Value of $\sum_{n=0}^{\infty} \exp(-bn^a)$ 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.
 A: 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.
A: 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
$$\sum_{i=1}^n f(i)=\int_0^{n}f(x)dx+B_1(f(n)-f(0))+\sum_{k=1}^p\frac{B_{2k}}{(2k)!}\left(f^{(2‌​k-1)}(n)-f^{(2k-1)}(0)\right)+R\tag{2}$$
A: 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.
