I want to show that
$\displaystyle \int\limits _{0}^{\infty } x\cdotp \tanh( 2x) \cdotp \ln(\coth x)\mathrm{d} x=\frac{\pi ^{2} \cdotp \ln( 2)}{2^{4}}\tag*{}$
I tried integration by parts, Feynman trick with no luck. It just gets complicated.
Definite integral $ \int _{0}^{\infty } x\cdotp \tanh( 2x) \cdotp \ln(\coth x)\mathrm{d} x$
calculusdefinite integralshyperbolic-functionsimproper-integralsintegration
Related Solutions
We'll evaluate the following auxiliary integral:
$$\int_0^\infty \sqrt{x^n + \alpha} - x^{\frac{n}{2}} \, \mathrm{d}x = \frac{\alpha^{\frac1n + \frac12} }{(n+2)\sqrt{\pi}}\Gamma\left(\frac{1}{2} - \frac{1}{n} \right)\Gamma\left( \frac{1}{n} \right)\qquad \text{for} \quad \alpha >0, \, n>2$$
Proof: Taking the substitution $\sqrt{x^n +\alpha}- \sqrt{x^n} = \sqrt{\alpha t}$ gives $x = \alpha^{\frac1n}2^{-\frac{2}{n}}t^{-\frac{1}{n}}\left(1-t\right)^{\frac{2}{n}}$ which then gives $\mathrm{d}x =-\frac{1}{n}\alpha^{\frac1n}2^{-\frac{2}{n}}\left[ t^{-\frac1n -1}(1-t)^{\frac2n -1} + t^{-\frac1n}(1-t)^{\frac2n -1}\right] \mathrm{d}t $. Combining everything we get: $\require{cancel}$ \begin{align*} \int_0^\infty \sqrt{x^n + \alpha} - x^{\frac{n}{2}} \, \mathrm{d}x & =\frac{\alpha^{\frac1n + \frac12}2^{-\frac{2}{n}}}{n}\left[\int_{0}^{1} t^{\frac12-\frac1n -1}(1-t)^{\frac2n -1}\mathrm{d}t + \int_{0}^{1} t^{\frac12-\frac1n +1-1}(1-t)^{\frac2n -1}\mathrm{d}t\right]\\ & =\frac{\alpha^{\frac1n + \frac12}2^{-\frac{2}{n}}}{n}\left[B\left(\frac12-\frac1n, \frac2n\right) + B\left(\frac12-\frac1n+1, \frac2n\right)\right]\\ & =\frac{\alpha^{\frac1n + \frac12}2^{-\frac{2}{n}}}{n}\left[1 + \frac{\frac12 - \frac1n}{\frac12+\frac1n}\right]B\left(\frac12-\frac1n, \frac2n\right)\\ & =\frac{\alpha^{\frac1n + \frac12}\cancel{2^{-\frac{2}{n}}}}{\cancel{n}}\frac{\cancel{2}\cancel{n}}{n+2}\frac{\cancel{2^{\frac2n-1}}\cancel{\Gamma\left(\frac1n + \frac12\right)}\cancel{\Gamma\left(\frac12\right)}}{\cancel{\sqrt{\pi}}\cancel{\Gamma\left(\frac12 + \frac1n\right)}}B\left(\frac12-\frac1n, \frac1n\right)\\ & =\frac{\alpha^{\frac1n + \frac12} }{n+2}\frac{\Gamma\left(\frac{1}{2} - \frac{1}{n} \right)\Gamma\left( \frac{1}{n} \right)}{\sqrt{\pi}} \end{align*} using that $B(x+1,y) = \frac{x}{x+y}B(x,y)$ and $B(x,2y) = \frac{2^{2y-1}\Gamma\left(y + \frac12\right)\Gamma\left(x+y\right)}{\sqrt{\pi} \,\Gamma\left(x+2y\right)}B(x,y)$. Note that the first result is a consequence of $\Gamma(z+1) = z\Gamma(z)$ and the second result is a consequence of Legendre's duplication formula.
