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Is it possible to evaluate this integral in a closed form? $$\int_0^\infty\ln x\cdot\ln\left(1+\frac1{2\cosh x}\right)dx=\int_0^\infty\ln x\cdot\ln\left(1+\frac1{e^{-x}+e^x}\right)dx$$

I tried to evaluate it with a CAS, and looked up in integral tables, but was not successful.

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    $\begingroup$ Do you have any reason to expect it to have a closed form? $\endgroup$ – robjohn Nov 8 '13 at 19:37
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    $\begingroup$ Nope. Neither I have a reason to expect it to not have a closed form. $\endgroup$ – OlegK Nov 8 '13 at 19:44
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    $\begingroup$ One should always believe there is a closed form. $\endgroup$ – Vladimir Reshetnikov Nov 8 '13 at 20:06
  • $\begingroup$ Try using the Taylor series expansion for $\ln(1-x)=\sum_1^\infty\frac{x^n}n$ . $\endgroup$ – Lucian Nov 8 '13 at 23:13
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Define

$$ I(\alpha) = \int_{0}^{\infty} \log x \log(1 - e^{-\alpha x}) \, dx. $$

Integrating by parts, followed by the substitution $\alpha x \mapsto x$, we have

\begin{align*} I(\alpha) &= \alpha \int_{0}^{\infty} \frac{x - x\log x}{e^{\alpha x} - 1} \, dx \\ &= \frac{1}{\alpha} \int_{0}^{\infty} \frac{(1+\log \alpha) x - x \log x}{e^{x} - 1} \, dx\\ &= \frac{1}{\alpha} \left\{ (1+\log\alpha)\zeta(2) - \left.\frac{d \zeta(s)\Gamma(s)}{s}\right|_{s=2} \right\}\\ &= \frac{1}{\alpha} \left\{ (\gamma+\log\alpha)\zeta(2) - \zeta'(2) \right\}. \end{align*}

Then it follows that

\begin{align*} \int_{0}^{\infty} \log x \log \left( 1 + \frac{1}{2\cosh x} \right) \, dx &= \int_{0}^{\infty} \log x \log \left( \frac{1 - e^{-3x}}{1 - e^{-x}} \cdot \frac{1 - e^{-2x}}{1 - e^{-4x}} \right) \, dx \\ &= I(2) + I(3) - I(1) - I(4) \\ &= \frac{5}{12} \zeta'(2) - \frac{5}{72}\gamma\pi^{2} + \frac{1}{18}\pi^{2} \log (3). \end{align*}

Plugging some identities relating $\zeta'(2)$ and the Glaisher-Kinkelin constant $A$, this reduces to Vladimir's answer.

Addendum - Something you might want to know:

The following identity played the key role in this proof.

$$ \int_{0}^{\infty} \frac{x^{s-1}}{e^{x} - 1} \, dx = \Gamma(s)\zeta(s), $$

which holds for $\Re s > 1$.

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    $\begingroup$ As an aside, $$|\zeta^{(n)}(2)|=\sum_{k=1}^\infty\frac{\ln^nk}{k^2}\simeq\int_1^\infty\frac{\ln^nx}{x^2}dx=n!$$ and generally $$|\zeta^{(n)}(a)|=\sum_{k=1}^\infty\frac{\ln^nk}{k^a}\simeq\int_1^\infty\frac{\ln^nx}{x^a}dx=\frac{n!}{(a-1)^{n+1}}$$ where the integral expression for the factorial function from the first approximation is obtained from Euler's formula for the $\Gamma$ function, $n!=\int_0^1(-\ln x)^ndx=\int_0^1\ln^n\frac1xdx$ through the $t=\frac1x$ substitution. $\endgroup$ – Lucian Nov 9 '13 at 22:43

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