1}{n+4m-1} + \frac{1}{n-4m+3} - \frac{1}{n - 4m+1}\right) \\ & = \; \sum_{m=1}^\infty \left(\frac{1}{m+\frac{n-3}{4}} - \frac{1}{m + \frac{n-1}{4}} - \frac{1}{m - \frac{n+3}{4}} + \frac{1}{m - \frac{n+1}{4}}\right) \\ & = \; -\psi\big(\tfrac{n+1}{4}\big) + \psi\big(\tfrac{n+3}{4}\big) + \psi\big(\tfrac{1-n}{4}\big) - \psi\big(\tfrac{3-n}{4}\big) \\ & = \; -\psi\big(\tfrac{n+1}{4}\big) + \psi\big(\tfrac{n+3}{4}\big) + \psi\big(1 - \tfrac{n+3}{4}\big) - \psi\big(1 - \tfrac{n+1}{4}\big) \\ & = \; 2\left[\psi\big(\tfrac{n+3}{4}\big) - \psi\big(\tfrac{n+1}{4}\big)\right] +\pi\left[\cot\big(\tfrac14(n+3)\pi\big) - \cot\big(\tfrac14(n+1)\pi\big)\right] \\ & = \; 2\left[\psi\big(\tfrac{n+3}{4}\big) - \psi\big(\tfrac{n+1}{4}\big)\right] - 2\pi\sec(\tfrac12n\pi) \end{aligned}$ so we deduce that $X(n) \; = \; \frac{1}{8n}\left[\psi\big(\tfrac{n+3}{4}\big) -\psi\big(\tfrac{n+1}{4}\big)\right]$ making the answer $8+3+4+1=\boxed{16}$ .

We have proved this formula for all $n > 0$ , except when $n$ is an odd positive integer (in that case the pole at $\tfrac12ni$ is a double pole, not simple, which would change the details of the above calculations. However this final formula for $X(n)$ can be extended to all $n > 0$ by continuity.