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I need to prove that the following bound is true. I thought this might follow from the inversion property of the theta function, as the infinite sum in the integrand is precisely $\theta_3(0,\mathrm{e}^{-\pi t/L^2})$, but I cannot seem to prove this. Any ideas would be appreciated! The bound that I'm trying to show is:

$$\int_0^\infty t^{\frac{1}{2}} \left(\sum_{k=1}^\infty \mathrm{e}^{-\frac{\pi^2}{L^2}k^2 t} \right)^2 dt \leq \int_0^\infty t^{\frac{1}{2}} \sum_{k=1}^\infty \mathrm{e}^{-\frac{\pi^2}{L^2}k^2 t} dt,$$

where $L$ is a positive real number (in fact, evaluating the integrals for a wide range of $L$ suggests that the ratio of the LHS to the RHS is constant and does not depend on $L$ at all).

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    $\begingroup$ You can just evaluate both sides. The ratio is a constant independent of $L$, and I think it equals $(\zeta(3/2)L(3/2,\chi_{-4})-\zeta(3))/\zeta(3)$ where $L(s,\chi_{-4}) = 1/1^s-1/3^s+1/5^s-1/7^s+\ldots$ is the Dirichlet $L$-function to the character $mod 4$. $\endgroup$
    – Lucia
    Commented Mar 27, 2014 at 2:14
  • $\begingroup$ Thanks Lucia. For the LHS, is the idea to transform it to a double sum, and do something like this: $\sum_{m=1}^\infty \sum_{n=1}^\infty \int_0^\infty \mathrm{e}^{-\frac{\pi^2}{L^2}k^2t}dt = \mathrm{Consts} L \sum_{k=1}^\infty\sum_{m=1}^\infty \left(\frac{1}{k^2 + m^2}\right)^{3/2}$? If so, please forgive my ignorance but I don't see how this relates to an L function (my knowledge of these is very limited). $\endgroup$
    – HeatKernel
    Commented Mar 27, 2014 at 8:02
  • $\begingroup$ That's exactly right. Now look up the number of ways of writing integers as a sum of two squares; that's where this $L$-function comes from. You need to be a tiny bit careful because the two squares both have to be positive (ie one can't be zero), and this is why you have to subtract $\zeta(3)$ in the numerator. $\endgroup$
    – Lucia
    Commented Mar 27, 2014 at 15:29

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