Proof of Ramanujan doubly exponential series identity

Apologies if this has been asked here already. On page 7 of this paper the following formula due to Ramanujan is presented:

$$\alpha \sum_{k=0}^\infty e^{n e^{k\alpha}}=\alpha\left(\frac{1}{2}+\sum_{k=1}^\infty\frac{(-1)^{k-1}n^k}{k!(e^{k\alpha}-1)}\right)-\gamma-\ln{n}+2\sum_{k=1}^\infty\varphi(k\beta)$$

where:

$$\varphi(\beta)=\frac{1}{\beta}\Im\left(n^{-i\beta}\Gamma(i\beta+1)\right)$$

and $\alpha, n>0$, for any $\beta>0$ such that $\alpha\beta=2\pi$. No proof is presented there although it is mentioned that Poisson's formula can be used, with the note that the proof is 'intricate' (although it seems to imply that Ramanujan used properties of the theta function to prove it). The theorem is also mentioned in this answer, which mentions that the proof can be found here, but like the author of that answer I do not have access to this paper. I have tried to prove this myself, but I have not had luck using the Poisson summation formula directly since I do not know how to find the Fourier transform of $e^{a e^{bt}}$ (although using functions like $e^{t}e^{-e^t}$ I was able to formally find some vaguely similar summation identities involving imaginary parts of gamma functions). I was wondering whether anyone knows how to prove this formula?

• possibly you have to use the Ramanujan master theorem. And the Laplace transform of $e^{-e^u}$ is the Gamma function – reuns Nov 27 '16 at 4:20
• @user1952009 Possibly. I am not sure how the Laplace transform would help though. – Anon Dec 1 '16 at 2:52
• You asked for the Fourier transform... You didn't know it was the same as the Laplace transform ? – reuns Dec 1 '16 at 7:24
• @user1952009 I know that it's very similar, but I thought that it wouldn't be so simple here for $e^{-e^u}$ because of the way that function behaves at $u\rightarrow-\infty$. Wolfram Alpha didn't help me here. – Anon Dec 9 '16 at 22:46