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.



Approximation for a series involving the derivative of a Jacobi theta function
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](https://en.wikipedia.org/wiki/Poisson_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](https://www.math.purdue.edu/~eremenko/dvi/ageofearth.pdf) 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](https://www.math.purdue.edu/~eremenko/dvi/ageofearth2.pdf). It is unclear from his papers on the subject, whether Kelvin knew the exact solution and Poisson's formula.


