Evaluating $\int\limits_0^\infty x\operatorname{sech}^3x\ln(\operatorname{sech}x)\ dx$

Let $\operatorname{sech}(x) = t$ and integrate by parts to obtain \begin{align} -I &= \int \limits_0^\infty \frac{x \log(\cosh(x))}{\cosh^3(x)} \, \mathrm{d}x = \int \limits_0^1 \frac{-\log(t) \operatorname{arsech}(t) t^2}{\sqrt{1-t^2}} \, \mathrm{d} t \\ &= \int \limits_0^1 \sqrt{1-t^2} \frac{\mathrm{d}}{\mathrm{d} t} \left[-\log(t) t \operatorname{arsech}(t)\right] \mathrm{d} t \\ &= \int \limits_0^1 \frac{-\log(t) \operatorname{arsech}(t) (1-t^2)}{\sqrt{1-t^2}} \, \mathrm{d} t - \int \limits_0^1 \left[\sqrt{1-t^2} \operatorname{arsech}(t) - \log(t)\right] \mathrm{d} t \, . \end{align} Averaging the second and the fourth expression yields $$- I = \frac{1}{2} \int \limits_0^1 \frac{-\log(t) \operatorname{arsech}(t)}{\sqrt{1-t^2}} \, \mathrm{d} t - \frac{1}{2}\int \limits_0^1 \sqrt{1-t^2} \operatorname{arsech}(t)\, \mathrm{d} t - \frac{1}{2} \equiv J - K - \frac{1}{2}\, .$$ $K$ can be computed by reversing the previous substitution: $$ K = \frac{1}{2} \int \limits_0^\infty \frac{x \sinh^2(x)}{\cosh^3(x)} \, \mathrm{d} x = \frac{1}{4} \int\limits_0^\infty \frac{\sinh(x) + x \cosh(x)}{\cosh^2(x)} \, \mathrm{d} x = \frac{1}{4}(1+2 \mathrm{G}) = \frac{1}{4} + \frac{\mathrm{G}}{2} \, . $$ For $J$ we can use $t = \frac{2u}{1+u^2}$ to find \begin{align} J &= \int \limits_0^1 \frac{\log(u) \log\left(\frac{2u}{1+u^2}\right)}{1+u^2} \, \mathrm{d} u \\ &= \int \limits_0^1 \frac{-\log(u) \log(1+u^2)}{1+u^2} \, \mathrm{d} u + \int \limits_0^1 \frac{\log^2(u)}{1+u^2} \, \mathrm{d} u - \log(2) \int \limits_0^1 \frac{-\log(u)}{1+u^2} \, \mathrm{d} u \\ &= 2 \operatorname{Im} \operatorname{Li}_3(1+\mathrm{i}) + \mathrm{G} \log(2) - \frac{\pi}{8} \log^2(2) - \frac{3 \pi^3}{32} + \frac{\pi^3}{16} - \mathrm{G} \log(2) \\ &= 2 \operatorname{Im} \operatorname{Li}_3(1+\mathrm{i}) - \frac{\pi}{8} \log^2(2) - \frac{\pi^3}{32} \, . \end{align} The first integral has been calculated here and the others are well-known special values of the Dirichlet beta function. Therefore, $$ -I = J - K - \frac{1}{2} = 2 \operatorname{Im} \operatorname{Li}_3(1+\mathrm{i}) - \frac{\pi}{8} \log^2(2) - \frac{\pi^3}{32} - \frac{\mathrm{G}}{2} - \frac{3}{4} \, . $$

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The problem with the original approach is that for $b=3$ the series only converges for $a > 1$, which leads to the two divergent series in the final answer. This can be avoided by computing the result for sufficiently large values of $a$ first and then taking the limit $a \to 1^+$, which can be justified by analytic continuation. $\Omega_3, \dots, \Omega_8$ are calculated as before after taking the limit inside the series, but the divergent terms are replaced by the regularised versions $$ \Omega_1 = \lim_{a \to 1^+} \sum \limits_{n=1}^\infty \frac{(-1)^{n-1}}{(2n+1)^{a-1}} = \lim_{a \to 1^+} [1 - \beta(a-1)] = 1 - \beta (0) = \frac{1}{2}$$ and \begin{align} \Omega_2 &= \lim_{a \to 1^+} \sum \limits_{n=1}^\infty \frac{(-1)^{n-1} H_n}{(2n+1)^{a-1}} = \lim_{a \to 1^+} \sum \limits_{n=1}^\infty \frac{(-1)^{n-1}}{(2n+1)^{a-1}} \int \limits_0^1 \frac{1 - x^n}{1-x} \, \mathrm{d} x \\ &= \lim_{a \to 1^+} \int \limits_0^1 \frac{\frac{\operatorname{Ti}_{a-1}(\sqrt{x})}{\sqrt{x}} - \beta(a-1)}{1-x} \, \mathrm{d} x = \int \limits_0^1 \frac{\frac{\operatorname{Ti}_{0}(\sqrt{x})}{\sqrt{x}} - \beta(0)}{1-x} \, \mathrm{d} x \\ &= \int \limits_0^1 \frac{\frac{1}{1+x} - \frac{1}{2}}{1-x} \, \mathrm{d} x = \frac{1}{2}\int \limits_0^1 \frac{\mathrm{d} x}{1+x} = \frac{1}{2} \log(2) \, . \end{align}