Problem
Evaluate $\displaystyle\int_0^{\infty} x^2\ln(\sinh x)\operatorname{sech}(3 x){\rm d}x .$
Someone writes as follows \begin{align*} &\int_0^{\infty} x^2\ln(\sinh x)\operatorname{sech}(3 x){\rm d}x\\ =&\frac{1}{27}\int_0^{\infty} x^2\ln\left(\sinh \frac{x}{3}\right)\operatorname{sech}(x){\rm d}x\\ =&\frac{2}{27}\int_0^{\infty} \frac{x^2e^{x}}{e^{2x}+1}\ln\left(\frac{e^{\frac{2x}{3}}-1}{2e^{\frac{x}{3}}}\right){\rm d}x\\ =&\frac{2}{27}\int_0^{\infty} \frac{x^2e^{x}}{e^{2x}+1}\left[\ln\left(e^{\frac{2x}{3}}-1\right)-\frac{x}{3}-\ln2\right]{\rm d}x. \end{align*} This will help?
Let $I$ denote the integral. Here, we prove:
Step 1. (Reduction) Let $I$ denote the integral and substitute $u=\sinh x$. Then by using the identities
$$ \mathrm{d}u = \cosh x \, \mathrm{d}x, \qquad \cosh x \cosh 3x = (\sinh^2 x + 1)(4\sinh^2 x + 1),$$
it follows that
$$ I = \int_{0}^{\infty} \frac{(\operatorname{arcsinh} u)^2 \log u}{(u^2+1)(4u^2 + 1)} \, \mathrm{d}u. $$
In order to compute this integral, we introduce three auxiliary functions:
$$ f(x) = (\operatorname{arcsinh} x)^2, \qquad A(\theta) = \int_{0}^{\infty} \frac{f(u\sin\theta)}{u^2+1} \, \mathrm{d}u, \qquad B(\theta) = \int_{0}^{\infty} \frac{f(u\sin\theta)\log u}{u^2+1} \, \mathrm{d}u. $$
Using this notation and applying the partial fraction decomposition, we are led to the following representation of $I$:
\begin{align*} I &= \int_{0}^{\infty} \left( \frac{2}{3} \cdot \frac{2}{4u^2+1} - \frac{1}{3} \cdot \frac{1}{u^2+1} \right) f(u) \log u \, \mathrm{d}u \\ &= \frac{2}{3} \int_{0}^{\infty} \frac{f(v/2) \log(v/2)}{v^2+1} \, \mathrm{d}u - \frac{1}{3} \int_{0}^{\infty} \frac{f(u) \log u}{u^2+1} \, \mathrm{d}u \tag{$v=2u$} \\ &= \frac{2}{3} B\left(\frac{\pi}{6}\right) - \frac{2}{3} A\left(\frac{\pi}{6}\right)\log 2 - \frac{1}{3} B\left(\frac{\pi}{2}\right) \tag{1} \end{align*}
Step 2. (General Formula for $A(\theta)$ and $B(\theta)$) Next, we identify 'closed forms' of the auxiliary functions $A(\theta)$ and $B(\theta)$. In this regard, we claim:
Proof. Note that both $A$ and $B$ are smooth on $(0, \pi)$ and continuous on $[0, \pi]$. We will study their second derivatives and use them to deduce the claim. In doing so, a key observation is the following identity:
$$ x f'(x) + (x^2+1)f''(x) = 2. \tag{2} $$
Indeed, this is easily verified by differentiating both sides of $f'(x)\sqrt{x^2+1} = 2\operatorname{arcsinh}(x)$. Then
\begin{align*} \frac{\partial^2}{\partial\theta^2} f(x\sin\theta) &= \frac{\partial}{\partial\theta} (x \cos \theta) f'(x\sin\theta) \\ &= - (x \sin\theta) f'(x\sin\theta) + (x \cos \theta)^2 f''(x\sin\theta) \\ &= (x^2 + 1)f''(x\sin\theta) - 2, \tag{3} \end{align*}
where the last step follows from $\text{(2)}$. From this, we find that
\begin{align*} A''(\theta) &= \int_{0}^{\infty} \frac{\partial^2}{\partial\theta^2} \frac{f(u\sin\theta)}{u^2+1} \, \mathrm{d}u \\ &= \int_{0}^{\infty} \left( f''(u\sin\theta) - \frac{2}{u^2+1} \right) \, \mathrm{d}u \\ &= \left[ \frac{f'(u\sin\theta)}{\sin\theta} - 2\arctan u \right]_{u=0}^{u=\infty} \\ &= -\pi. \end{align*}
