Finding a closed form, or reasonably simple series representation, for $ \lim_{x\to 1^-} (1-x)\sum_{n\geq 1}\log(1-x^n)$ When dealing with the computation/approximation of $\int_{0}^{1}\sqrt{\prod_{n\geq 1}(1-x^n)}\,dx$ I faced the following problem:

Find a closed form, or a reasonably simple series representation, for
$$ \lim_{x\to 1^-} (1-x)\sum_{n\geq 1}\log(1-x^n) $$

The tricky parts comes from the fact that we are not allowed to exchange $\lim$ and $\sum$. Here we have a sketch of the convergence of $(1-x)\sum_{n=1}^{N}\log(1-x^n)$ to $f(x)=(1-x)\sum_{n\geq 1}\log(1-x^n)$:

I tried to compute the above limit as
$$ \lim_{m\to +\infty} \int_{0}^{1} (m+1) x^m f(x)\,dx $$
and I got stuck in the evaluation of series involving harmonic numbers (sadly).
I guess the Jacobi triple product can be helpful. Numerically the limit is $\approx -1.64493$.

Post-mortem addendum: the answers below are both great, and I checked that also my idea leads to $-\psi'(1)=-\zeta(2)$ as a value of the limit. As a by-product we have the approximation
$$ \prod_{n\geq 1}(1-x^n) \approx \exp\left(\frac{-\pi^2 x}{(1-x)(\pi^2-x(\pi^2-6))}\right) $$
over $[0,1]$, whose absolute error is less than $0.003$.
 A: For $x \in (0,1)$, one has \begin{align*} (1-x)\sum_{n= 1}^{+\infty}\log(1-x^n) & = (1-x)\sum_{n=1}^{+\infty}\sum_{k=1}^{+\infty} - \dfrac{x^{nk}}{k}\\ 
& = (x-1)\sum_{k=1}^{+\infty} \dfrac{1}{k}\sum_{n=1}^{+\infty} x^{nk}\\
&=(x-1)\sum_{k=1}^{+\infty} \dfrac{x^k}{k(1-x^k)} \\
& = -\sum_{k=1}^{+\infty} \dfrac{x^k}{k \sum_{j=0}^{k-1} x^j}\end{align*}
Now you can intervert the limit with the summation :
\begin{align*} \lim_{x \rightarrow 1^-} (1-x)\sum_{n= 1}^{+\infty}\log(1-x^n) & = \lim_{x \rightarrow 1^-} -\sum_{k=1}^{+\infty} \dfrac{x^k}{k \sum_{j=0}^{k-1} x^j}\\ &= -\sum_{k=1}^{+\infty} \lim_{x \rightarrow 1^-} \dfrac{x^k}{k \sum_{j=0}^{k-1} x^j} \\
&= - \sum_{k=1}^{+\infty} \dfrac{1}{k^2}\end{align*}
i.e. $$\boxed{\lim_{x \rightarrow 1^-} (1-x)\sum_{n= 1}^{+\infty}\log(1-x^n) = - \dfrac{\pi^2}{6}}$$
A: In general, let $f : [0, 1] \to \mathbb{C}$ be continuous. Then
$$ \lim_{x \to 1^-} \sum_{n=1}^{\infty} f(x^n) x^n (1 - x) = \int_{0}^{1} f(x) \, \mathrm{d}x. $$
This is easily seen by noting that the sum $\sum_{n=1}^{\infty} f(x^n) (x^n - x^{n+1})$ can be regarded as a Riemann sum. This may possibly open up a possibility of invoking Euler–Maclaurin formula to obtain an asymptotic expansion, but let me not delve into this.
Anyway, applying this observation to $f(x) = \frac{\log (1-x)}{x}$ (with the singularity at the origin removed), it follows that
$$ \lim_{x \to 1^-} \sum_{n=1}^{\infty} \log(1 - x^n) (1 - x) = \int_{0}^{1} \frac{\log(1 - x)}{x} \, \mathrm{d}x = -\zeta(2). $$
A: You can use the known functional relation of the Dedekind eta function. Thus for $t>0$,
$$
\prod\limits_{n = 1}^\infty  {(1 - e^{ - nt} )}  = e^{\frac{t}{{24}}} \eta \left( { - \frac{t}{{2\pi i}}} \right) = e^{\frac{t}{{24}}} \sqrt {\frac{{2\pi }}{t}} \eta \left( {\frac{{2\pi i}}{t}} \right) = e^{ - \frac{{\pi ^2 }}{{6t}} + \frac{t}{{24}}} \sqrt {\frac{{2\pi }}{t}} \prod\limits_{n = 1}^\infty  {(1 - e^{ - \frac{{4\pi ^2 }}{t}n} )} .
$$
Accordingly,
$$
\prod\limits_{n = 1}^\infty  {(1 - x^n )}  = x^{ - \frac{1}{{24}}} e^{\frac{{\pi ^2 }}{{6\log x}}} \sqrt {\frac{{ - 2\pi }}{{\log x}}} \prod\limits_{n = 1}^\infty  {(1 - e^{\frac{{4\pi ^2 }}{{\log x}}n} )} 
$$
for $0<x<1$.
