How to integrate$\int_0^1 \frac{\ln x}{x-1}dx$ without power series expansion

Solution 1:

Here is to integrate without resorting to power series. Note \begin{align} \int_0^1\frac{\ln x}{1-x}dx& =\frac43\int_0^1 \frac{\ln x }{1-x}dx -\frac13\int_0^1 { \frac{\ln x }{1-x} } \overset{x\to x^2}{dx} \\ &= \frac43\int_0^1 \frac{\ln x}{1-x^2}dx = \frac23\int_0^\infty \frac{\ln x}{1-x^2}dx=\frac23J(1) \end{align}

where $ J(\alpha) =-\frac 12 \int_0^\infty \frac{\ln (1-\alpha^2 + \alpha^2 x^2)}{x^2-1}dx $

$$ J'(\alpha) =-\int_0^\infty \frac{\alpha dx}{1-\alpha^2 + \alpha^2 x^2} = -\frac{\pi/2}{\sqrt{1-\alpha^2}}$$

Thus

$$ \int_0^1\frac{\ln x}{1-x}dx = \frac23\int_0^1 J'(\alpha) d\alpha =-\frac{\pi}{3}\int_0^1 \frac{d\alpha}{\sqrt{1-\alpha^2}}= -\frac{\pi^2}{6}$$

Solution 2:

Here is another way of showing that $\int_0^1 \frac{\log(x)}{1-x}\,dx=\pi^2/6$. First, enforce the substitution $x\mapsto 1-x$ to obtain

$$I=-\int_0^1 \frac{\log(1-x)}{x}\,dx$$

Then, noting that $\int_0^1 \frac{1}{1-xy}\,dy=-\frac{\log(1-x)}{x}$ we write $I$ as

$$I=\int_0^1 \int_0^1 \frac{1}{1-xy}\,dx\,dy\tag1$$

In THIS ANSWER, I used the transformation $x=s+t$, $y=s-t$ to write the double integral in $(1)$ as

$$\begin{align} \int_0^1\int_0^1 \frac{1}{1-xy}\,dx\,dy&=\int_0^{1/2}\int_{-s}^{s}\frac{2}{(1-s^2)+t^2}\,dt\,ds+\int_{1/2}^{1}\int_{s-1}^{1-s}\frac{2}{(1-s^2)+t^2}\,dt\,ds\\\\ &=\int_0^{1/2}\frac{4}{\sqrt{1-s^2}}\arctan\left(\frac{s}{\sqrt{1-s^2}}\right)\,ds\\\\&+\int_{1/2}^{1}\frac{4}{\sqrt{1-s^2}}\arctan\left(\sqrt{\frac{1-s}{1+s}}\right)\,ds\\\\ &=4\int_0^{1/2}\frac{\arcsin(s)}{\sqrt{1-s^2}}\,ds+4\int_{1/2}^1\frac{\arccos(s)}{2\sqrt{1-s^2}}\,ds\\\\ &=2\arcsin^2(1/2)+\arccos^2(1/2)\\\\ &=2\left(\frac{\pi}{6}\right)^2+\left(\frac{\pi}{3}\right)^2\\\\ &=\frac{\pi^2}{6} \end{align}$$

And we are done!