# Prove that $\int_0^1 \frac{\ln x}{x-1} dx$ converge.

Prove that $\int_0^1 \frac{\ln x}{x-1} dx$ converges.

We cannot apply Abel's/Dirichliet's tests here (For example, Dirichliet's test demands that for $g(x)=\ln x$, $\int_0^1 g(x)dx < \infty$ which isn't true).

I also tried to compare the integral to another;

Since $x>\ln x$ I tried to look at $\int_0^1 \frac{x}{x-1} dx$ but this integral diverges.

What else can I do?

EDIT:
Apparently I also need to show that the integral equals $\sum_{n=1}^\infty \frac{1}{n^2}$.

I used WolframAlpha and figured out that the expansion of $\ln x$ at $x=1$ is $\sum_{k=1}^\infty \frac{(-1)^k(-1+x)^k}{k}$. Would that be helpful?

\begin{align} \int_0^1 \frac{\ln x}{x-1} dx &=\int_0^1 \frac{\sum_{k=1}^\infty \frac{-(-1)^k(-1+x)^k}{k}}{x-1} dx \\~\\ &=\int_0^1 \sum_{k=1}^\infty \frac{-(-1)^k(-1+x)^{k-1}}{k} dx \\~\\ &= \sum_{k=1}^\infty \frac{-(-1)^k}{k}\int_0^1(-1+x)^{k-1} dx \\~\\ &= \sum_{k=1}^\infty \frac{(-1)^k}{k}\dfrac{(-1)^k}{k} \\~\\ &= \sum_{k=1}^\infty \frac{1}{k^2} \\~\\ \end{align}

• Can you explain why does the sum and the integral can be interchanged? Commented Jan 28, 2015 at 14:38
• Thats the power of power series representation. With power series, we can just integrate the terms of the series to get the integral of the series itself provided we stay within the radius of convergence. Please see tutorial.math.lamar.edu/Classes/CalcII/… Commented Jan 28, 2015 at 14:47

HINT:

use $$\sum_{n=0}^{\infty}x^n=\frac{1}{1-x}$$

$$\int_0^1\frac{\ln x}{x-1}dx=-\sum_{n=0}^{\infty}\int_0^1x^n\ln x dx=\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$$

• Could you add more details to this hint please? Commented Jan 28, 2015 at 14:48
• @Elimination see the edit
– mnsh
Commented Jan 28, 2015 at 15:01
• That's awesome. Thanks Commented Jan 28, 2015 at 15:33

Note that the power series for $\ln$ may be integrated termwise within $(0,1)$.

$$\int_{\epsilon}^{1-\delta}\frac{\ln x}{x-1}dx=-\int_{\delta}^{1-\epsilon}\frac{\ln (1-x)}{x}dx \\= \int_{\delta}^{1-\epsilon}\sum_{k=1}^{\infty}\frac{x^{k-1}}{k}dx\\=\sum_{k=1}^{\infty}\int_{\delta}^{1-\epsilon}\frac{x^{k-1}}{k}dx \\=\sum_{k=1}^{\infty}\frac{1}{k^2}[(1-\epsilon)^k-\delta^k].$$

Since $\sum 1/k^2$ is convergent, then taking one-sided limits as $\delta, \epsilon \to 0$ is justified by the Abel limit theorem.

• Why is the first equality true? Commented Jan 28, 2015 at 14:42
• @Elimination: I fixed the sign error. I used a change of variables $x \to 1-x$. You can integrate a power series termwise on any compact interval inside of the radius of convergence -- since it is uniformly convergent there.
– RRL
Commented Jan 28, 2015 at 14:46
• You are welcome. Since this is a "proof" of convergence, the limit as the the upper and lower limits go to $1$ and $0$ must be justified as per Abel's theorem.
– RRL
Commented Jan 28, 2015 at 14:50

Since: $$\int_{0}^{1}\frac{dt}{1-xt}=-\frac{\log(1-x)}{x}$$ and: $$I=\int_{0}^{1}\frac{\log x}{x-1}\,dx = -\int_{0}^{1}\frac{\log(1-x)}{x}\,dx$$ we have: $$I = \iint_{(0,1)^2}\frac{1}{1-xt}\,dt\,dx = \sum_{n\geq 0}\iint_{(0,1)^2}(xt)^n\,dt\,dx = \sum_{n\geq 0}\frac{1}{(n+1)^2}=\color{red}{\zeta(2)}$$ as wanted.

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align}&\overbrace{% \color{#66f}{\large\int_{0}^{1}\frac{\ln\pars{x}}{x - 1}\,\dd x}} ^{\ds{\dsc{x}\ \mapsto\ \dsc{1 - x}}}\ =\ =\int_{0}^{1}\ \overbrace{\frac{-\ln\pars{1 - x}}{x}} ^{\ds{=\ \dsc{\Li{2}'\pars{x}}}}\,\dd x\ =\ \int_{0}^{1}\Li{2}'\pars{x}\,\dd x=\ \overbrace{\Li{2}\pars{1}}^{\ds{=\ \dsc{\frac{\pi^{2}}{6}}}}\ -\ \overbrace{\Li{2}\pars{0}}^{\ds{=\ \dsc{0}}} \\[5mm]&=\color{#66f}{\large\frac{\pi^{2}}{6}} \end{align}

$\ds{\Li{\rm s}}$ is the PolyLogarithm Function where we used $\ds{\Li{\rm s}'\pars{x}=\frac{\Li{\rm s - 1}\pars{x}}{x}}$ and $\ds{\Li{1}\pars{x} = -\ln\pars{1 - x}}$. Note that $\ds{\Li{1}\pars{1}=\sum_{n=1}^{\infty}{1^{n} \over n^{2}} =\sum_{n=1}^{\infty}{1 \over n^{2}}=\frac{\pi^{2}}{6}}$.