Then the conditions $A'(\frac{\pi}{2}) = 0$ (which easily follows from the symmetry $A(\theta) = A(\pi-\theta)$) and $A(0) = 0$ determines $A(\theta)$ as in the claim. Similarly,
\begin{align*} \require{cancel} B''(\theta) &= \int_{0}^{\infty} f''(u\sin\theta)\log u \, \mathrm{d}u - \cancel{\int_{0}^{\infty} \frac{2 \log u}{u^2+1} \, \mathrm{d}u} \\ &= \cancel{\left[ \frac{f'(u\sin\theta) \log u}{\sin\theta} \right]_{u=0}^{u=\infty}} - \int_{0}^{\infty} \frac{f'(u\sin\theta)}{u\sin\theta} \, \mathrm{d}u. \end{align*}
Substituting $u \sin \theta = \sinh y$, we get
\begin{align*} B''(\theta) &= - \frac{1}{\sin\theta} \int_{0}^{\infty} \frac{2y}{\sinh y} \, \mathrm{d}y \\ &= - \frac{4}{\sin\theta} \sum_{n=0}^{\infty} \int_{0}^{\infty} y e^{-(2n+1)y} \, \mathrm{d}y \\ &= - \frac{4}{\sin\theta} \sum_{n=0}^{\infty} \frac{1}{(2n+1)^2} \\ &= - \frac{\pi^2}{2\sin\theta}. \end{align*}
Again, together with $B'(\frac{\pi}{2}) = 0$ and $B(0) = 0$ proves the desired claim.
Step 3. (Computation of $B(\frac{\pi}{6})$ and $B(\frac{\pi}{2})$) Now it remains to identify the closed forms of $B(\frac{\pi}{6})$ and $B(\frac{\pi}{2})$. To this end, we derive a Fourier series of $B(\theta)$. A key ingredient is the following computation: if $0 < t < \pi$, then
\begin{align*} \log \cot \left( \frac{t}{2} \right) &= \log \left| \frac{1 + e^{it}}{1 - e^{it}} \right| = \operatorname{Re} \bigl[ \log ( 1 + e^{it} ) - \log (1 - e^{it}) \bigr] \\ &= 2 \operatorname{Re} \Biggl[ \sum_{n=0}^{\infty} \frac{e^{i(2n+1)t}}{2n+1} \Biggr] = 2 \sum_{n=0}^{\infty} \frac{\cos((2n+1)t)}{2n+1}. \end{align*}
Plugging this back,
\begin{align*} B(\theta) = \pi^2 \sum_{n=0}^{\infty} \int_{0}^{\theta} \frac{\cos((2n+1)t)}{2n+1} \, \mathrm{d}t = \pi^2 \sum_{n=0}^{\infty} \frac{\sin((2n+1)\theta)}{(2n+1)^2}. \end{align*}
This immediately determines that $B(\frac{\pi}{2}) = \pi^2 G$. Moreover,
\begin{align*} B(\tfrac{\pi}{6}) &= \pi^2 \sum_{n=0}^{\infty} \frac{\sin((2n+1)\pi/6)}{(2n+1)^2} \\ &= \frac{\pi^2}{2} \Biggl[ \Biggl( \frac{1}{1^2} + \frac{2}{3^2} + \frac{1}{5^2} - \frac{1}{7^2} - \frac{2}{9^2} - \frac{1}{11^2} \Biggr) \\ &\hspace{3em} + \Biggl( \frac{1}{13^2} + \frac{2}{15^2} + \frac{1}{17^2} - \frac{1}{19^2} - \frac{2}{21^2} - \frac{1}{23^2} \Biggr) \\ &\hspace{3em} + \dots \Biggr] \\ &= \frac{\pi^2}{2} \Biggl[ \Biggl( \frac{1}{1^2} - \frac{1}{3^2} + \frac{1}{5^2} - \frac{1}{7^2} + \frac{1}{9^2} - \frac{1}{11^2} + \dots \Biggr) \\ &\hspace{3em} + 3 \Biggl( \frac{1}{3^2} - \frac{1}{9^2} + \frac{1}{15^2} - \frac{1}{21^2} + \dots \Biggr) \Biggr] \\ &= \frac{2\pi^2 G}{3}. \end{align*}
4. (Conclusion) Combining all the efforts altogether, $\text{(1)}$ yields
\begin{align*} I &= \frac{2}{3} \left( \frac{2\pi^2 G}{3} \right) - \frac{2}{3} \left( \frac{5\pi^3}{72} \right)\log 2 - \frac{1}{3} (\pi^2 G) \\ &= \frac{\pi^2 G}{9} - \frac{5\pi^3}{108} \log 2 \end{align*}
as desired.