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Evaluate

$$ \lim_{h \to 0} h\sum_{n=0}^{\infty} e^{-n^2h^2}$$

I think it is somehow related to Riemann Sums, but I'm not sure.
Please help.

user21820
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Robin
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2 Answers2

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I think a good way is to use integral test for convergence.

This shows that for all $h > 0$, you have : $$|\sum_{n=0}^\infty e^{-n^2 h^2} - \int_0^\infty e^{-x^2 h^2} dx| \leq 1$$. Using this you can see that $$\lim_{h \to 0} h \sum_{n=0}^\infty e^{-n^2 h^2} = \lim_{h \to 0} h \int_0^\infty e^{-x^2 h^2} dx$$. This last integral is a Gaussian integral, and can be evaluated for any $h > 0$:
$$ \int_0^\infty e^{-x^2 h^2} dx = \frac{\sqrt{\pi}}{2} \frac{1}{h}$$

So finally $$\lim_{h \to 0} h \sum_{n=0}^\infty e^{-n^2 h^2} = \frac{\sqrt{\pi}}{2}$$

seamp
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If a function $f$ is Riemann-integrable on $[0,\infty)$, then its integral $\int_0^\infty f(x)\,dx$ is equal to $$\lim_{h\rightarrow 0}\,h\sum_{n=0}^\infty f(nh)$$ Now you just have to set $f$ appropriately. See this link if you can't do the resulting integration.

Edited to add: As commenters have pointed out, I have over-simplified here. What I should have said was this:

Let $f:[0,\infty)\rightarrow \Bbb R$ be a function of bounded variation, and suppose that it is Riemann-integrable over $[0,M]$ for all $M\ge 0$, and that $\lim_{M\rightarrow\infty}\int_0^M f(x)\;dx$ exists. Then this limit, which we denote by $\int_0^\infty f(x)\,dx$, is equal to $$\lim_{h\rightarrow 0}\,h\sum_{n=0}^\infty f(nh)$$

TonyK
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  • What theorem did you use? Can you give a short proof or a reference? – Robin Oct 24 '18 at 10:32
  • I used the definition of the Riemann integral. – TonyK Oct 24 '18 at 10:33
  • I only know these definitions and properties listed here. – Robin Oct 24 '18 at 10:35
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    This requires some additional hypotheses like $f$ being monotone and is not really a definition but rather a theorem with not so easy proof. – Paramanand Singh Oct 24 '18 at 23:31
  • @ParamanandSingh: I am simply choosing my partition to be $0<h<2h<\ldots N$ for some provosional limit $N$, letting $h$ tend to zero to get the Riemann integral over $[0,N]$, and then letting $N$ tend t0 infinity to get the integral over $[0,\infty)$. Very straightforward stuff. (BTW, $f$ doesn't have to be monotone. It just has to be Riemann-integrable, as I said.) – TonyK Oct 25 '18 at 01:11
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    Well an improper integral is limit of a proper integral which is a limit of a sum. This implies that an imtegral is an iterated limit of two variables and in general one can't replace that with a single variable limit unless some extra conditions are met. For the case of improper integrals with bounded domain and unbounded functions here is a proof which uses monotone nature. A similar analysis is needed here also. – Paramanand Singh Oct 25 '18 at 08:25
  • @ParamanandSingh I have posted a question based on that here. – Robin Oct 25 '18 at 14:48
  • @TonyK I have posted a question based on that here. Can you elaborate your comment as an answer? – Robin Oct 25 '18 at 14:49
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    There's a simple counterexample here: https://math.stackexchange.com/questions/2970721/a-property-of-a-riemann-integrable-function/2970767#2970767 – David C. Ullrich Oct 25 '18 at 15:19
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    "Very straightforward stuff.": But totally wrong. Your error is assuming that $\lim_{N\to\infty}\lim_{h\to0}=\lim_{h\to0}\lim_{N\to\infty}$. Limits don't automatically commute like that, in fact many theorems in analysis amount to showing that you can sometimes switch the order of two limits. – David C. Ullrich Oct 25 '18 at 15:53
  • I think you meant to use this Theorem, which doesn't hold true for all Riemann Integrable functions necessarily. – Robin Oct 25 '18 at 15:59