I am looking for a nice slick way to show

$$\int^1_0 \frac{\tanh^{-1} x}{x} \ln [(1 + x)^3 (1 - x)] \, dx = 0.$$

So far I can only show the result using brute force as follows. Let $$I = \int^1_0 \frac{\tanh^{-1} x}{x} \ln [(1 + x)^3 (1 - x)] \, dx.$$ Since $$\tanh^{-1} x = \frac{1}{2} \ln \left (\frac{1 + x}{1 - x} \right ),$$ the above integral, after rearranging, can be rewritten as $$I = \frac{3}{2} \int^1_0 \frac{\ln^2 (1 + x)}{x} \, dx - \int^1_0 \frac{\ln (1 - x) \ln (1 + x)}{x} \, dx - \frac{1}{2} \int^1_0 \frac{\ln^2 (1 - x)}{x} \, dx.\tag1$$ Each of the above three integrals can be found. The results are: $$\int^1_0 \frac{\ln^2 (1 + x)}{x} \, dx = \frac{1}{4} \zeta (3).$$ For a proof, see here or here. $$\int^1_0 \frac{\ln (1 - x) \ln (1 + x)}{x} \, dx = -\frac{5}{8} \zeta (3).$$ For a proof, see here. And $$\int^1_0 \frac{\ln^2 (1 - x)}{x} \, dx = 2 \zeta (3).$$ For a proof of this last one, see here.

Thus (1) becomes $$\int^1_0 \frac{\tanh^{-1} x}{x} \ln [(1 + x)^3 (1 - x)] \, dx = \frac{3}{8 } \zeta (3) + \frac{5}{8} \zeta (3) - \zeta (3) = 0,$$ as expected.

  • $\begingroup$ what about the change of variable $y=\dfrac{1-x}{1+x}$? $\endgroup$ – FDP Jun 12 at 8:26
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    $\begingroup$ Your integral is \begin{align}\int^1_0\frac{\ln\left(\frac{16x}{(1+x)^4}\right)\ln x}{1-x^2}\, dx\end{align}. \begin{align}\int^1_0\frac{\ln(1+x)\ln x}{1-x^2}\, dx\end{align} is not so easy to compute. $\endgroup$ – FDP Jun 12 at 11:11
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    $\begingroup$ The integrand here in fact has an antiderivative that can be expressed in terms of trilogs, dilogs and elementary functions using the integration formula found in this question. This expression isn't quite what most people would describe as "nice", but it certainly could be much worse. The alternative ways of solving the definite integral used in the answers provided below are admittedly slicker. OTOH, having the antiderivative opens the door to computing more general integrals. $\endgroup$ – David H Jun 22 at 9:25
  • $\begingroup$ What's the point for the bounty? $\endgroup$ – FDP Jun 25 at 10:35
  • $\begingroup$ I tested something actually and in the same time to bring attention, as it was mentioned. $\endgroup$ – Zacky Jun 28 at 23:03

$$\sf I = \frac12 \int^1_0 \frac{\ln\left(\frac{1+x}{1-x}\right)}{x} \ln ((1 + x)^3 (1 - x)) dx=\frac12 \int_0^1 \frac{(a-b)(3a+b)}{x}dx$$ Where we denoted $\sf a=\ln(1+x)$ and $\sf b=\ln(1-x)$. Now we're going to use the following expression: $$\sf (a-b)(3a+b)=(a+b)^2+2(a-b)^2 -4b^2$$

