Computing the exact value of the integral $∫_0^∞ \tanh(2x)\ln(\tanh x)dx.$

Question

LATEST EDIT Thanks to @FDP’s alternative method and @Claude Leibovici’s generalisation on the integral. Meanwhile, I had found a formula for the integral with power n in general. $$\boxed{I_n=∫_0^∞ \tanh(2x)\ln^n(\tanh x)dx =\frac{(-1)^n n !}{2^n}\left(1-\frac{1}{2^{n+1}}\right)\zeta(n+1)} $$ and its proof is shown below.

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Recently, I investigate the integral

$$∫_0^∞ \tanh(2x)\ln (\tanh x)dx,$$ using the substitution $y=\tanh x$.

$$ \begin{aligned} I &=\int_0^{\infty} \frac{2 \tanh x}{1+\tanh ^2 x} \ln (\tanh x) d x \\ &=\int_0^{\infty} \frac{2 y \ln y}{1+y^2} \cdot \frac{d y}{2\left(1-y^2\right)} \\ &=\int_0^{\infty} \frac{y \ln y d y}{1-y^4} \\ &\stackrel{y^2\mapsto y}{=} \frac{1}{4} \int_0^{\infty} \frac{\ln y}{1-y^2} d y \end{aligned} $$

By my post, $$ \begin{aligned} &\int_0^{\infty} \frac{\ln y}{1-y^2} d y=-\frac{\pi^2}{4} \end{aligned} $$ We now conclude that $$\boxed{∫_0^∞ \tanh(2x)\ln (\tanh x)dx= -\frac{\pi^2}{16}}$$

Is there any other method to evaluate the integral? Your comments and alternative methods are highly appreciated.

Answer 1

After submitting the question, I was reminded by @ Claude Leibovici that I had added $n\in N$ originally. I am sorry for that and I want to generalise the integral to $$I_n=∫_0^∞ \tanh(2x)\ln^n(\tanh x)dx,$$ where $n\in N.$ Using the same substitution $y=\tanh x$, we have

$$ \begin{aligned} I_n &=\int_0^{\infty} \frac{y \ln ^n y d y}{1-y^4} \stackrel{y^2\mapsto y}{=} \frac{1}{2^{n+1}} \int_0^{\infty} \frac{\ln ^n y}{1-y^2} d y= \frac{1}{2^n} \int_0^1 \frac{\ln ^n y}{1-y^2} d y \end{aligned} $$ For the last integral, expanding the denominator yields $$\int_0^1 \frac{\ln ^n y}{1-y^2} d y =\sum_{k=0}^{\infty} \int_0^1 y^{2 k} \ln ^n y d y =\left.\frac{\partial^n}{\partial a^n} \int_0^1 y^a d y\right|_{a=2 k} \\= \sum_{k=0}^{\infty}\frac{(-1)^n n !}{(2 k+1)^{n+1}} =n!\left(1-\frac{1}{2^{n+1}}\right) \zeta(n+1) $$

Hence we can conclude that $$\boxed{∫_0^∞ \tanh(2x)\ln^n(\tanh x)dx =\frac{(-1)^n n !}{2^n}\left(1-\frac{1}{2^{n+1}}\right)\zeta(n+1)} $$

For examples, $$I_5= -\frac{5!}{2^5}\left(1-\frac{1}{2^6}\right) \zeta(6)=-\frac{\pi^6}{256}$$ which is checked by WA.

Answer 2

\begin{align}J&=\int_0^\infty \tanh(2x)\ln (\tanh x)dx\\ &=\int_0^\infty \frac{\text{e}^{2x}-\text{e}^{-2x}}{\text{e}^{2x}+\text{e}^{-2x}}\ln\left(\frac{\text{e}^{x}-\text{e}^{-x}}{\text{e}^{x}+\text{e}^{-x}}\right)dx\\ &=\int_0^\infty \frac{1-\text{e}^{-4x}}{1+\text{e}^{-4x}}\ln\left(\frac{1-\text{e}^{-2x}}{1+\text{e}^{-2x}}\right)dx\\ &\overset{u=-\text{e}^{-2x}}=\frac{1}{2}\int_0^1 \frac{1-u^2}{(1+u^2)u}\ln\left(\frac{1-u}{1+u}\right)du\\ &\overset{z=\frac{1-u}{1+u}}=\int_0^1 \frac{2z\ln z}{1-z^4}dz\\ &\overset{t=z^2}=\frac{1}{2}\int_0^1 \frac{\ln t}{1-t^2}dt\\ &=\frac{1}{2}\int_0^1 \frac{\ln t}{1-t}dt-\frac{1}{2}\underbrace{\int_0^1 \frac{t\ln t}{1-t^2}dt}_{y=t^2}\\ &=\frac{3}{8}\int_0^1 \frac{\ln t}{1-t}dt\\ &=\frac{3}{8}\times -\frac{\pi^2}{6}=\boxed{-\frac{\pi^2}{16}} \end{align}

Answer 3

If we consider $$I_n=\int \tanh(nx)\,\log(\tanh(x))\,dx$$ there is a (nasty) antiderivative involving logarithms of complex arguments and polylogarithms.

Doing the same as you did and using multiple angles formulae $$I_1=\int_0^1 \frac y{1-y^2}\log(y)\,dy=-\frac{\pi ^2}{24}$$

$$I_2=\int_0^1 \frac {2y}{1-y^4}\log(y)\,dy=-\frac{\pi ^2}{16}$$ $$I_3=\int_0^1 \frac{y \left(y^2+3\right) }{1-2y^2-3 y^4}\log (y)\,dy=-\frac{5 \pi ^2}{72}-\frac{\text{Li}_2\left(-\frac{1}{3}\right)}{6}-\frac{\log ^2(3)}{12}$$ $$I_4=\int_0^1 \frac{4 y \left(y^2+1\right) }{\left(1-y^2\right) \left(1+6 y^2+y^4\right) }\log (y)\,dy= -\frac{\pi ^2}{16}-\frac{1}{16} \log ^2\left(3-2 \sqrt{2}\right)$$

$$I_5=\int_0^1 \frac{y \left(y^4+10 y^2+5\right) }{1-9y^2+5y^4+5 y^6}\log (y)\,dy=$$ $$\frac{1}{120} \left(-6 \left(2 \text{Li}_2\left(-1+\frac{2}{\sqrt{5}}\right)-2 \text{Li}_2\left(-5+2 \sqrt{5}\right)+\log ^2\left(5+2 \sqrt{5}\right)\right)-7 \pi ^2\right)$$