Evaluating $\int\limits_0^\infty x\operatorname{sech}^3x\ln(\operatorname{sech}x)\ dx$

Question

How to prove that

$$I=\int_0^\infty x \operatorname{sech}^3x\ln{(\operatorname{sech}x)}\ dx=\frac{\pi^3}{32}+\frac{\pi}{8}\ln^22+\frac14(3+2G)-2\ \operatorname{Im}\operatorname{Li}_3(1+i)\ ?$$

This problem was proposed by a friend and here is his solution

Using the identity

$$\int_0^\infty x^a\operatorname{sech}^bx\ dx=\frac{2^b\Gamma(a+1)}{\Gamma(b)}\sum_{n=0}^\infty\frac{\Gamma(n+b)}{\Gamma(n+1)}\frac{(-1)^n}{(2n+b)^{a+1}}\tag1$$

which he proved by writing $$\operatorname{sech}^bx=\left(\frac{2e^{-x}}{1+e^{-2x}}\right)^b=2^b\sum_{n=0}^\infty {b+n-1\choose n}(-1)^n e^{-(2n+b)x}$$

and differentiating $(1)$ with respect to $b$ then setting $a=1$ and $b=3$ we get

$$I=4\sum_{n=0}^\infty(n+2)(n+1)\frac{(-1)^n}{(2n+3)^2}\left(\ln2+H_{n+2}-H_2-\frac{2}{2n+3}\right)$$

$$=4\sum_{n=1}^\infty\frac{n(n+1)(-1)^{n-1}}{(2n+1)^2}\left(\ln2+H_{n+1}-H_2-\frac{2}{2n+1}\right)$$

$$=\sum_{n=1}^\infty\left((-1)^{n-1}+\frac{(-1)^n}{(2n+1)^2}\right)\left(\ln2+H_{n}+\frac{1}{n+1}-\frac32-\frac{2}{2n+1}\right)$$

$$=\left(\ln2-\frac32\right)(\Omega_1+\Omega_5)+\Omega_2+\Omega_3+\Omega_6+\Omega_7+2(\Omega_4+\Omega_8)$$

where

$$\Omega_1=\sum_{n=1}^\infty (-1)^{n-1}=\frac12$$

$$\Omega_2\sum_{n=1}^\infty(-1)^{n-1}H_n=\frac{\ln2}{2}$$

$$\Omega_3=\sum_{n=2}^\infty\frac{(-1)^n}{n}=1-\ln2$$

$$\Omega_4=\sum_{n=1}^\infty\frac{(-1)^n}{2n+1}=\frac{\pi}{4}-1$$

$$\Omega_5=\sum_{n=1}^\infty\frac{(-1)^n}{(2n+1)^2}=G-1$$

$$\Omega_6=\sum_{n=1}^\infty\frac{(-1)^nH_{n+1}}{(2n+1)^2}=\frac{3}{32}\pi^3+\frac{\pi}{8}\ln^22-G\ln2+2\ \operatorname{Im}\operatorname{Li}_3(1+i)$$

$$\Omega_7=\sum_{n=1}^\infty=\frac{(-1)^n}{(2n+1)^2(n+1)}=2G-\frac{\pi}{2}+\ln2-1$$

$$\Omega_8=\sum_{n=1}^\infty\frac{(-1)^n}{(2n+1)^2}=\beta(3)-1=\frac{\pi^3}{32}-1$$

By combining theses results, the closed form of $I$ follows.

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My friend ( the answerer) and I are not happy with this approach as $\Omega_1$ and $\Omega_2$ are divergent series, so the question is how to void this issue? Still, he got the right closed form.

