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The limit of infinite sum to the integral is considered usually on the compact interval. However, I am curious about unbounded cases.

Let $f(x)=e^{-x^2}$, which is a very good function. Then, does \begin{equation} \frac{1}{2^n} \sum_{k=1}^{n\cdot 2^n} e^{-k^2/4^n} \to \int_0^\infty e^{-x^2}dx \end{equation} hold as $n \to \infty$?

I tried to show this but it seems trickier than expected since the domain of integration is not bounded.. Could anyone please help me?

If the above limit does not hold, how should I set the partition in order to have a sum that converges to $\int_0^\infty e^{-x^2}dx$?

Keith
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    Infinity cannot be divided into partitions by one limit as you describe. That's why the definition of improper integrals $$\int_a^\infty f(x)dx \equiv \lim_{b\to\infty}\int_a^bf(x)dx$$ exists. You would need to approach the problem that way, as a nested limit. – Ninad Munshi Feb 03 '22 at 00:50
  • Yes, but the given function is known to have the limit you stated. So, can't I choose just any partition? – Keith Feb 03 '22 at 00:59
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    No , a partition of an infinite interval isn't defined : the definition of Riemann integration is restricted to finite intervals. Therefore, you need to ensure that you first work on finite intervals, and then use the definition of the infinite integral as the limit (as mentioned in the comment above), to make the move towards $\infty$. An example of this is shown here. I believe there is a reason why one doesn't use partitions to define the infinite integral as well, but I'll have to look that up. – Sarvesh Ravichandran Iyer Feb 03 '22 at 01:08
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    You may also like this post, where one can convert Riemann-type sums to integrals under certain conditions. It doesn't translate verbatim to your situation, but it's still useful to know that the direction taken by you was potentially fruitful. – Sarvesh Ravichandran Iyer Feb 03 '22 at 01:16

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Since $t\mapsto e^{ - t^2 /4^n } $ is monotonically decreasing for $t\geq 1$ and any fixed $n\geq 1$, we have $$ \frac{1}{{2^n }}\int_1^{n2^n } {e^{ - t^2 /4^n } dt} \le \frac{1}{{2^n }}\sum\limits_{k = 1}^{n2^n } {e^{ - k^2 /4^n } } \le \frac{{e^{ - 1/4^n } }}{{2^{n} }} + \frac{1}{{2^n }}\int_1^{n2^n } {e^{ - t^2 /4^n } dt} . $$ Noting that $$ \frac{1}{{2^n }}\int_1^{n2^n } {e^{ - t^2 /4^n } dt} = \int_{2^{ - n} }^n {e^{ - x^2 } dx} $$ and using the squeeze theorem, yields the required result.

Gary
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