# Proving $\int_0^1\frac{\log (1-x)}{x}\mathrm dx=-\frac{\pi^2}6$ [duplicate]

It is a well known fact that $\displaystyle\sum_{k=1}^{\infty}\frac1{k^2}=\frac{\pi^2}6$. I wanted to prove this using elementary techniques. By doing some easy algebra, I found it was sufficient to prove that $$\int_0^1\frac{\log (1-x)}{x}\mathrm dx=-\frac{\pi^2}6.$$ (By defining a function $f(x)=\displaystyle\sum_{k=1}^{\infty}\frac{x^k}{k^2}$ and doing some integration.)

I have been trying to evaluate this integral using secondary school techniques (i.e., substitution or partial integration) but I didn't achieve much.

Some approaches:

1. It's not hard to see that $$\int_0^1\frac{\log (1-x)}{x}\mathrm dx=\int_0^1\frac{\log x}{1-x}\mathrm dx$$ It's not much, but the logarithm looks nicer now.

2. Also, since we want $\pi$ to appear, a trigonometric substitution might be helpful. I tried subsituting $x=\sin^2t$ which gives $$\int_0^{\frac\pi2}4\log(\cos t)\cdot \cot t\;\mathrm dt$$ or, using the second form, $$\int_0^{\frac\pi2}4\log(\sin t)\cdot \tan t\;\mathrm dt$$ Another remark: $\int\tan a=-\log|\cos a|$ might help, because $\log|\cos a|$ appeared in the integral. But I dont's see immediately how to use this. (Absolute values within the logarithm are unnecessary because $\sin t,\cos t\geq0$ in this case.)

3. Setting $x=\cos 2t$ gives $$-2\int_0^{\frac\pi2}(\log2+2\log(\sin t))\tan 2t\;\mathrm dt$$ by using the identity $\frac{1-\cos2a}{2}=\sin^2a$. And since, by symmetry, $$\int_0^{\frac\pi2}\tan 2t\;\mathrm dt=0$$ the integral is just $$-4\int_0^{\frac\pi2}\log(\sin t)\tan 2t\;\mathrm dt$$ which looks quite nice.

## marked as duplicate by Norbert, TZakrevskiy, Amzoti, Dan Rust, Cameron BuieAug 19 '13 at 12:28

• This is "$-\mathrm{Li}_2(1)$", see Dilogarithm, I'm sure it's evaluated somewhere. Some of your moves are also to be found in the related links. – Nikolaj-K Aug 19 '13 at 11:01