The other question is about my following approach,

$$I=\int_0^\infty x\operatorname{sech}^3x\ln(\operatorname{sech}x)\ dx=\int_0^\infty x\left(\frac{2e^{-x}}{1+e^{-2x}}\right)^3\ln\left(\frac{2e^{-x}}{1+e^{-2x}}\right)\ dx$$

$$\overset{e^{-x}=u}{=}-\int_0^1\frac{\ln u}{u}\left(\frac{2u}{1+u^2}\right)^3\ln\left(\frac{2u}{1+u^2}\right)\ du$$

$$\overset{u=\tan\theta}{=}-2\int_0^{\pi/4}\ln(\tan\theta)\sin^2(2\theta)\ln(\sin(2\theta))\ d\theta$$

$$\overset{2\theta\to\theta}{=}-\int_0^{\pi/2}\ln\left(\tan\frac{\theta}{2}\right)\sin^2(\theta)\ln(\sin(\theta))\ d\theta$$

From here, I took two different paths, the first one is using the Fourier series of $\ln\left(\tan\frac{\theta}{2}\right)=-2\sum_{n=0}^\infty\frac{\cos((2n+1)\theta)}{2n+1}$ and the second one is using $\tan\frac{\theta}{2}=\frac{\sin\theta}{1+\cos\theta}=\frac{1-\cos\theta}{\sin\theta}$ but both didn't work for me. Continuing my work or providing different ideas would be appreciated. Thank you

Answer 1

Let $\operatorname{sech}(x) = t$ and integrate by parts to obtain \begin{align} -I &= \int \limits_0^\infty \frac{x \log(\cosh(x))}{\cosh^3(x)} \, \mathrm{d}x = \int \limits_0^1 \frac{-\log(t) \operatorname{arsech}(t) t^2}{\sqrt{1-t^2}} \, \mathrm{d} t \\ &= \int \limits_0^1 \sqrt{1-t^2} \frac{\mathrm{d}}{\mathrm{d} t} \left[-\log(t) t \operatorname{arsech}(t)\right] \mathrm{d} t \\ &= \int \limits_0^1 \frac{-\log(t) \operatorname{arsech}(t) (1-t^2)}{\sqrt{1-t^2}} \, \mathrm{d} t - \int \limits_0^1 \left[\sqrt{1-t^2} \operatorname{arsech}(t) - \log(t)\right] \mathrm{d} t \, . \end{align} Averaging the second and the fourth expression yields $$- I = \frac{1}{2} \int \limits_0^1 \frac{-\log(t) \operatorname{arsech}(t)}{\sqrt{1-t^2}} \, \mathrm{d} t - \frac{1}{2}\int \limits_0^1 \sqrt{1-t^2} \operatorname{arsech}(t)\, \mathrm{d} t - \frac{1}{2} \equiv J - K - \frac{1}{2}\, .$$ $K$ can be computed by reversing the previous substitution: $$ K = \frac{1}{2} \int \limits_0^\infty \frac{x \sinh^2(x)}{\cosh^3(x)} \, \mathrm{d} x = \frac{1}{4} \int\limits_0^\infty \frac{\sinh(x) + x \cosh(x)}{\cosh^2(x)} \, \mathrm{d} x = \frac{1}{4}(1+2 \mathrm{G}) = \frac{1}{4} + \frac{\mathrm{G}}{2} \, . $$ For $J$ we can use $t = \frac{2u}{1+u^2}$ to find \begin{align} J &= \int \limits_0^1 \frac{\log(u) \log\left(\frac{2u}{1+u^2}\right)}{1+u^2} \, \mathrm{d} u \\ &= \int \limits_0^1 \frac{-\log(u) \log(1+u^2)}{1+u^2} \, \mathrm{d} u + \int \limits_0^1 \frac{\log^2(u)}{1+u^2} \, \mathrm{d} u - \log(2) \int \limits_0^1 \frac{-\log(u)}{1+u^2} \, \mathrm{d} u \\ &= 2 \operatorname{Im} \operatorname{Li}_3(1+\mathrm{i}) + \mathrm{G} \log(2) - \frac{\pi}{8} \log^2(2) - \frac{3 \pi^3}{32} + \frac{\pi^3}{16} - \mathrm{G} \log(2) \\ &= 2 \operatorname{Im} \operatorname{Li}_3(1+\mathrm{i}) - \frac{\pi}{8} \log^2(2) - \frac{\pi^3}{32} \, . \end{align} The first integral has been calculated here and the others are well-known special values of the Dirichlet beta function. Therefore, $$ -I = J - K - \frac{1}{2} = 2 \operatorname{Im} \operatorname{Li}_3(1+\mathrm{i}) - \frac{\pi}{8} \log^2(2) - \frac{\pi^3}{32} - \frac{\mathrm{G}}{2} - \frac{3}{4} \, . $$