With the previous result we can conclude the problem as follows: \begin{align*} I &=\frac{1}{a-b} \int_{0}^{\infty} \sqrt{x^n +a} -\sqrt{x^n +b} \, \mathrm{d}x\\ & = \frac{1}{a-b} \left[\int_{0}^{\infty} \sqrt{x^n +a}\color{blue}{-x^{\frac{n}{2}}} \, \mathrm{d}x-\int_{0}^{\infty}\sqrt{x^n +b}\color{blue}{-x^{\frac{n}{2}}} \, \mathrm{d}x\right]\\ & = \boxed{\frac{a^{\frac1n + \frac12 } - b^{\frac1n + \frac12 } }{(a-b)(n+2)\sqrt{\pi}} \Gamma \left(\frac12 - \frac1n\right) \Gamma \left( \frac1n\right)} \end{align*}
Best Answer
With the variable change $t=\tanh x$ \begin{align} I= &\int_{0}^{\infty } x\tanh( 2x) \ln(\coth x)\ dx\\ =&\int_0^1 \frac {t \ln t \ln \frac{1-t}{1+t}}{1-t^4}\ dt \overset{t\to\frac{1-t}{1+t}} =\frac14 \int_0^1 \ln t \ln \frac{1-t}{1+t}\left(\frac1t-\frac{2t}{1+t^2}\right)dt\tag1 \end{align} Note that \begin{align} J=\int_0^1 \frac{t\ln t \ln \frac{1-t}{1+t}}{1+t^2}dt \overset{t\to\frac{1-t}{1+t}} = \int_0^1 \frac{\ln t \ln \frac{1-t}{1+t}}{1+t}dt-J=\frac12 \int_0^1 \frac{\ln t \ln \frac{1-t}{1+t}}{1+t}dt \end{align} and substitute $J$ into (1) to get \begin{align} I=& \ \frac14 \int_0^1 \ln t \ln \frac{1-t}{1+t}\left(\frac1t-\frac{1}{1+t}\right) \overset{t\to\frac{1-t}{1+t}}{dt}\\ =& \ \frac14 \bigg( \int_0^1 \frac{\ln t \ln (1-t)}{1-t}dt - \int_0^1 \frac{\ln t \ln (1+t)}{1-t}dt\bigg)\tag2 \end{align} Next, let \begin{align} K(a) &= \int_0^1\frac{\ln t}{1+t}\left(\ln(1-t) + \frac{ 1}t\ln(1-at) \right)dt\\ K’(a) &=-\int_0^1\frac{\ln t}{(1+t)(1-at)}dt= -\frac{1}{1+a}\int_0^1 \bigg(\frac{\ln t}{1+t}+ \overset{y=at}{\frac{a\ln t}{1-at}} \bigg){dt} \\ &= \frac{\zeta(2)}{2(1+a)} - \frac{\ln a\ln(1-a)}{1+a}- \frac1{1+a}\int_0^a\frac{\ln y}{1-y}dy\\ \end{align} Then \begin{align} &\int_0^1 \frac{\ln t\ln(1-t)}{1-t}dt = \int_0^1 \frac{\ln t\ln(1-t)}{t}dt \\ =& \ K(1)=K(0)+ \int_0^1 K’(a)da = \frac{\zeta(2)}2\ln2-\int_0^1 d[\ln(1+a)]\int_0^a \frac{\ln y}{1-y}dy \\ \overset{ibp}= &\ \frac{3}2\ln2\ \zeta(2)+ \int_0^1 \frac{\ln a \ln(1+a)}{1-a}da \tag3\\\end{align}
Substitute (3) into (2) to obtain
$$I=\frac14\cdot \frac{3}2\ln2\ \zeta(2)= \frac{\pi^2}{16}\ln2$$