I want to evaluate the following Laplace transform:
$$\int_0^{\infty}\frac{e^{-st}}{\sqrt{{1-e^{-{\sqrt{t}}}}}}dt$$
I tried using the substitution method, but it led me nowhere. I think the problem is arising because of the square root function.
I even tried integrating the function
$$\int\frac{1}{\sqrt{{1-e^{-\sqrt{t}}}}}dt$$
but it was of no luck. I am not able to understand how to proceed with the problem.
I have been using Mathematica for awhile now, but it returns the same expression. I can get a numerical answer for it using Mathematica, but I was wondering if there exists a functional form of the Laplace transform. Is it possible to get the expression in the Laplace domain for this particular function?
I was looking up for the conditions that make the Laplace transform exist. But, the conditions are satisfied by this particular function.
We may start with the formal power series of $\frac{1}{\sqrt{1-x}}$: $$\frac{1}{\sqrt{1-x}}=\sum_{n\geq 0}\frac{1}{4^n}\binom{2n}{n}x^n$$ in order to get $$\frac{1}{\sqrt{1-e^{-\sqrt{t}}}}=\sum_{n\geq 0}\frac{1}{4^n}\binom{2n}{n}e^{-n\sqrt{t}}$$ and $$\mathcal{L}\left(\frac{1}{\sqrt{1-e^{-\sqrt{t}}}}\right)(s) = \sum_{n\geq 0}\frac{1}{4^n}\binom{2n}{n}\int_{0}^{+\infty}2u e^{-(nu+su^2)}\,du $$ which equals $$\sum_{n\geq 0}\frac{1}{4^n}\binom{2n}{n}\left( \frac{1}{s}-\frac{\sqrt{\pi } n e^{\frac{n^2}{4 s}} \text{erfc}\left(\frac{n}{2 \sqrt{s}}\right)}{2 s^{3/2}}\right)$$ with a similar behaviour on $\mathbb{R}^+$ to the one of $s^{-3/4}$.