Problem showing dilogarithm integral is -pi^2/6

[aheight]

Homework Statement

I am having problems showing an integral converges to -Pi^2/6

Relevant Equations

Li_2(1)

I am working with the Dilogarithm function and am having problems showing the following and was wondering if someone could help:
$$
\int_0^1\int_0^y\left(\frac{1}{x-1}\right)\left(\frac{1}{y}\right)dxdy=-\frac{\pi^2}{6}
$$
This is what I have so far:

Iterating the first level:
$$
\begin{align*}
&=\int_0^1\frac{1}{y}\biggr(\text{Log}(x-1)\biggr|_0^y\biggr)dy\\
&=\int_0^1\frac{1}{y}\biggr(\frac{\text{Log}(y-1)-\pi i}{y}\biggr)dy
\end{align*}
$$
which is now improper so I can write:
$$
\begin{align*}
&=\lim_{\epsilon\to 0^+}\int_{\epsilon}^{1-\epsilon}\frac{\text{Log}(y-1)-\pi i}{y}dy\\
&=\lim_{\epsilon\to 0^+}\biggr\{\int_{\epsilon}^{1-\epsilon}\frac{\text{Log}(y-1)}{y}dy+\pi i\text{Log}(\epsilon)\biggr\}
\end{align*}
$$
and an this point I'm stuck. I realize the integral is equal to the special function $-\text{Li}_2(1)$ but I'd like to prove it by evaluating the limit above unless there is a more standard approach of doing so. Perhaps a more straight-forward question would be "How does one show $\text{Li}_2(1)=-\frac{\pi^2}{6}$?"

MENTOR NOTE: see post #6 for a correction to this post.

 

[anuttarasammyak]

The integral is calculated to
$$-\int_0^1 \frac{\log(1-y)}{y}dy$$
Why do not you exchange log(1-y) with its Maclaurin expansion?

 

[aheight]

The integral is calculated to
$$-\int_0^1 \frac{\log(1-y)}{y}dy$$
Why do not you exchange log(1-y) with its Maclaurin expansion?

Thanks. I can convert it to the expansion but before I do, I don't understand how you're equating the two integrals. As I see it,
$$
\int_0^1 \frac{\log(y-1)}{y}dy=\int_0^1 \frac{\log(1-y)+\pi i}{y}dy
$$

Could you please explain how you got your expression?

 

[fresh_42]

Given that $\sum n^{-2}= \pi^2/6$ I would try to use Riemann sums and the Cauchy product of series, or Cauchy's integral formula with a contour that avoids $1$.

 

[etotheipi]

The integral is calculated to
$$-\int_0^1 \frac{\log(1-y)}{y}dy$$
Why do not you exchange log(1-y) with its Maclaurin expansion?

One small thing... why is there a minus sign here? I thought considering that $|y-1| = (1-y)$ for $y \in [0,1]$, we'd have
$$
\begin{align*}
\int_0^1 \int_0^y \left(\frac{1}{x-1}\right) \left(\frac{1}{y} \right) \mathrm{d}x \mathrm{d}y = \int_0^1 \frac{\ln|y-1|}{y} \mathrm{d}y = \int_0^1 \frac{\ln(1-y)}{y} \mathrm{d}y
\end{align*}
$$Then as you say we could just re-write $\ln(1-y) = -\sum_{k=1}^{\infty} \frac{y^k}{k}$ to get$$\int_0^1 \frac{\ln(1-y)}{y} \mathrm{d}y = - \sum_{k=1} \int_0^1 \frac{y^{k-1}}{k} \mathrm{d}y = - \sum_{k=1} \left[ \frac{y^{k}}{k^2} \right]_0^1 = - \sum_{k=1} \frac{1}{k^2} = - \frac{\pi^2}{6}$$

 

[aheight]

I think I understand how the $\log(y-1)$ gets converted to $\log(1-y)$. Considering the expression:
$$
\begin{align*}
&=\lim_{\epsilon\to 0^+}\int_{\epsilon}^{1-\epsilon}\frac{\text{Log}(y-1)-\pi i}{y}dy\\
&=\lim_{\epsilon\to 0^+}\int_{\epsilon}^{1-\epsilon}\frac{\text{Log}(1-y)+\pi i-\pi i}{y}dy\\
\end{align*}
$$
and then use the series expansion as per etotheipi above where I'm distinguishing the principal branch of $\log$ by Log.