A Nearly Impossible Integral

First note that $\begin{aligned} \sum_{j=1}^J \frac{1}{(2j-1)(4j-1)(4j-3)} & = \; \sum_{j=1}^J \left(\frac{1}{4j-1} + \frac{1}{4j-3} - \frac{1}{2j-1}\right) \; = \; \sum_{j=1}^{2J}\frac{1}{2j-1} - \sum_{j=1}^J \frac{1}{2j-1} \\ & = \; \sum_{j=1}^{4J} \frac{1}{j} - \sum_{j=1}^{2J} \frac{1}{2j} - \sum_{j=1}^{2J} \frac{1}{j} + \sum_{j=1}^J \frac{1}{2j} \\ & = \; \big(S_{4J} + \ln(4J)\big) - \tfrac32\big(S_{2J} + \ln(2J)\big) + \tfrac12\big(S_J + \ln J\big) \; = \; S_{4J} - \tfrac32S_{2J} + \tfrac12S_J + \tfrac12\ln2 \end{aligned}$ where $S_J \; = \; \sum_{j=1}^J\frac{1}{j} - \ln J$ is well-known to converge to the Euler-Mascheroni constant $\gamma$ . Thus we deduce that $\sum_{j=1}^\infty \frac{1}{(2j-1)(4j-1)(4j-3)} \; = \; \tfrac12\ln2$ Consider the desired integral $X \; = \; \int_0^\infty \tanh(\pi x) \left(\frac{1}{x} - \frac{x}{\frac{1}{16} + x^2}\right)\,dx \; = \; \int_0^\infty \frac{\tanh(\pi x)}{x(1 + 16x^2)}\,dx \; = \; \tfrac12\int_{-\infty}^\infty \frac{\tanh(\pi x)}{x(1 + 16x^2)}\,dx$ The function $\frac{\tanh(\pi z)}{z(1 + 16z^2)}$ is meromorphic, with a removable singularity at $z=0$ , and its simple poles in the upper half plane are at $z = (j-\tfrac12) i$ for all $j \in \mathbb{N}$ .

If $C_N$ is the square with vertices $N+Ni$ , $-N+Ni$ , $-N-Ni$ , $N-Ni$ , it is standard theory that $|\tanh(\pi z)| \le \coth \pi$ for all $z \in C_N$ and all $N \in \mathbb{N}$ . Thus, if $R_N$ is the positively oriented rectangle with vertices $N$ , $N+Ni$ , $-N+Ni$ , $-N$ , then $\lim_{N \to \infty} \int_{R_N} \frac{\tanh(\pi z)}{z(1 + 16z^2)}\,dz \; = \; 2X$ On the other hand, residue calculus tells us that $\begin{aligned} \int_{R_N} \frac{\tanh(\pi z)}{z(1 + 16z^2)}\,dz & = \; 2\pi i\left\{ \mathrm{Res}_{z=\frac14i} + \sum_{j=1}^N \mathrm{Res}_{z=(j-\frac12)i}\right\} \frac{\tanh(\pi z)}{z(1 + 16z^2)} \; = \; 2\pi i\left\{ -\frac12i + \sum_{j=1}^N \frac{1}{\pi (j-\frac12)i[1 - 4(2j-1)^2]}\right\} \\ & = \; \pi - \sum_{j=1}^N \frac{4}{(2j-1)(4j-1)(4j-3)} \end{aligned}$ Letting $N \to \infty$ we deduce that $X \; = \; \tfrac12\pi - \ln 2$ making the answer $1 + 2 + 2 = \boxed{5}$ .

Note that the above integral can be expressed as $\sum_{n\geq 0}\frac{2}{(2n+1)(4n+3)} =\psi^0\left(\frac{3}{4}\right)-\psi^0\left(\frac{1}{2}\right)=\frac{\pi}{2}-\ln2$ Further this integral can be generalized and it has close relation with this problem.