As my previous post remains unanswered, I thought I would post a more complete form of the problem in case it would be more practical to work on/ solve. I am trying to compute the following integral \begin{align} I_{\pm\pm} (x,y) = \int_0^1 dz \int_0^\infty ds \, \frac{e^{-\alpha/s}}{s} \,\partial_s \left[ \theta_3 \left( \frac{\pi(x \pm z)}{2},e^{-s} \right) \,\theta_3 \left( \frac{\pi(y \pm z)}{2},e^{-s} \right) \right] \end{align} where $x,y \in (0,1)$ and $\alpha > 0$. To give a bit of context, I have come across this in a setting of heat propagation with Neumann boundary conditions and it does not seem to be something that can be calculated analytically. I have indeed looked at the numerical results for various values of $x$ and $y$, but I still like to have an expression that approximates $I(x,y)$ (or various expression for different regimes). Any suggestions are therefore appreciated.
My own (hopeless) attempt is based on approximating the $\theta_3$ functions first, for exmaple using Abel's summation formula following this post. Let's define \begin{align} f (z,s) = \sum_{n=0}^\infty e^{-n^2 \pi^2 s} \cos (n \pi z) \end{align} which in terms of Jacobi theta function could be represented as $ f (z,s) = ( \theta_3 ( \pi z/2 , e^{-\pi^2 s}) + 1 )/2. $ To use Abel's formula, we can define $ \phi (h) = e^{-h^2 \pi^2 s} \cos(h \pi z) $ and write \begin{align} \sum_{n=0}^\infty \phi(n) = \lim_{h \to \infty} \left( \lfloor h+1 \rfloor \phi(h) \right) - \int_0^\infty \lfloor h+1 \rfloor \phi'(h) \, dh \end{align} The first term on the r.h.s vanishes because of the exponential factor. We have \begin{align} \phi'(h) = -\pi e^{-h^2 \pi^2 s} \left( 2\pi h s \cos(h\pi z) + z \sin(h\pi z) \right) \end{align} and $\lfloor h+1 \rfloor = h+1 -\{h+1\} = 1 + h - \{h\}$. Moreover, we can use the integral results \begin{align} \int_0^\infty dh \, 2 h \pi^2 s \, e^{-h^2 \pi^2 s} \, h \, \cos(h \pi z) &= 1 - \frac{z}{\sqrt{s}} D_+(\frac{z}{2\sqrt{s}}) \\ % \int_0^\infty dh \, 2 h^2 \pi^2 s \, e^{-h^2 \pi^2 s} \, \cos(h \pi z) &= \frac{ e^{-\frac{z^2}{4s}}}{\sqrt{4\pi s}} \left( 1 - \frac{z^2}{2 s} \right) \\ % \int_0^\infty dh \, z \pi \, e^{-h^2 \pi^2 s} \, \sin(h \pi z) &= \frac{z}{\sqrt{s}} D_+(\frac{z}{2\sqrt{s}})\\ % \int_0^\infty dh \, z \pi h \, e^{-h^2 \pi^2 s} \, \sin(h \pi z) &= \frac{ e^{-\frac{z^2}{4s}}}{\sqrt{4\pi s}} \, \frac{z^2}{2 s} \end{align} where $D_+(z)$ is the Dawson function. Then we'll have (ps: possibly easier to get this just using integrations by part...) \begin{align} \sum_{n=0}^\infty \phi(n) = 1 + \frac{e^{-\frac{z^2}{4s}}}{\sqrt{4\pi s}} + \int_0^\infty \{h\} \phi'(h) dh \end{align} I have no idea how to estimate/find bounds of the remaining integral. In case that integral is something that could be neglected (I doubt), then we can just take the rest of the r.h.s and plug it into the expression for $I$, compute the $s$ derivative and perform the integrations (which I think will give some error functions etc).
Update We can work out the following bounds: as $ 0 \leq \{h\} <1 $ we have \begin{align} \int_0^\infty \{h\} \phi'(h) \, dh &\leq \int_0^\infty \{h\} |\phi'(h)| \, dh \\ & \leq \int_0^\infty |\phi'(h)| dh \\ &\leq \int_0^\infty \pi e^{-h^2 \pi^2 s}\sqrt{4\pi^2h^2s^2+z^2} dh\\ &= \frac{ z^2 e^{\frac{z^2}{8s}}}{8s} \left( K_0(\frac{z^2}{8s}) + K_1 (\frac{z^2}{8s} ) \right), \end{align} where $K_\alpha$ is the modified Bessel function of the second kind. I am not sure if this bound is actually useful.
