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

Question

I happen to watch the video here, which gives a solution to the definite integral below using the power series approach. Then answer is $\frac{\pi^2}{6}$, given by:

$$\int_0^1 \frac{\ln x}{x-1}dx=\int_{-1}^0 \frac{\ln(1+u)}{u}du=\sum_{n=0}^{\infty}\frac{1}{(n+1)^2}=\frac{\pi^2}{6},$$

where the power seires expansion of the function $\ln(1+u)$ is used.

I tried for some time, but could not find another approach. Does anyone know any alternative methods to evaluate above definite integral without using the infinite series expansion?

Any comments, or ideas, are really appreciated.

Answer 1

Here is to integrate without resorting to power series. \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 =\frac43 J(1) \end{align}

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

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

Thus

$$ \int_0^1\frac{\ln x}{1-x}dx = \frac43\int_0^1 J'(\alpha) d\alpha =-\frac43\int_0^1 \frac{\sin^{-1}\alpha}{\sqrt{1-\alpha^2}}d\alpha= -\frac{\pi^2}{6}$$

Answer 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!

Answer 3

I have $2$ approaches, both of which rely on the result

$$\int_0^1\frac{\ln x}{x^2-1}~\mathrm dx=\frac{\pi^2}8$$

\begin{align}\int_0^1\frac{\ln x}{x^2-1}~\mathrm dx&\overset{x\mapsto\tanh x}=\int_0^\infty\ln(\coth x)~\mathrm dx\\&=\frac12\int_0^\infty\ln\left(\frac{2\cosh^2x}{2\sinh^2x}\right)~\mathrm dx\\&=\frac12\int_0^\infty\ln\left(\frac{\cosh2x+1}{\cosh2x-1}\right)~\mathrm dx\\&\overset{2x\mapsto x}=\frac14\int_0^\infty\ln\left(\frac{\cosh x+1}{\cosh x-1}\right)\mathrm dx\\&=\frac14\int_0^\infty\int_{-1}^1\frac{\mathrm dy}{\cosh x+y}~\mathrm dx\\&=\frac14\int_0^\infty\left(\int_0^1\frac1{\cosh x+y}~\mathrm dy+\underbrace{\int_{-1}^0\frac1{\cosh x+y}~\mathrm dy}_{y\mapsto-y}\right)~\mathrm dx\\&=\frac14\int_0^\infty\int_0^1\frac1{\cosh x+y}+\frac1{\cosh x-y}~\mathrm dy~\mathrm dx\\&=\frac12\int_0^1\int_0^\infty\frac{\cosh x}{\sinh^2x+1-y^2}~\mathrm dx~\mathrm dy\\&=\frac\pi4\int_0^1\frac{\mathrm dy}{\sqrt{1-y^2}}\\&=\frac{\pi^2}8\end{align}

$\textbf{Method 1:}$ Consider $I_\pm=\int_0^1\frac{\ln x}{1\pm x}~\mathrm dx$. Then, IBP,

$$I_\pm=\pm\int_0^1\frac{\ln(1\pm x)}x~\mathrm dx$$

Now, utilizing the identity $(1+x)(1-x)=1-x^2$, \begin{align}I_+-I_-&=\int_0^1\frac{\ln(1-x^2)}x~\mathrm dx\\&\overset{x^2\mapsto x}=-\frac{I_-}2\\\implies I_+&=\frac{I_-}2\end{align}

And since \begin{align}\int_0^1\frac{\ln x}{x^2-1}~\mathrm dx&=\frac12\int_0^1\ln x\left(\frac1{1+x}+\frac1{1-x}\right)~\mathrm dx\\&=\frac{I_++I_-}2\\&=\frac{3I_-}4\\\implies I_-&=\frac{\pi^2}6\end{align}

as desired.

$\textbf{Method 2:}$ Consider $I=\int_0^1\frac{x\ln x}{x^2-1}~\mathrm dx$, $J=\int_0^1\frac{\ln x}{x^2-1}~\mathrm dx$.

Note that on adding both these integrals, a factor of $(x+1)$ gets cancelled:

$$I+J=\int_0^1\frac{\ln x}{x-1}\mathrm dx$$

But substituting $x^2\mapsto x$ in $I$ gives:

$$I=\frac14\int_0^1\frac{\ln x}{x-1}~\mathrm dx$$

Hence,

$$I=\frac{J}3=\frac{\pi^2}{24}$$ $$\implies\int_0^1\frac{\ln x}{x-1}~\mathrm dx=4I=\frac{\pi^2}6$$

as desired.

Answer 4

• Using diagamma function

$$\begin{align*} & \color{blue}{I = \int_0^1 \frac {\ln x}{x-1}dx}\\ \end{align*}$$

Now, $$\color{red}{\psi_0 (z) = -\gamma + \int_0^{1} \frac {x^{z-1}-1}{x-1}dx}$$ We'll differentiate and that is trigamma function $\psi_1(z)$

$$\implies \frac {\partial\psi_0}{\partial z}= \psi_1 (z) = \int_0^1 \frac {x^{z-1}\ln x}{x-1}dx$$ $$\color{green}{I = \left[\psi_1(z)\right]_{z=1} = \frac {\pi^2}{6}}$$

I have copied this answer from my old answer