Consider the following function :
$$G(x) = \sum_{n=1}^\infty (e^{x/n²}-1)$$
I know that the sum converges .
Also ,$$G(x) =(1/2π) \int_{-π}^{π} f(xe^{-it})e^{e^{it}} dt$$
Where, $f(x)= π\sqrt{x}cot(π\sqrt{x}) -1$
Question : Can a closed form possible for this function (other than mentioned above)?
So writing $$ e^{\frac{x}{n^2}} -1 = \sum_{k=1}^\infty \frac{(x/n^2)^k}{k!} $$ we can write $$ G(x) = \sum_{k=1}^\infty \sum_{n=1}^\infty \frac{x^k}{n^{2k}k!}=\sum_{k=1}^\infty \frac{\zeta(2k)x^k}{k!} $$
Just some deeper thoughts here will update as I go. Consider $$ \zeta(2k) = \frac{1}{\Gamma(2k)}\int_0^\infty \frac{t^{2k-1}}{e^t-1} \; dt $$ then $$ G(x) = \sum_{k=1}^\infty \frac{x^k}{\Gamma(k+1)\Gamma(2k)}\int_0^\infty \frac{t^{2k-1}}{e^t-1} \; dx $$ and possibly by exchanging the integral and sum $$ G(x) = \int_0^\infty t x \frac{_0F_2(;\frac{3}{2},2;\frac{t^2 x}{4})}{e^t-1} \;dt $$
Idea 2
According to OEIS A002432 $$ \zeta(2n) = (-1)^{n-1} B_{2n} 2^{2n-1} \frac{\pi^{2n}}{(2n)!} $$ so $$ G(x) = \sum_{n=1}^\infty (-1)^{n-1} B_{2n} 2^{2n-1} \frac{\pi^{2n} x^n}{(2n)! n!} $$ $$ H(x) = \frac{G(x)}{2 \pi^2 x} = \sum_{k=0}^\infty \frac{(-1)^{k}}{k!} B_{2k+2} \frac{(2\pi)^{2k} x^{k}}{\Gamma(2k+3)(k+1)} $$ then according to Ramanujan Master Theorem the Mellin transform of $H(x)$ can be written $$ \mathcal{M}[H](s) = \Gamma(s)B_{2-2s} \frac{(2\pi)^{-2s} }{\Gamma(3-2s)(1-s)} $$ my next thoughts would be to substitute the analytic continuation for the Bernoulli numbers and see if an inverse Mellin transform or sum over residues makes any sense here to get back to the function...