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The problem with the original approach is that for $b=3$ the series only converges for $a > 1$, which leads to the two divergent series in the final answer. This can be avoided by computing the result for sufficiently large values of $a$ first and then taking the limit $a \to 1^+$, which can be justified by analytic continuation. $\Omega_3, \dots, \Omega_8$ are calculated as before after taking the limit inside the series, but the divergent terms are replaced by the regularised versions $$ \Omega_1 = \lim_{a \to 1^+} \sum \limits_{n=1}^\infty \frac{(-1)^{n-1}}{(2n+1)^{a-1}} = \lim_{a \to 1^+} [1 - \beta(a-1)] = 1 - \beta (0) = \frac{1}{2}$$ and \begin{align} \Omega_2 &= \lim_{a \to 1^+} \sum \limits_{n=1}^\infty \frac{(-1)^{n-1} H_n}{(2n+1)^{a-1}} = \lim_{a \to 1^+} \sum \limits_{n=1}^\infty \frac{(-1)^{n-1}}{(2n+1)^{a-1}} \int \limits_0^1 \frac{1 - x^n}{1-x} \, \mathrm{d} x \\ &= \lim_{a \to 1^+} \int \limits_0^1 \frac{\frac{\operatorname{Ti}_{a-1}(\sqrt{x})}{\sqrt{x}} - \beta(a-1)}{1-x} \, \mathrm{d} x = \int \limits_0^1 \frac{\frac{\operatorname{Ti}_{0}(\sqrt{x})}{\sqrt{x}} - \beta(0)}{1-x} \, \mathrm{d} x \\ &= \int \limits_0^1 \frac{\frac{1}{1+x} - \frac{1}{2}}{1-x} \, \mathrm{d} x = \frac{1}{2}\int \limits_0^1 \frac{\mathrm{d} x}{1+x} = \frac{1}{2} \log(2) \, . \end{align}

Answer 2

Bonus

In the question body, we have

$$\int_0^\infty x^a\operatorname{sech}^bx\ dx=\frac{2^b\Gamma(a+1)}{\Gamma(b)}\sum_{n=0}^\infty\frac{\Gamma(n+b)}{\Gamma(n+1)}\frac{(-1)^n}{(2n+b)^{a+1}}\tag1$$ On the other hand $$\int_0^\infty x^a\operatorname{sech}^bx\ dx=\int_0^\infty x^a\left(\frac{2e^{-x}}{1+e^{-2x}}\right)^b\ dx$$ $$=(-1)^a\int_0^1\frac{\ln^a u}{u}\left(\frac{2u}{1+u^2}\right)^b\ du=(-1)^a \int_0^1 f(u)\ du\tag2$$

where

$$\int_0^1 f(u) \ du=\int_0^\infty f(u) \ du-\underbrace{\int_1^\infty f(u) \ du}_{u\to 1/u}$$

$$=\int_0^\infty f(u) \ du-(-1)^a\int_0^1 f(u) \ du$$

so

$$\int_0^1 f(u) \ du=\frac{1}{1+(-1)^a}\int_0^\infty f(u) \ du\tag3$$

plug $(3)$ in $(2)$ we get

$$\int_0^\infty \frac{\ln^ax}{x}\left(\frac{2x}{1+x^2}\right)^b \ dx=(1+(-1)^a)\int_0^\infty x^a\operatorname{sech}^bx\ dx$$

Clearly, for odd $a$, the integral is $0$ and for even $a$ we have

$$\int_0^\infty \frac{\ln^ax}{x}\left(\frac{2x}{1+x^2}\right)^b \ dx=2\int_0^\infty x^a\operatorname{sech}^bx\ dx\tag4$$

From $(1)$ and $(4)$ we have

$$\int_0^\infty \frac{\ln^ax}{x}\left(\frac{2x}{1+x^2}\right)^b \ dx=\frac{2^{b+1}\Gamma(a+1)}{\Gamma(b)}\sum_{n=0}^\infty\frac{\Gamma(n+b)}{\Gamma(n+1)}\frac{(-1)^n}{(2n+b)^{a+1}}$$