Can $\int\limits_0^\infty\frac{e^{-t}}{e^{c\sqrt t}-1}dt$ be evaluated in closed form? 
Can the integral
  $$\int\limits_0^\infty\frac{e^{-t}}{e^{c\sqrt t}-1}dt$$
  be evaluated in closed form using some known special functions? ($c \in \Bbb C$)

This is taken from a question on MathOverflow, which aims a finding a function based on its asymptotic expansion. It appears to boil down to solving the above integral. 
 A: Not a closed form, but it's quite straightforward to find a series representation of the integral.
Consider the case $\text{Re} (c)>0$, the opposite case can be dealt with in the same manner.
We simply expand the denominator as geometric series:
$$\int_0^\infty \frac{e^{-t}}{e^{c \sqrt{t}}-1}dt=\sum_{k=0}^\infty \int_0^\infty e^{-t-c(k+1) \sqrt{t}}dt$$
The integrals look quite familiar:
$$\int_0^\infty e^{-t-a \sqrt{t}}dt=2 \int_0^\infty u e^{-u^2-a u}du  $$
We consider a more simple integral:
$$\int_0^\infty e^{-u^2-a u}du=e^{a^2/4} \int_0^\infty e^{-\left(u+a/2 \right)^2}du= \frac{\sqrt{\pi}}{2} e^{a^2/4} \text{erfc} \left( \frac{a}{2} \right)$$
Differentiating w.r.t. $a$ we obtain:
$$\int_0^\infty e^{-t-a \sqrt{t}}dt=1- \frac{\sqrt{\pi}}{2} a e^{a^2/4} \text{erfc} \left( \frac{a}{2} \right)$$
Finally we have a (slow converging) series for our original integral:

$$\int_0^\infty \frac{e^{-t}}{e^{c \sqrt{t}}-1}dt=\sum_{k=0}^\infty \left(1- \frac{\sqrt{\pi}}{2} c (k+1) e^{c^2(k+1)^2/4} \text{erfc} \left( \frac{c(k+1)}{2} \right) \right)$$

