# 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..

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

-
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 $$I=I'(0)=-\sum_{n=0}^\infty\frac{1}{(2n+1)^2}=-\frac{\pi^2}{8}.$$

-


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}

-

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}

-