If $ \int_{0}^{1} \frac{\ln x}{1-x^2} dx = -\frac{π^2}{\lambda} $ find $\lambda$ given that $\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{π^2}{6} $

Question

If $$\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{π^2}{6} $$ then $$ \int_{0}^{1} \frac{\ln x}{1-x^2} dx = -\frac{π^2}{\lambda} $$ then the value of $\lambda$ equals?

My attempt- I tried using integrating by parts to the integral. As a result, I'm left with $$-\frac{1}{2} \int_{0}^{1} \frac{\ln\left(\frac{1+x}{1-x}\right)}{x} dx $$ Now I'm stuck in here! I don't know how to move on from here! Or maybe integrating by parts was a bad option? If it is, please guide me to a solution or Please help me on how to continue from here! Any help would be appreciated.

Answer 1

Naaah. Since $\frac{1}{1-x^2}=1+x^2+x^4+\ldots$ over $(0,1)$ and $\int_{0}^{1}x^{2n}\log(x)\,dx = -\frac{1}{(2n+1)^2}$,

$$ \int_{0}^{1}\frac{\log x}{1-x^2}\,dx = -\sum_{n\geq 0}\frac{1}{(2n+1)^2} = -\left[\zeta(2)-\tfrac{1}{4}\zeta(2)\right] $$ and $\lambda=\color{red}{8}$.

Answer 2

Use the expansion $$\ln\frac{1+x}{1-x}=2\left(x+\frac{x^3}{3}+\frac{x^5}{5}+\cdots\right)$$ then \begin{align} -\frac{1}{2} \int_{0}^{1} \frac{\ln\frac{1+x}{1-x}}{x} dx &= -\frac{1}{2} \int_{0}^{1} \dfrac1x2\left(x+\frac{x^3}{3}+\frac{x^5}{5}+\cdots\right) dx \\ &= -\left(1+\frac{1}{3^2}+\frac{1}{5^2}+\cdots\right) \\ &= \color{blue}{-\dfrac{\pi^2}{8}} \end{align}

Answer 3

For another approach, take $f(x)=x^2$ on $[0,2\pi]$ and extend periodically. Then the Fourier series for $f$ converges to $\frac{1}{2}(\text{jump})$ where "jump" is the value of the difference of the left-and right-handed limit as $x\to 2\pi.$ For this $f$, jump=$4\pi^2$. Therefore, we have, after computing the Fourier series for $f$ and substituting $x=2\pi,$

$\frac{4\pi^{2}}{3}+\sum^{\infty}_{n=1}\frac{4\cos(2\pi n)}{n^2}=2\pi^2\Rightarrow\sum^{\infty}_{n=1}\frac{1}{n^2}=\frac{\pi^{2}}{6}$

Answer 4

First we use partial fraction expansion to write

$$\begin{align} \int_0^1 \frac{\log(x)}{1-x^2}\,dx&=\frac12\int_0^1 \frac{\log(x)}{1-x}\,dx-\frac12\int_0^1 \frac{\log(x)}{1+x}\,dx\tag1 \end{align}$$

Next, enforcing the substitution $x\mapsto x^2$ reveals

$$\begin{align} \int_0^1 \frac{\log(x)}{1-x}\,dx&=\int_0^1 \frac{\log(x^2)}{1-x^2}\,2x\,dx\\\\ &=2\int_0^1 \frac{\log(x)}{1-x}\,dx+2\int_0^1 \frac{\log(x)}{1+x}\,dx \tag2 \end{align}$$

whence we find from $(2)$ that

$$\int_0^1 \frac{\log(x)}{1+x}\,dx=-\frac12\int_0^1 \frac{\log(x)}{1-x}\,dx\tag3$$

Substituting $(3)$ into $(1)$ we obtain

$$\begin{align} \int_0^1 \frac{\log(x)}{1-x^2}\,dx&=\frac34 \int_0^1 \frac{\log(x)}{1-x}\,dx\\\\ &=\frac34 \int_0^1 \frac{\log(1-x)}{x}\,dx\\\\ &=\frac34\int_0^1 \frac{-\sum_{n=1}^\infty \frac{x^n}{n}}{x}\,dx\\\\ &=-\frac34\sum_{n=1}^\infty\frac1n \int_0^1 x^{n-1}\,dx\\\\ &=-\frac34\sum_{n=1}^\infty\frac1{n^2}\\\\ &-\frac{\pi^2}{8} \end{align}$$