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

Question

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.

Answer 1

$$\sf I = \frac12 \int^1_0 \frac{\ln\left(\frac{1+x}{1-x}\right)}{x} \ln ((1 + x)^3 (1 - x)) dx=\frac12 \int_0^1 \frac{(a-b)(3a+b)}{x}dx$$ Where we denoted $\sf a=\ln(1+x)$ and $\sf b=\ln(1-x)$. Now we're going to use the following algebraic expression: $$\sf (a-b)(3a+b)=(a+b)^2+2(a-b)^2 -4b^2$$ Which is obtainable easily by combining the following expressions: $$\sf a^2=\frac12(a+b)^2+\frac12(a-b)^2 -b^2,\quad ab=\frac14(a+b)^2-\frac14(a-b)^2\tag 1$$

$$\Rightarrow \sf 2I=\int_0^1 \frac{\ln^2\left(1-x^2\right)}{x}dx +2\int_0^1 \frac{\ln^2\left(\frac{1+x}{1-x}\right)}{x}dx-4\int_0^1 \frac{\ln^2\left(1-x\right)}{x}dx$$ 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: $$\sf 2I=\frac12 \int_0^1 \frac{\ln^2(1-t)}{t}dt+4\int_0^1 \frac{\ln^2 t }{1-t^2}dt-4\int_0^1 \frac{\ln^2(1-t)}{t}dt$$ $$\sf =-\frac72\int_0^1 \frac{\ln^2 t}{1-t}dt+4\int_0^1 \frac{\ln^2 t }{1-t^2}dt=-\frac72\int_0^1 \frac{\ln^2 t}{1-t}dt+\frac72\int_0^1 \frac{\ln^2 t}{1-t}dt=0$$

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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: $$\sf \int_0^1 \frac{\ln^2 x}{1-x^2}dx=\frac12 \int_0^1 \frac{\ln^2 x}{1-x}dx+\frac12 \int_0^1 \frac{\ln^2 x}{1+x} dx $$ $$\sf =\frac12\int_0^1 \frac{\ln^2 x}{1-x}dx+ \frac38\int_0^1 \frac{\ln^2 x}{1-x}dx=\frac78 \int_0^1 \frac{\ln^2 x}{1-x}dx$$

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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: $$\sf f=(ma+nb)(qa+pb)=mqa^2+(mp+nq)ab+npb^2$$ Using $(1)$ again we obtain: $$\sf f=\left(\frac{mq}{2}+\frac{mp+nq}{4}\right)(a+b)^2+\left(\frac{mq}{2}-\frac{mp+nq}{4}\right)(a-b)^2+(np-mq)b^2$$ Furthermore we can write: $$\sf \int_0^1 \frac{(a+b)^2}{x}dx=\int_0^1 \frac{\ln^2(1-x^2)}{x}dx=\frac12 \int_0^1 \frac{\ln^2 x}{1-x}dx$$ $$\sf \int_0^1 \frac{(a-b)^2}{x}dx=\int_0^1 \frac{\ln^2\left(\frac{1+x}{1-x}\right)}{x}dx=2\int_0^1 \frac{\ln^2 x}{1-x^2}dx=\frac74\int_0^1 \frac{\ln^2 x}{1-x}dx$$ $$\sf \Rightarrow I(m,n,q,p)=\left(\frac{mq}{8}-\frac{5}{16}(mp+nq)+np\right)\int_0^1 \frac{\ln^2 x}{1-x}dx$$ $$=\boxed{\sf \left(\frac{mq}{4}-\frac{5}{8}(mp+nq)+2np\right)\zeta(3)}$$

Answer 2

\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}