How to prove that $ \int_0^1 \sqrt{\ln(\frac{1}{x})} \ \frac{\vartheta _3(0,x)-1}{x} dx = \sqrt{\pi}\zeta(3) $? Where $ \vartheta _3(0,x) $ is the elliptic theta function.
I first tried to expand the series since the integrand is within the radius of convergence of the series to no avail.
Also, I'm not sure an exponential substitution would make the Theta function any simpler though it would simplify the $\ln$ part.
Any suggestions would be appreciated.

For simplicity, define the function
$$\psi(x)=\sum_{n=1}^{\infty}e^{-n^2\pi x}:=\frac{1}{2}(\theta_3(0,e^{-\pi x})-1)$$
Then the integral can be equivalently rewritten as
$$I=2\int_0^1\frac{dx}{x}\sqrt{\ln \frac{1}{x}}\psi\left(\frac{\ln 1/x}{\pi}\right)$$
which, after substituting $u=\ln 1/x$ can be shown to be
\begin{align}
I&=2\int_0^\infty\sqrt{u}\psi(u/\pi)du \\
&=2\sum_{n=1}^\infty\int_0^\infty\sqrt{u}e^{-n^2 u} \\
&=2\Gamma(3/2)\sum_{n=1}^\infty\frac{1}{n^3}\\&=\sqrt{\pi}\zeta(3)
\end{align}
There are no special Jacobi identities that need to be used for this result, but that doesn't mean it isn't cute!
