# Find the Value of Integral

Find the Value of \begin{align}I=\int_{0}^{1}\frac{\ln(x)\,dx}{1-x^2}\end{align}

I have tried like this: We have \begin{align}2I=\int_{0}^{1}\frac{\ln(x^2)\,dx}{1-x^2}=\int_{0}^{1}\frac{\ln(1-(1-x^2))\,dx}{1-x^2}\end{align} So

2I=\begin{align}\int_{0}^{1}\frac{-(1-x^2)-\frac{(1-x^2)^2}{2}-\frac{(1-x^2)^3}{3}-\cdots }{1-x^2}\end{align}=

I need help from here..

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The solution should be $-\frac{\pi^2}{8}$, cannot help further unfortunatly. btw, congrats to 666 MSE friend-points :) – NicoDean Jul 8 '14 at 13:47

Use the series expansion for $\dfrac{1}{1-x^2}$ i.e

$\displaystyle \frac{1}{1-x^2}=\sum_{k=0}^{\infty} x^{2k}\,dx$

Hence, $$I=\int_0^1 \ln(x)\left(\sum_{k=0}^{\infty} x^{2k}\right)\,dx=\sum_{k=0}^{\infty} \int_0^1 x^{2k}\ln x\,dx=-\sum_{k=0}^{\infty}\frac{1}{(2k+1)^2}$$ where the final integral can be evaluated using the substitution $\ln x=-t$.

Since, $$\sum_{k=1}^{\infty} \frac{1}{k^2}=\sum_{k=1}^{\infty} \frac{1}{(2k)^2}+\sum_{k=0}^{\infty} \frac{1}{(2k+1)^2} \Rightarrow \sum_{k=0}^{\infty} \frac{1}{(2k+1)^2}=\frac{\pi^2}{6}-\frac{1}{4}\frac{\pi^2}{6}$$ $$\Rightarrow \sum_{k=0}^{\infty} \frac{1}{(2k+1)^2}=\frac{3\pi^2}{24}=\frac{\pi^2}{8}$$

Hence, $$\boxed{I=-\dfrac{\pi^2}{8}}$$

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in the 2nd to last line, doesn't the sum start from k=0? – Eul Can Jul 8 '14 at 13:55
@oliveeuler: No. :) – Pranav Arora Jul 8 '14 at 13:58
but it's different to the sum from I in the 3rd line down? – Eul Can Jul 8 '14 at 13:59
@oliveeuler: I edited the post, does it look ok now? :) – Pranav Arora Jul 8 '14 at 14:02
great! (sorry about nitpicking but I wanted to make sure :) ) – Eul Can Jul 8 '14 at 14:05

Let $$I(\alpha)=\int_{0}^{1}\frac{x^{\alpha}dx}{1-x^2}.$$ Then $I'(0)=I$. Now \begin{eqnarray*} I(\alpha)=\lim_{a\to 1^-}\int_{0}^{a}\sum_{n=0}^\infty x^{\alpha+2n}dx=\lim_{a\to 1^-}\sum_{n=0}^\infty \frac{1}{\alpha+2n+1}a^{\alpha+2n+1} \end{eqnarray*} and hence \begin{eqnarray*} I'(\alpha)&=&\lim_{a\to 1^-}\sum_{n=0}^\infty \left(\frac{-1}{(\alpha+2n+1)^2}a^{\alpha+2n+1}+\frac{1}{\alpha+2n+1}a^{\alpha+2n+1}\ln a\right)\\ &=&-\sum_{n=0}^\infty\frac{1}{(\alpha+2n+1)^2} \end{eqnarray*} Thus $$I=I'(0)=-\sum_{n=0}^\infty\frac{1}{(2n+1)^2}=-\frac{\pi^2}{8}.$$

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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ \begin{align} I&=\color{#66f}{\Large\int_{0}^{1}{\ln\pars{x} \over 1 - x^{2}}\,\dd x} =\half\int_{0}^{1}{\ln\pars{x} \over 1 - x}\,\dd x +\half\int_{0}^{1}{\ln\pars{x} \over 1 + x}\,\dd x \\[3mm]&=\half\int_{0}^{1}{\ln\pars{x} \over 1 - x}\,\dd x -\half\int_{0}^{-1}{\ln\pars{-x} \over 1 - x}\,\dd x =\half\int_{0}^{1}{\ln\pars{1 - x} \over x}\,\dd x -\half\int_{0}^{-1}{\ln\pars{1 - x} \over x}\,\dd x \\[3mm]&=-\half\sum_{n = 1}^{\infty}{1 \over n^{2}} +\half\sum_{n = 1}^{\infty}{\pars{-1}^{n} \over n^{2}} =-\sum_{n = 1}^{\infty}{1 \over \pars{2n - 1}^{2}} =-\sum_{n = 1}^{\infty}{1 \over n^{2}} +\sum_{n = 1}^{\infty}{1 \over \pars{2n}^{2}} \\[3mm]&=-\,{3 \over 4}\ \underbrace{\sum_{n = 1}^{\infty}{1 \over n^{2}}} _{\ds{\color{#c00000}{\zeta\pars{2} = {\pi^{2} \over 6}}}} =-\,{3 \over 4}\,{\pi^{2} \over 6} = \color{#66f}{\Large -\,{\pi^{2} \over 8}} \end{align}

since \begin{align} \int_{0}^{a}{\ln\pars{1 - x} \over x}\,\dd x\, &= \int_{0}^{a}{1 \over x}\pars{-\sum_{n = 1}^{\infty}{x^{n} \over n}}\,\dd x =-\sum_{n = 1}^{\infty}{a^{n} \over n^{2}}\,,\qquad \verts{a}\ \leq\ 1 \end{align}

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Integrating by parts,

\begin{align} \int \frac{\ln (x)}{1-x^{2}} \ dx &= \ln(x) \ \text{arctanh} (x) - \int \frac{\text{arctanh}(x)}{x} \ dx \\ &= \ln (x) \ \text{arctanh} (x) - \frac{1}{2} \int \frac{\ln (1+x)}{x} \ dx + \frac{1}{2} \int \frac{\ln (1-x)}{x} \ dx \\ &= \ln(x) \ \text{arctanh}(x) + \frac{1}{2} \text{Li}_{2}(-x) - \frac{1}{2} \text{Li}_{2}(x) + C \end{align}

where $\text{Li}_{2}(x)$ is the dilogarithm function.

Then

\begin{align} \int_{0}^{1} \frac{\ln (x)}{1-x^{2}} \ dx &= \frac{1}{2} \Big( \text{Li}_{2}(-1) - \text{Li}_{2}(1) \Big) \\ &= \frac{1}{2} \Big( - \frac{\zeta(2)}{2} - \zeta(2) \Big) \\ &= \frac{1}{2} \left(- \frac{\pi^{2}}{12}-\frac{\pi^{2}}{6} \right) \\ &= -\frac{\pi^{2}}{8} . \end{align}

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