Value of $\sum_{n=0}^{\infty} \exp(-bn^a)$

Question

What is the value of $$ S = \sum_{n=0}^{\infty} \exp(-bn^a) $$ where $a>0$ and $b>0$?

I know that $$ \int_0^{\infty} \exp(-bx^a) \, \mathrm{d}x=\frac{1}{ab^{1/a}}\Gamma(1/a) $$ but cannot find an appropriate expression for the sum. Any suggestions are appreciated. Thanks.

Answer 1

For $a=0$, it diverges. For $a=1$, we have a geometric series. For $a=2$, we have the elliptic theta function. Otherwise, there are no other known closed forms.

Answer 2

You may use Euler-MacLaurin Sum Formula to get an approximate solution.

$$\sum_{i=m+1}^n f(i)=\int_m^{n}f(x)dx+B_1(f(n)-f(m))+\sum_{k=1}^p\frac{B_{2k}}{(2k)!}\left(f^{(2‌​k-1)}(n)-f^{(2k-1)}(m)\right)+R\tag{1}$$

Setting $m=0,f(n)=\exp(-bn^a)$, we have

$$\sum_{i=1}^n f(i)=\int_0^{n}f(x)dx+B_1(f(n)-f(0))+\sum_{k=1}^p\frac{B_{2k}}{(2k)!}\left(f^{(2‌​k-1)}(n)-f^{(2k-1)}(0)\right)+R\tag{2}$$

Answer 3

Define $S(b)$ the function $$S(b)=\sum_{n=0}^\infty \mathrm e^{-bn^a}$$ and consider its Mellin transform $$\hat S(z)=\int_0^\infty S(b)b^{z-1}\,\mathrm db=\sum_{n=0}^\infty (n^a)^{-z}\Gamma(z)=\zeta(az)\Gamma(z)$$ Now looking up in the table of Mellin transforms (for instance Erdélyi, Table of integral transforms volume 1), I can only found the inverse Mellin transform in tha case $a=1$ and $a=2$. So I can't give any full answer, but it seems that the functions for different values of $a$ have not yet been given a name.