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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.

Cornelius
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1 Answers1

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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.

Claude Leibovici
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