Intereting Posts

Using Fermat's little theorem to find remainders.
Proof that the product of two differentiable functions is also differentiable
In how many ways can we colour $n$ baskets with $r$ colours?
Prove that a power series $\sum_{n=0}^{\infty}{{a_n}{z^n}}$ which converges for any $z \in \mathbb{N}$, converges for any $z \in \mathbb{C}$.
Normal Subgroup Counterexample
Hatcher problem 1.2.3 – technicality in proof of simply connectedness
$A+B=AB$ does it follows that $AB=BA$?
A proof of the identity $ \sum_{k = 0}^{n} \frac{(-1)^{k} \binom{n}{k}}{x + k} = \frac{n!}{(x + 0) (x + 1) \cdots (x + n)} $.
Question about Benford's law
What are “Lazard” sheaves?
If a product is normal, are all of its partial products also normal?
Showing $f_n(x):=\frac{x}{1+n^2x^2}$ uniformly convergent in $\mathbb R$ using $\epsilon-n_0$
How to solve $x^3\equiv 10 \pmod{990}$?
Examples of group extension $G/N=Q$ with continuous $G$ and $Q$, but finite $N$
Prove $\lim_{n\to\infty}x_n=2$ Given $\lim_{n \to \infty} x_n^{x_n} = 4$

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.

- Integral $\int_0^1\frac{\ln x}{x-1}\ln\left(1+\frac1{\ln^2x}\right)dx$
- Log integrals I
- Show that, for $t>0$, $\log t$ is not a polynomial.
- Intuition behind logarithm inequality: $1 - \frac1x \leq \log x \leq x-1$
- Easy way to compute logarithms without a calculator?
- A contradiction or a wrong calculation? $3\lt\lim_{n\to\infty}log_n(p)\lt3$, $\forall n$ (sufficiently large) where $n^3\lt p\in\Bbb P\lt(n+1)^3$?

- Why is this function a really good asymptotic for $\exp(x)\sqrt{x}$
- $\ln(x^2)$ vs $2\ln x$
- How can I solve $8n^2 = 64n\,\log_2(n)$
- How is $ \sum_{n=1}^{\infty}\left(\psi(\alpha n)-\log(\alpha n)+\frac{1}{2\alpha n}\right)$ when $\alpha$ is great?
- Product of logarithms, prove this identity.
- Proving: $\frac{2x}{2+x}\le \ln (1+x)\le \frac{x}{2}\frac{2+x}{1+x}, \quad x>0.$
- ln(z) as antiderivative of 1/z
- Number raised to log expression

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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- Generalized Cross Product
- A proof of $(\forall x P(x)) \to A) \Rightarrow \exists x (P(x) \to A)$
- Suppose $\phi$ is a weak solution of $\Delta \phi = f \in \mathcal{H}^1$. Then $\phi\in W^{2,1}$
- Finite sum $\sum_{n=2}^N\frac{1}{n^2}\sin^2(\pi x)\csc^2(\frac{\pi x}{n})$
- ELI5: Riemann-integrable vs Lebesgue-integrable