Bilogarithmic function

I have shown that the power series $$\sum\limits_{i=1}^\infty \frac{x^n}{n^2} = -\int\limits_{0}^x \frac{\log(1-t)}{t}dt$$ for $x\in (-1,1)$. How can I show that this is also true for the boundary: $$\sum\limits_{k=1}^\infty \frac{1}{n^2}=-\int\limits_{0}^1 \frac{\log(1-t)}{t}dt$$

I hope somebody can help :)

Best regards

-
You are making a tacit use of Abel's theorem. – ncmathsadist May 22 '12 at 17:59

I assume this is what you did

$$-\log(1-x)=\sum_{n=1}^{\infty}\frac{x^n}{n}$$

Then

$$-\frac{\log(1-x)} x=\sum_{n=1}^{\infty}\frac{x^{n-1}}{n}$$

The series is clearly valid for $[-1,1)$, since for $x=-1$ we get the alternating harmonic series, which converges to $\log 2$. We now integrate termwise in the interval $(0,x)$, getting

$$-\int_0^x\frac{\log(1-t)} tdt=\sum_{n=1}^{\infty}\frac{x^{n}}{n^2}$$

What happens if we let $x=1$?

$$-\int_0^1\frac{\log(1-t)} tdt=\sum_{n=1}^{\infty}\frac{1}{n^2}$$

The series converges and is $\dfrac{\pi^2}{6}$, so the integral can be assigned this value. You can also try changing some variables to get a "nicer" integral:

$$1 - t = {e^{ - h}}$$

$$\int_0^\infty {\frac{h}{{{e^h} - 1}}} dh$$

When $h \to \infty$ the integral goes rapidly to $0$, and when $h \to 0$ the integral goes to $1$, so there is no "bad" singularity to worry about.

-