Is there a closed form for $\sum_{n=1}^\infty \frac{q^{-n^2}}{n}$?

Question

I found that $\sum_{n=1}^\infty q^{-n^2}$ can be expressed using a theta function. Is there also a closed form for the following series? $$\sum_{n=1}^\infty \frac{q^{-n^2}}{n}$$

Answer 1

We have $$\sum_{n\geq1}q^{-n^{2}}\cos\left(2nz\right)=\frac{1}{2}\left(\theta_{3}\left(z,\frac{1}{q}\right)-1\right)$$ then if we integrate with respect to $z$ we get $$\sum_{n\geq1}q^{-n^{2}}\frac{\sin\left(2nz\right)}{2n}=\frac{1}{2}\int\theta_{3}\left(z,\frac{1}{q}\right)dz-\frac{z}{2}.$$ Now you can make an estimation, because a I think there isn't a closed form of your series.