An integral $\int^\infty_0\frac{\tanh{x}}{x(1-2\cosh{2x})^2}{\rm d}x$

I would like to enquire about the possible methods of computing the following integral $$ \color{blue}{% \int^{\infty}_{0}\frac{\tanh\left(\, x\,\right)} {x\left[\, 1 - 2\cosh\left(\, 2x\,\right)\,\right]^{2}}\,{\rm d}x =\ ?}$$ A possible way I see of doing this is to apply the substitution $x\mapsto-\ln{x}$, which yields $$ -\int^{1}_{0}\frac{x^{3}\left(\, 1 - x^{2}\,\right)} {\left(\, 1 + x^{2}\,\right)\left(\, 1 - x^{2} + x^{4}\,\right)^{2}}\, \frac{{\rm d}x}{\ln\left(\, x\,\right)} $$ We can then introduce an extra parameter and differentiate to get rid of the $\ln$ in the denominator. However, things seem to get rather messy after this, hence this is unlikely to be the most ideal way of tackling the integral.

Therefore, may I ask if anyone has cleverer ideas ?. I am particularly interested in knowing if this integral can be solved via contour integration ( which I am not very good at as of now ). Help will certainly be appreciated. Thank you.

Solution 1:

The value of the integral has a very nice closed form in terms of the trigamma function:

$$\mathcal{I}=\color{blue}{-\frac{\ln{3}}{6}-\frac{5\pi}{9\sqrt{3}}+\frac{5\psi^{(1)}{\left(\frac13\right)}}{6\sqrt{3}\,\pi}}.$$

Starting from the second form of the integral given by the OP,

$$\begin{align} \mathcal{I} &=-\int_{0}^{1}\frac{x^3(1-x^2)}{(1+x^2)(1-x^2+x^4)^2}\frac{\mathrm{d}x}{\ln{x}}\\ &=-\int_{0}^{1}\frac{x^3(1-x^2)}{(1+x^6)(1-x^2+x^4)}\frac{\mathrm{d}x}{\ln{x}}\\ &=-\int_{0}^{1}\frac{x^3(1-x^4)}{(1+x^6)^2}\frac{\mathrm{d}x}{\ln{x}}\\ &=-\int_{0}^{1}\frac{z(1-z^{4/3})}{(1+z^2)^2}\frac{\mathrm{d}z}{3z^{2/3}\ln{\left(z^{1/3}\right)}}\\ &=\int_{0}^{1}\mathrm{d}z\,\frac{1}{(1+z^2)^2}\frac{z^{5/3}-z^{1/3}}{\ln{z}}\\ &=\int_{0}^{1}\mathrm{d}z\,\frac{1}{(1+z^2)^2}\int_{1/3}^{5/3}\mathrm{d}\mu\,z^{\mu}\\ &=\int_{1/3}^{5/3}\mathrm{d}\mu\int_{0}^{1}\mathrm{d}z\,\frac{z^{\mu}}{(1+z^2)^2}\\ &=\int_{1/3}^{5/3}\mathrm{d}\mu\left[-\frac14+\frac{\mu-1}{4}\beta{\left(\frac{\mu-1}{2}\right)}\right]\\ &=-\frac13+\int_{1/3}^{5/3}\mathrm{d}\mu\left[\frac{\mu-1}{4}\beta{\left(\frac{\mu-1}{2}\right)}\right]\\ &=-\frac13+\int_{-1/3}^{1/3}\mathrm{d}t\,t\beta{\left(t\right)}\\ &=-\frac13+\int_{-1/3}^{1/3}\mathrm{d}t\,\frac{t}{2}\left[\psi{\left(\frac{t+1}{2}\right)}-\psi{\left(\frac{t}{2}\right)}\right]\\ &=-\frac13+\int_{-1/3}^{1/3}\mathrm{d}t\,\frac{t}{2}\psi{\left(\frac{t+1}{2}\right)}-\int_{-1/3}^{1/3}\mathrm{d}t\,\frac{t}{2}\psi{\left(\frac{t}{2}\right)}\\ &=-\frac13+\int_{1/3}^{2/3}\mathrm{d}u\,(2u-1)\psi{\left(u\right)}-2\int_{-1/6}^{1/6}\mathrm{d}u\,u\psi{\left(u\right)}\\ &=-\frac13-\int_{1/3}^{2/3}\mathrm{d}u\,\psi{\left(u\right)}+2\int_{1/3}^{2/3}\mathrm{d}u\,u\psi{\left(u\right)}-2\int_{-1/6}^{1/6}\mathrm{d}u\,u\psi{\left(u\right)}\\ &=-\frac13+\ln{\left(\frac{\Gamma{\left(\frac13\right)}}{\Gamma{\left(\frac23\right)}}\right)}+2\int_{1/3}^{2/3}\mathrm{d}u\,u\psi{\left(u\right)}-2\int_{-1/6}^{1/6}\mathrm{d}u\,u\psi{\left(u\right)}\\ &=-\frac13+\ln{\left(\frac{\Gamma{\left(\frac13\right)}}{\Gamma{\left(\frac23\right)}}\right)}+2\int_{1/3}^{2/3}\mathrm{d}u\,u\psi{\left(u\right)}-2\int_{5/6}^{7/6}\mathrm{d}v\,(1-v)\psi{\left(1-v\right)}\\ &=-\frac13+\ln{\left(\frac{\Gamma{\left(\frac13\right)}}{\Gamma{\left(\frac23\right)}}\right)}+2\left[u\ln{\Gamma\left(u\right)}-\psi^{(-2)}{\left(u\right)}\right]_{1/3}^{2/3}\\ &~~~~~ +2\left[(1-v)\ln{\Gamma\left(1-v\right)}-\psi^{(-2)}{\left(1-v\right)}\right]_{5/6}^{7/6}\\ &=\ln{\left(\frac{\Gamma{\left(\frac13\right)}}{\Gamma{\left(\frac23\right)}}\right)}-\frac{5\pi}{9\sqrt{3}}-\frac{\ln{\left(2\pi\right)}}{3}+\frac23\ln{\left(\frac{\Gamma{\left(\frac23\right)}^2}{\Gamma{\left(\frac13\right)}}\right)}+\frac{5\psi^{(1)}{\left(\frac13\right)}}{6\sqrt{3}\,\pi}\\ &=-\frac{5\pi}{9\sqrt{3}}-\frac{\ln{3}}{6}+\frac{5\psi^{(1)}{\left(\frac13\right)}}{6\sqrt{3}\,\pi}.~\blacksquare\\ \end{align}$$