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Beginning with the form

$$\begin{align} &\int_0^1\int_0^{\infty}\frac{\tanh\left(x\right)}{2x\cosh\left(2kx\right)}dxdk\\ \overset{x\to-(\ln t)/2}{=}&-\int_0^1\int_0^1\frac{\left(\frac{1-t}{1+t}\right)}{t\left(\ln t\right)\left(t^{-k}+t^k\right)}dtdk\\ =&-\int_0^1\int_0^1\frac{1}{\left(1+t\right)\left(t^{-k}+t^k\right)}\left(\frac{t^{-1}-t^0}{\ln t}\right)dtdk\\ =&-\int_{0}^1\int_0^1\frac{1}{\left(1+t\right)\left(t^{-k}+t^k\right)}\left(\int_0^{-1}t^udu\right)dtdk\\ =&\int_{0}^1\int_{-1}^{0}\int_0^1\frac{t^u}{\left(1+t\right)\left(t^{-k}+t^k\right)}dtdudk \end{align}$$

Integrating by parts with respect to $t$ results in the use of the incomplete Beta function and the Polygamma function, which after much arranging yields $$\begin{align} &\int_{0}^1\int_{-1}^{0}\frac{1}{4k}\sum_{n=0}^{\infty}\left(-1\right)^n\left(\psi^{(0)}\left(\frac{u+n+1}{4k}+\frac{3}{4}\right)-\psi^{(0)}\left(\frac{u+n+1}{4k}+\frac{1}{4}\right)\right)dudk \end{align}$$

Integrating with respect to $u$ gives $$\int_{0}^1\sum_{n=0}^{\infty}\left(-1\right)^n\left[\ln\left(\Gamma\left(\frac{n+1}{4k}+\frac34\right)\right)-\ln\left(\Gamma\left(\frac{n}{4k}+\frac34\right)\right)\\+\ln\left(\Gamma\left(\frac{n}{4k}+\frac14\right)\right)-\ln\left(\Gamma\left(\frac{n+1}{4k}+\frac14\right)\right)\right]dk$$

Which by using the series definition of the log gamma function is $$\int_{0}^1\sum_{n=0}^{\infty}\left(-1\right)^n\sum_{i=0}^{\infty}\left(\ln\left(1-\frac{2}{\frac{n+1}{k}+4i+3}\right)-\ln\left(1-\frac{2}{\frac{n}{k}+4i+3}\right)\right)dk$$

Integrating then gives $$\sum_{n=0}^{\infty}\left(-1\right)^n\sum_{i=0}^{\infty}\left(\frac{\left(n+1\right)\ln\left(\frac{4i+1}{n+1}+1\right)-n\ln\left(\frac{4i+1}{n}+1\right)}{4i+1}-\frac{\left(n+1\right)\ln\left(\frac{4i+3}{n+1}+1\right)-n\ln\left(\frac{4i+3}{n}+1\right)}{4i+3}+\ln\left(\frac{4i+n+2}{4i+n+4}\right)-\ln\left(\frac{4i+n+1}{4i+n+3}\right)\right)$$

This is clearly not any cleaner. How do I continue? Or what is a better way to solve this integral? (Sorry in advance for some of the jumps in my steps. The jumps are all just simplifying.)

Edit:

Similar parts of the sums above can be combined and manipulated to obtain $$\sum_{n=0}^{\infty}\left(-1\right)^n\sum_{i=0}^{\infty}\left(2\left(-1\right)^i\frac{n+1}{2i+1}\ln\left(\frac{2i+1}{n+1}+1\right)+\ln\left(\frac{4i+n+2}{4i+n+4}\right)-\ln\left(\frac{4i+n+1}{4i+n+3}\right)\right)$$

The sum of the last two logarithmic terms can be found to equal $\ln 2 / 2$ so we have

$$\frac{\ln 2}{2}+2\sum_{n=0}^{\infty}\left(-1\right)^n(n+1)\sum_{i=0}^{\infty}\frac{(-1)^i}{2i+1}\ln\left(\frac{2i+1}{n+1}+1\right)$$

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  • $\begingroup$ Have you tried to fix the $u,k$ variables and integrate with respect to $t$ by using the residue theorem? I don't know if this gives an explicit answer, just asking. $\endgroup$ – Klaramun Oct 20 '17 at 2:21
  • $\begingroup$ No, I don't know the residue theorem. $\endgroup$ – tyobrien Oct 20 '17 at 2:24
  • $\begingroup$ en.wikipedia.org/wiki/Residue_theorem Hope it helps. $\endgroup$ – Klaramun Oct 20 '17 at 4:38
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By integrating with respect to $k$ first we are left with $$\int_{0}^{+\infty}\frac{\tanh(x)\arctan(\tanh x)}{2x^2}\,dx= \frac{1}{4}\sum_{n\geq 0}\frac{(-1)^n}{2n+1}\underbrace{\int_\mathbb{R}\frac{\tanh(x)^{2n+2}}{x^2}\,dx}_{\mathfrak{I}_n}$$ and $\mathfrak{I}_n$ can be computed through the residue theorem. In particular $$\mathfrak{I}_0 = \frac{28\,\zeta(3)}{\pi^2},\qquad \mathfrak{I}_1 = \frac{102\,\pi^2\,\zeta(3)-744\,\zeta(5)}{3\,\pi^4}$$ but by enforcing the substitution $\tanh x\mapsto u$, then the substitution $u\mapsto\frac{z-1}{z+1}$, we get: $$ \int_{1}^{+\infty}\frac{(z-1)\left(\arctan z-\tfrac{\pi}{4}\right)}{z(z+1)\log^2(z)}\,dz=\int_{0}^{1}\frac{(1-z)\left(\tfrac{\pi}{4}-\arctan z\right)}{z(1+z)\log^2(z)}\,dz $$ or, by integration by parts $$ \int_{0}^{1}\frac{(2-\pi)z^2-(\pi+2)+4(1+z^2)\arctan z}{2(1+z)^2(1+z^2)\log z}\,dz$$ where $$ \int_{0}^{1}\frac{(2-\pi)(z^2-1)}{2(1+z)^2(1+z^2)\log z}\,dz\stackrel{\text{Frullani}}{=}\frac{2-\pi}{4}\log 2$$ leaves us with just $$ \int_{0}^{1}\frac{2(1+z^2)\arctan z-\pi}{(1+z)^2(1+z^2)\log z}\,dz$$ and the original integral, always by Frullani's theorem, simplifies into $$ \frac{\log 2}{2}+2\int_{0}^{1}\frac{\arctan z-\tfrac{\pi}{4}}{(z+1)^2\log z}\,dz=\frac{\log 2}{2}+\int_{0}^{+\infty}\frac{\arctan z-\tfrac{\pi}{4}}{(z+1)^2\log z}\,dz. $$ At this point it is enough to compute the Taylor series of $\arctan z-\frac{\pi}{4}$ at $z=1$, then to recall $$ \zeta(s)\Gamma(s)\left(1-\frac{2}{2^s}\right) = \int_{0}^{+\infty}\frac{t^{s-1}}{e^t+1}\,dt $$

$$ \zeta(s)\Gamma(s+1)\left(1-\frac{2}{2^s}\right)=\int_{0}^{+\infty}\frac{t^s e^t}{(e^t+1)^2}\,dt $$ to have an explicit representation of the original integral in terms of values of $\zeta$ and $\zeta'$.

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