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I was evaluating Evaluate $\displaystyle\int_0^{\infty} x^2\ln(\sinh x)\operatorname{sech}(3 x){\rm d}x .$
On the path of integrating the main function, I am stuck at this integral. I don't know how to evaluate it. $$\int_1^{\infty}\frac{t^2\ln^2 t\ln(t^2-1)}{1+t^6}{\rm d}t $$

here is my try to evaluate main integral

\begin{align*} I:=&\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}}{1+e^{-2x}}\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}}{1+e^{-2x}}\left[\ln\left(e^{\frac{2x}{3}}-1\right)-\frac{x}{3}-\ln2\right]{\rm d}x\\ =&\frac{2}{27}\int_0^{\infty}\frac{x^2e^{-x}}{1+e^{-2x}}\ln\left(e^{\frac{2x}{3}}-1\right){\rm d}x-\frac{2}{81}\int_0^{\infty}\frac{x^3e^{-x}}{1+e^{-2x}}{\rm d}x-\frac{2\ln2}{27}\int_0^{\infty}\frac{x^2e^{-x}}{1+e^{-2x}}{\rm d}x. \end{align*}

\begin{align*} J:&=\int_0^{\infty}\frac{x^3e^{-x}}{1+e^{-2x}}{\rm d}x=\int_0^{\infty}x^3e^{-x}\sum_{n=0}^{\infty}(-1)^ne^{-2nx}{\rm d}x\\ &=\sum_{n=0}^{\infty}(-1)^n\int_0^{+\infty}x^3e^{-(2n+1)x}{\rm d}x=6\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^4}=6\beta(4). \end{align*} \begin{align*} K:&=\int_0^{\infty}\frac{x^2e^{-x}}{1+e^{-2x}}{\rm d}x=\int_0^{\infty}x^2e^{-x}\sum_{n=0}^{\infty}(-1)^ne^{-2nx}{\rm d}x\\ &=\sum_{n=0}^{\infty}(-1)^n\int_0^{+\infty}x^2e^{-(2n+1)x}{\rm d}x=2\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^3}=2\beta(3)=\frac{\pi^3}{16}. \end{align*}

\begin{align*} L:&=\int_0^{\infty}\frac{x^2e^{-x}}{1+e^{-2x}}\ln\left(e^{\frac{2x}{3}}-1\right){\rm d}x=27\int_1^{\infty}\frac{t^2\ln^2 t\ln(t^2-1)}{1+t^6}{\rm d}t . \end{align*}

Martin.s
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1 Answers1

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A lazy solution

Using $$\frac 1{1+t^6}=\sum_{n=0}^\infty (-1)^n\, t^{-6 (n+1)}$$ $$\int \frac{t^2 \log ^2(t) \log \left(t^2-1\right)}{t^{6 n}+1}\,dt=\sum_{n=0}^\infty (-1)^n\,\int t^{2-6 n} \log ^2(t) \log \left(t^2-1\right)\,dt$$

$$I_n=\int t^{2-6 n} \log ^2(t) \log \left(t^2-1\right)\,dt$$ is given in terms of the Gaussian hypergeometric function. Using the bounds $$J_n=\int_1^\infty t^{2-6 n} \log ^2(t) \log \left(t^2-1\right)\,dt$$ $$J_n=-\frac{9 (1-2 n)^2 \psi ^{(2)}\left(3 n-\frac{3}{2}\right)+8 \psi ^{(0)}\left(3 n-\frac{3}{2}\right)+12(1-2 n) \psi ^{(1)}\left(3 n-\frac{3}{2}\right)+8 \gamma }{108 (2 n-1)^3}$$

The summation converges slowly to $0.1277413$

Claude Leibovici
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