This trick is quite heuristic, but easily solves the problem since:

$$\sf 2I=\int_0^1 \frac{\ln^2\left(1-x^2\right)}{x}dx +2\int_0^1 \frac{\ln^2\left(\frac{1+x}{1-x}\right)}{x}dx-4\int_0^1 \frac{\ln^2\left(1-x\right)}{x}dx$$ Now we put $\sf x^2=t$ in the first part and $\sf \frac{1-x}{1+x}=t$ for the second one to get: $$\sf 2I=\frac12 \int_0^1 \frac{\ln^2(1-t)}{t}dt+4\int_0^1 \frac{\ln^2 t }{1-t^2}dt-4\int_0^1 \frac{\ln^2(1-t)}{t}dt$$ $$\sf =-\frac72\int_0^1 \frac{\ln^2 t}{1-t}dt+4\int_0^1 \frac{\ln^2 t }{1-t^2}dx=-\frac72\int_0^1 \frac{\ln^2 t}{1-t}dt+\frac72\int_0^1 \frac{\ln^2 t}{1-t}dt=0$$

Above we used that: $$\boxed{\sf \int_0^1 \frac{\ln^2 x}{1-x^2}dx=\frac78 \int_0^1 \frac{\ln^2 x}{1-x}dx}$$ And we can show this in two steps. First: $$\sf {\int_0^1 \frac{\ln^2 x}{1-x}dx}\overset{x\to x^2}=8\int_0^1 \frac{x\ln^2 x}{1-x^2}dx=4{\int_0^1 \frac{\ln^2 x}{1-x}dx}-4\int_0^1 \frac{\ln^2 x}{1+x}dx$$ $$\sf \Rightarrow (1-4){\int_0^1 \frac{\ln^2 x}{1-x}dx}=-4\int_0^1 \frac{\ln^2 x}{1+x}dx\Rightarrow \boxed{\int_0^1 \frac{\ln^2 x}{1+x}dx=\frac34 \int_0^1 \frac{\ln^2 x}{1-x}dx}$$ But we also have that: $$\sf \int_0^1 \frac{\ln^2 x}{1-x^2}dx=\frac12 \int_0^1 \frac{\ln^2 x}{1-x}dx+\frac12 \int_0^1 \frac{\ln^2 x}{1+x} dx $$ $$\sf =\frac12\int_0^1 \frac{\ln^2 x}{1-x}dx+ \frac38\int_0^1 \frac{\ln^2 x}{1-x}dx=\frac78 \int_0^1 \frac{\ln^2 x}{1-x}dx$$

Generalization. In a similar fashion we can deal with the following integral: $$\sf I(m,n,q,p)=\int_0^1 \frac{[m\ln(1+x)+n\ln(1-x)][q\ln(1+x)+p\ln(1-x)]}{x}dx$$ Like from above we will keep $\sf a=\ln(1+x)$ and $\sf b=\ln(1-x)$. Thus we can write the numarator as: $$\sf f=(ma+nb)(qa+pb)=mqa^2+(mp+nq)ab+npb^2$$ We also have the following two expressions: $$\sf a^2=\frac12(a+b)^2+\frac12(a-b)^2 -b^2,\quad ab=\frac14(a+b)^2-\frac14(a-b)^2$$ $$\sf \Rightarrow f=\left(\frac{mq}{2}+\frac{mp+nq}{4}\right)(a+b)^2+\left(\frac{mq}{2}-\frac{mp+nq}{4}\right)(a-b)^2+(np-mq)b^2$$ Furthermore we can write: $$\sf \int_0^1 \frac{(a+b)^2}{x}dx=\int_0^1 \frac{\ln^2(1-x^2)}{x}dx=\frac12 \int_0^1 \frac{\ln^2 x}{1-x}dx$$ $$\sf \int_0^1 \frac{(a-b)^2}{x}dx=\int_0^1 \frac{\ln^2\left(\frac{1+x}{1-x}\right)}{x}dx=2\int_0^1 \frac{\ln^2 x}{1-x^2}dx=\frac74\int_0^1 \frac{\ln^2 x}{1-x}dx$$ $$\sf \Rightarrow I(m,n,q,p)=\left(\frac{mq}{8}-\frac{5}{16}(mp+nq)+np\right)\int_0^1 \frac{\ln^2 x}{1-x}dx$$ $$ =\boxed{\sf \left(\frac{mq}{4}-\frac{5}{8}(mp+nq)+2np\right)\zeta(3)}$$ So now it's trivial to see that $\sf I(1,-1,3,1)=0$ plugging into the above expression.

