How find this sum $f(x)=\sum_{n=1}^{\infty}\frac{x^{n^2}}{n}$

First, Merry Christmas everyone!

Find this sum $$f(x)=\sum_{n=1}^{\infty}\dfrac{x^{n^2}}{n},1>x\ge 0 \tag{1}$$

This problem is creat by Laurentiu Modan.and I can't see this solution.

I know this sum $$\sum_{n=1}^{\infty}\dfrac{x^n}{n}=-\ln{(1-x)},-1\le x<1$$

and I know this $$\sum_{n=1}^{\infty}x^{n^2}\approx \dfrac{\sqrt{\pi}}{2\sqrt{1-x}},x\to 1^{-}$$

But for $(1)$,I can't find it,Thank you.

This problem is from this

Solution 1:

This is not a solution, but by playing around with the problem I found the following nasty looking identity:

$$\sum_{n=1}^{\infty}\frac{x^{n^2}}{n} = -\sqrt{\frac{1}{\pi\log(1/x)}}\int_{-\infty}^{\infty}e^{-\frac{y^2}{\log(1/x)}}\log(2|\sin(y)|)dy$$

for all $0<x<1$. Note that the function we integrate over is singular at $y=\pm n\pi$ for $n\in Z$, but the integral itself is well defined. However, this integral seems even harder to evaluate as the sum. The derivation is quite simple: $f(x)$ can be though of as a solution to the heat equation $u(t,z)$ at $z=0$ and time $t=\log(1/x)/\pi$ with the initial condition $u(0,z) = -\log(2|\sin(\pi z)|)$.