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I'm wondering that $\int_{x \in\mathbb{R}}e^{-\pi \frac{x^2}{m^2}} \ \mathrm{d}x $ is always less than $\sum_{x\in \mathbb{Z}} e^{-\pi \frac{x^2}{m^2}}$ for all real number $m > 0 $.

I know that the former is the Gaussian integral and $\int_{x \in\mathbb{R}}e^{-\pi \frac{x^2}{m^2}} \ \mathrm{d}x= m$, but the latter is the discrete gaussian summation on integer. I'm not sure if it converges to some function of $m$ or lager than $m$.

I tend to believe the former is less than the latter, for i test a set of value of $m$ in $Maple$ and found that the gap get smaller when $m$ grows.

besides, i found in wikipedia $\sum_{x\in \mathbb{Z}} e^{-\pi {x^2}} = \frac{\pi^{1/4}}{\Gamma{(3/4)}} $https://en.wikipedia.org/wiki/List_of_mathematical_series, but i could not find the proof.

Thanks!

Dai xiaokang
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1 Answers1

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Hint: Suppose $f$ is positive and strictly decreasing on $[a,b],$ with $0<a<b.$ Then

$$\int_a^b f(x)\,dx < f(a)(b-a).$$

zhw.
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  • I know what you mean, but the rectangle cannot always be above the curve, otherwise, the height of the rectangle would not be equal to the value of the function corresponding to the midpoint of the rectangle – Dai xiaokang Sep 15 '23 at 16:55
  • "Always above" is not necessary. What we have here is $f<f(a)$ on $(a,b],$ and that's enough. – zhw. Sep 15 '23 at 18:40
  • If so, then you need two rectangles of height $f(0)$, but the summation only have one. Maybe I misunderstood it, but I would appreciate it if you could write the detailed proof. – Dai xiaokang Sep 16 '23 at 08:51