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    $\begingroup$ The generalization is really interesting. $\endgroup$ – Claude Leibovici Jun 25 at 5:24

\begin{align}J=\int^1_0 \frac{\tanh^{-1} x}{x} \ln [(1 + x)^3 (1 - x)] \, dx\end{align} Perform the change of variable $y=\dfrac{1-x}{1+x}$, \begin{align}J&=\int^1_0\frac{\ln\left(\frac{16x}{(1+x)^4}\right)\ln x}{1-x^2}\, dx\\ &=4\ln 2\int_0^1\frac{\ln x}{1-x^2}\,dx+\int_0^1\frac{\ln^2 x}{1-x^2}\,dx-4\int_0^1\frac{\ln x\ln(1+x)}{1-x^2}\,dx\\ \end{align}

Define on $[0;1]$ the function $R$ by, \begin{align}R(x)&=\int_0^x \frac{\ln t}{1-t^2}\,dt\\ &=\int_0^1 \frac{x\ln(tx)}{1-t^2x^2}\,dt \end{align} Therefore, \begin{align}K&=\int_0^1\frac{\ln x\ln(1+x)}{1-x^2}\,dx\\ &=\Big[R(x)\ln(1+x)\Big]_0^1-\int_0^1\int_0^1 \frac{x\ln(tx)}{(1-t^2x^2)(1+x)}\,dt\,dx\\ &=\int_0^1 \frac{\ln 2\ln t}{1-t^2}\,dt-\int_0^1\left(\int_0^1 \frac{x\ln t}{(1-t^2x^2)(1+x)}\,dx\right)\,dt-\int_0^1\left(\int_0^1 \frac{x\ln x}{(1-t^2x^2)(1+x)}\,dt\right)\,dx\\ &=\ln 2\int_0^1 \frac{\ln t}{1-t^2}\,dt-\\ &\frac{1}{2}\left(\int_0^1 \frac{\ln t\ln(1+t)}{1-t}\,dt-\int_0^1 \frac{\ln t\ln\left(\frac{1-t}{1+t}\right)}{t}\,dt-\int_0^1 \frac{2\ln 2\ln t}{1-t^2}\,dt+\int_0^1 \frac{\ln(1-t)\ln t}{1+t}\,dt\right)-\\ &\frac{1}{2}\left(\int_0^1 \frac{\ln x\ln(1+x)}{1+x}\,dx-\int_0^1 \frac{\ln x\ln(1-x)}{1+x}\,dx\right)\\ &=2\ln 2\int_0^1 \frac{\ln t}{1-t^2}\,dt-K+\frac{1}{2}\int_0^1 \frac{\ln t\ln\left(\frac{1-t}{1+t}\right)}{t}\,dt \end{align} Therefore, \begin{align}K&=\ln 2\int_0^1 \frac{\ln t}{1-t^2}\,dt+\frac{1}{4}\int_0^1 \frac{\ln t\ln\left(\frac{1-t}{1+t}\right)}{t}\,dt\\ &=\ln 2\int_0^1 \frac{\ln t}{1-t^2}\,dt+\frac{1}{8}\left[\ln^2 t\ln\left(\frac{1-t}{1+t}\right)\right]_0^1+\frac{1}{4}\int_0^1 \frac{\ln^2 t}{1-t^2}\,dt\\ &=\ln 2\int_0^1 \frac{\ln t}{1-t^2}\,dt+\frac{1}{4}\int_0^1 \frac{\ln^2 t}{1-t^2}\,dt\\ \end{align} Therefore, \begin{align}\boxed{J=0}\end{align} NB:

It's easy to deduce that, \begin{align}\int_0^1\frac{\ln x\ln(1+x)}{1-x^2}\,dx=\frac{7}{16}\zeta(3)-\frac{1}{8}\pi^2\ln 2\end{align}


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