Hot Integrals

The first integral is easy: $\int_0^1 \frac{\ln x \ln(1-x)}{x}\,dx \; = \; -\sum_{n\ge 1}\frac{1}{n}\int_0^1 \ln x x^{n-1}\,dx \; = \; \sum_{n\ge1} \frac{1}{n^3} \; =\; \zeta(3)$ A double integration by parts shows that $\int_0^\pi u(\pi-u)e^{-(2k+1)ui}\,du \; = \; -\frac{4i}{(2k+1)^3} \hspace{2cm} k \in \mathbb{N}\cup\{0\}$ Thus $\int_0^\pi \frac{u(\pi-u)}{\sin u}\big(1 - e^{-2Kui}\big)\,du \; = \; 2i\sum_{k=0}^{K-1}\int_0^\pi u(\pi-u)e^{-(2k+1)ui}\,du \; = \; 8\sum_{k=0}^{K-1}\frac{1}{(2k+1)^3}$ for all $K \ge 0$ , and since $\frac{u(\pi-u)}{\sin u}$ is Lebesgue integrable over $(0,\pi)$ , we deduce, on letting $K \to \infty$ , that $\int_0^\pi \frac{u(\pi-u)}{\sin u}\,du \; = \; 8\sum_{k \ge 0}\frac{1}{(2k+1)^3} \; = \; 8\left(1-\tfrac18\right)\zeta(3) \; = \; 7\zeta(3)$ Then the successive substitutions $y = e^x$ , $y = \tan t$ , $u = 2t$ give $\begin{aligned} \int_{-\infty}^\infty \tan^{-1}(e^t)\tan^{-1}(e^{-t})\,dt & =\; \int_0^\infty \frac{\tan^{-1}(y) \tan^{-1}(y^{-1})}{y}\,dy \; = \; \int_0^{\frac12\pi}\frac{t(\frac12\pi-t)}{\sin t \cos t}\,dt \\ & = \; \frac14\int_0^\pi \frac{u(\pi-u)}{\sin u}\,du \; = \; \tfrac74\zeta(3) \end{aligned}$ so we deduce that $A=7$ , $B=4$ , and hence $A+B = \boxed{11}$ .