Approximation for a series involving the derivative of a Jacobi theta function I’ve considered the diffusion equation $$\frac{\partial f(x,t)}{\partial t}=\frac12 \frac{\partial^2 f(x,t)}{\partial x^2}$$ with the conditions $f(x,0)=\delta(x)$ and $f(-1,t)=f(1,t)=0\ \forall t>0$ and I’ve found the solution $$f(x,t)=\sum_{n=0}^\infty \cos\left[ \left( n+\frac12 \right)\pi x \right] e^{-\frac12 (n+1/2)^2\pi^2 t}$$then I’ve considered the function$$\Lambda(t)=-\frac d{dt}\int_{-1}^1f(x,t)dx=\sum_{n=0}^\infty (-1)^n \left(n+\frac12\right)\pi\ e^{-(n+1/2)^2\pi^2 t/2}=\\=\frac\pi2\left(e^{-\frac{\pi^2} 8t}-3\,e^{-\frac{9\pi^2} 8t}+5\,e^{-\frac{25\pi^2} 8t}-\ldots\right)=\frac \pi 4 \vartheta_1'(0,e^{-\pi^2 t/2})$$where $\vartheta_1'(u,q)=\dfrac{\partial}{\partial u}\vartheta_1(u,q)$ and $$\vartheta_1(u,q)=2\,q^{1/4}\sum_{n=0}^\infty (-1)^n q^{n(n+1)}\sin\left[\left(2n+1\right)u \right]$$is a Jacobi theta function.
It turns out that $$\lim_{t\to0^+}\frac{\Lambda(t)}{g(t)}=1$$ where $g(t)=\sqrt{\dfrac2{\pi t^3}}e^{-\frac1{2t}}$ (so Mathematica says and the approximation is very good as the values of $\Lambda$ and $g$ differ for less than $1\%$ for $0<t<0.7$), but I don’t know how to prove it.
Thanks in advance for your help.
 A: Write your $\Lambda$ as
$$\Lambda(t)=\frac{1}{2}\sum_{-\infty}^\infty f(n),$$
where
$$f(x)=\pi(x+1/2)\sin\pi(x+1/2)\exp\left(-\pi^2(x+1/2)^2t/2\right)=y\sin y\,e^{-ty^2/2},$$
where $y=\pi(x+1/2),$ and $t>0$.
Then use Poisson's summation formula
$$\sum_{-\infty}^\infty f(n)=\sum_{-\infty}^\infty \hat{f}(2\pi n),$$
where
$$\hat{f}(s)=\int_{-\infty}^\infty f(x)e^{-isx}dx$$
is the Fourier transform. This Fourier transform can be explicitly computed:
$$\hat{f}(s)=\frac{1}{\sqrt{2\pi}}t^{-3/2}e^{is/2}\left((s/\pi+1)e^{-(s/\pi+1)^2/(2t)}-(s/\pi-1)e^{-(s/\pi-1)^2/(2t)}\right).$$
The trick is that in the Fourier transform your parameter $t$ will stand in the denominator of the exponent, so the series $\sum\hat{f}(n)$ will be asymptotic to the sum of $3$ terms (with $n=0,\pm1$), z when $t\to 0$,
$$\sum_{-\infty}^\infty\hat{f}(2\pi n)\sim \hat{f}(0)+\hat{f(2\pi)}+\hat{f}(-2\pi)\sim 2\hat{f}(0)=2\sqrt{\frac{2}{\pi t^3}}e^{-1/(2t)},\quad t\to 0+,$$
and since $\Lambda(t)$ is $1/2$ of this sum, we obtain
the answer that wrote.
(In the computation of Fourier transform, I started with the $e^{-y^2/(2t)}$, and then used the transformation rules
of Fourier transform: multiplied my function on $y$, then on $\sin y$, and then scaled by $\pi$ and added $1/2$ to the argument).
Remark. This approximation is related to the famous computation of the age of Earth by Lord Kelvin. Roughly speaking, the series corresponds to the exact solution for the heat equation inside the spherical Earth, while the asymptotics corresponds to the
flat Earth approximation. It is unclear from his papers on the subject, whether Kelvin knew the exact solution and Poisson's formula.
