Evaluate $\int_0^{\infty} x^2\ln(\sinh x)\operatorname{sech}(3 x){\rm d}x $.

Question

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?

Answer 1

Let $I$ denote the integral. Here, we prove:

Claim 1. $$ I = \frac{\pi^2 G}{9} - \frac{5\pi^3}{108} \log 2, $$ where $G$ is the Catalan's constant.

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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*}

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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:

Claim 2. For $0 \leq \theta \leq \pi$, we have $$ A(\theta) = \frac{\pi\theta(\pi-\theta)}{2} \qquad \text{and} \qquad B(\theta) = \frac{\pi^2}{2}\int_{0}^{\theta} \log \cot \left( \frac{t}{2} \right) \, \mathrm{d}t. $$

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.

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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*}

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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.

Answer 2

Express the hyperbolic trig terms as their exponential form:$$I=\int_0^{\infty} \frac{x^2 \ln{\left(\frac{e^{2x}-1}{2e^x}\right)}}{\frac{e^{6x}+1}{2e^{3x}}} \; dx$$ $$= 2\int_0^{\infty}\frac{e^{3x}x^2}{e^{6x}+1}\left(\ln{\left(1-e^{-2x}\right)}+x-\ln{2}\right) \; dx$$ $$=2\int_0^{\infty}\frac{e^{3x}x^2}{e^{6x}+1} \sum_{k=1}^{\infty} \frac{- e^{-2xk}}{k} \; dx+2\int_0^{\infty}\frac{e^{3x}x^3}{e^{6x}+1} \; dx-2\ln{2}\int_0^{\infty}\frac{e^{3x}x^2}{e^{6x}+1} \; dx$$ Using Fubini's theorem/dominated convergence theorem, we can interchange the summation and integral sign for the first two integrals: $$I=-2\sum_{k=1}^{\infty} \frac{1}{k} \underbrace{\int_0^{\infty}\frac{e^{x(3-2k)}x^2}{e^{6x}+1}\; dx}_{I_1}+2\underbrace{\int_0^{\infty}\frac{e^{3x}x^3}{e^{6x}+1} \; dx}_{I_2}-2\ln{2}\underbrace{\int_0^{\infty}\frac{e^{3x}x^2}{e^{6x}+1} \; dx}_{I_3}$$

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First, for $I_2$, let $u=3x$ then divide by $e^{2u}$: $$I_3=\frac{1}{81} \int_0^{\infty} \frac{u^3e^{-u}}{1+e^{-2u}} \; du$$ Notice that this can be converted into an infinite geometric series with $r=e^{-2u}$: $$I_3=\frac{1}{81} \int_0^{\infty} u^3 \sum_{n=0}^{\infty} {(-1)}^n e^{-u(2n+1)} \; du$$ $$I_3=\frac{1}{81} \sum_{n=0}^{\infty} {(-1)}^n\int_0^{\infty} u^3 e^{-u(2n+1)} \; du$$ $$I_3=\frac{6}{81} \sum_{n=0}^{\infty} \frac{{(-1)}^n}{{(2n+1)}^4}=\frac{\zeta\left(4,\frac{1}{4}\right)-\zeta\left(4,\frac{3}{4}\right)}{3456}=\frac{2 \beta(4)}{27}$$ Where $\zeta\left(s,a\right)$ is the Hurwitz zeta function, and $\beta(z)$ is the Dirichlet Beta Function.

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Repeat the method for evaluating $I_3$ for $I_4$ and you get: $$I_4= \frac{1}{27}\sum_{k=0}^{\infty}{(-1)}^k \int_0^{\infty} e^{-u(2n+1)}u^2 \; du=\frac{\zeta\left(3,\frac{1}{4}\right)-\zeta\left(3,\frac{3}{4}\right)}{864}=\frac{\pi^3}{432}$$

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$I_1$ can be found in similarly: $$I_1=\sum_{j=0}^{\infty} {(-1)}^j\int_0^{\infty} e^{x(-6j-3-2k)}x^2 \; dx$$ $$=\sum_{j=0}^{\infty} \frac{2{(-1)}^j}{{(6j+2k+3)}^3}$$ $$=\frac{1}{864}\left(\zeta\left(3,\frac{k}{6}+\frac{1}{4}\right)-\zeta\left(3,\frac{k}{6}+\frac{3}{4}\right)\right)$$

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Putting this all together, the integral evaluates to: $$\boxed{I=\frac{4 \beta(4)}{27}-\frac{\pi^3\ln{(2)}}{216}-\sum_{k=1}^{\infty} \frac{\zeta\left(3,\frac{k}{6}+\frac{1}{4}\right)-\zeta\left(3,\frac{k}{6}+\frac{3}{4}\right)}{432k}}$$

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In response to @mengdie1982's comment, it is very interesting that there is indeed an elementary closed form expression. I'm still working on this answer, but I know that the Beta function is highly related to Catalan's constant. For the answer I provided, according to Wolfram Alpha, the summation converges to approximately $-0.037538$, and putting the three expressions together yields $I \approx 0.00947269$, which agrees with Wolfram Alpha's approximation of the original integral.