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I'm currently reading through a paper where the relation

$\dfrac{V}{1+V} = \dfrac{2}{\sqrt{\pi t}} \displaystyle{\sum_{n=0}^{\infty} \exp(-(2n+1)^2/4t)}$

(where $V$ is a constant value) is used to deduce the asymptotic equation:

$\dfrac{1}{t} \sim 4\ln(1/V) + 2\ln(\ln(1/V)) + 2\ln(16/\pi)$,

as $V \to 0$.

The above series converges to $\dfrac{\theta_2(0,e^{-1/t})}{\sqrt{\pi t}}$, where $\theta_2$ is the Jacobi theta function, which has a factor of $2e^{-1/4t}$ in its explicit form , so I can see how terms such as $2\ln(16/\pi)$ are involved here. However I am not at all sure where the remaining terms come from, especially $2\ln(\ln(1/V))$.

I would like to know how this asymptotic equation is derived.

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  • $\begingroup$ Is that your question at the end there? It's difficult to pick out. Perhaps you could rephrase it and separate it out so that it is more clear what you want to know. $\endgroup$
    – abiessu
    May 20 at 22:46
  • $\begingroup$ I would like to know how to derive the asymptotic equation. I've now added this statement to the end of my question. $\endgroup$ May 20 at 22:49
  • $\begingroup$ Sorry ! I lost a fator $2$. Fixed and edited. $\endgroup$ May 22 at 2:29
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As you wrote, we have $$\frac V{1+V}=\frac{\vartheta _2\left(0,e^{-1/t}\right)}{\sqrt{\pi t} }$$ Now, Using $\vartheta _2\left(0,e^{-1/t}\right)\sim 2e^{-\frac{1}{4 t}}$ we have $$t\sim -\frac{1}{2 W_{-1}\left(-\frac{\pi V^2}{8 (1+V)^2}\right)}\implies \frac 1t \sim -2 W_{-1}\left(-\frac{\pi V^2}{8 (1+V)^2}\right)\tag 1$$ where $W_{-1}(.)$ is the secondary branch of Lambert function.

Since $V \ll 1$ we have $$\frac 1t \sim -2 W_{-1}\left(-\frac{\pi V^2}{8}\right)\tag2$$

Close to $a=0$, we have $$W_{-1}(a)=L_1-L_2+\frac {L_2}{L_1}+\cdots$$ where $L_1=\log(-a)$ and $L_2=\log(-L_1)$.

In the case of $(2)$ $$a=\frac{\pi V^2}{8} \implies L_1=4 \log \left(\frac{1}{V}\right)+2 \log \left(\frac{8}{\pi }\right)$$ If I did not make any mistake, I do not see where the $16$ comes from in $(3)$ $$\frac 1t \sim 4 \log \left(\frac{1}{V}\right)+2 \log \left(\log \left(\frac{1}{V}\right)\right)+2 \log \left(\frac{16}{\pi }\right)\tag 3$$

This is how we arrive to this asymptotics.

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