5
$\begingroup$

The Riemann sum over an interval $[a,b]$ is usually defined as $$\lim\limits_{N\to\infty}\sum\limits_{k=0}^Nf\left(a+k\cdot\frac{b-a}{N}\right)\frac{b-a}{N}$$

Thus if we encounter a sum of the form $$\lim\limits_{N\to\infty}\sum\limits_{k=0}^Nf\left(k\cdot\frac{1}{N}\right)\frac{1}{N}$$ we can conclude that it is equal to an integral over the interval $[0,1]$. $$\lim\limits_{N\to\infty}\sum\limits_{k=0}^Nf\left(k\cdot\frac{1}{N}\right)\frac{1}{N}=\int_0^1f(x)dx\tag{1}\label{1}$$

What can we conclude about the following sum

$$\lim\limits_{N\to\infty}\lim\limits_{M\to\infty}\sum\limits_{k=0}^M f\left(k\cdot\frac{1}{N}\right)\frac{1}{N}\tag{2}\label{2}$$

To clarify, this is an infinite sum \eqref{2}, that differs from the Riemann sum \eqref{1}, in the upper limit of the sum. In the Riemann sum \eqref{1}, there is a relation between $M$ and $N$, namely $N=M$, while there is no such relation specified in \eqref{2}. If we can equate it to an integral, how are we to determine the limits of integration?

The equation \eqref{2} is to be taken, that the $M\to\infty$, we thus have an infinite sum (suppose it is convergent). Than we form a sequence of infinite sums, where $N$ increases for each element of the sequence. That is $$S_N=\lim\limits_{M\to\infty}\sum\limits_{k=0}^M f\left(k\cdot\frac{1}{N}\right)\frac{1}{N}$$ What does this sequence tend to?

Is it true that (or when is it true) $$\lim\limits_{N\to\infty}S_N=\int_0^\infty f(x)dx$$

Also the general term in \eqref{2} is $C_k=f\left(k\cdot\frac{1}{N}\right)$. How does it behave in the limit, namely $$\lim\limits_{N\to \infty}\lim\limits_{M\to \infty}f\left(M\cdot\frac{1}{N}\right)$$

$\endgroup$
7
  • $\begingroup$ Your sum $(2)$ is not well defined for $M>N.$ $\endgroup$
    – zhw.
    Feb 4, 2017 at 20:36
  • $\begingroup$ All derivations (I encountered) of the inverse fourier transform as a limiting fourier series use a sum of this form. I have similar objections as you, but seeing how so many texts use it, I though that perhaps I don't understand something. Looking at my question here I recreate the argument from those texts, where this sum shows up. I asked a separate question to seperate it from the fourier environment. $\endgroup$ Feb 4, 2017 at 20:44
  • 1
    $\begingroup$ If $f(x)$ is defined on $[0,\infty)$ then $\sum_{k=0}^\infty f(k/N) \frac{1}{N}$ is an approximation for $\int_0^\infty f(x){\rm d}x$ with step-size $\frac{1}{N}$. $\endgroup$
    – Winther
    Feb 4, 2017 at 20:44
  • $\begingroup$ Could you please derive this for me? It would also resolve my unanswered question here. I have been thinking about this for few days, and I would really be grateful if you can derive this result for me. $\endgroup$ Feb 4, 2017 at 20:46
  • $\begingroup$ So when we take the limit as N--> infinity, the approximation becomes the integral? I would really be grateful for a derivation and justification of the statement. $\endgroup$ Feb 4, 2017 at 20:49

1 Answer 1

4
$\begingroup$

Suppose $f$ is Riemann integrable on $[0,b]$ for every $b > 0$ and the improper integral over $[0, \infty)$ is convergent.

We first consider the case where $f$ is nonnegative and non-increasing, as suggested by @Winther, where we have

$$\frac{f((k+1)/N)}{N} \leqslant \int_{k/N}^{(k+1)/N} f(x) \, dx \leqslant \frac{f(k/N)}{N}. $$

This implies

$$\int_0^{(M+1)/N} f(x) \, dx \leqslant \frac{1}{N} \sum_{k=0}^{M} f(k/N) \leqslant \frac{f(0)}{N} + \int_0^{M/N} f(x) \, dx. $$

The sequence of partial sums is increasing and bounded , hence convergent as $M \to \infty$, with

$$\int_0^{\infty} f(x) \, dx \leqslant\frac{1}{N} \sum_{k=0}^{\infty} f(k/N) \leqslant \frac{f(0)}{N} + \int_0^{\infty} f(x) \, dx. $$

Therefore,

$$\lim_{N \to \infty} \lim_{ M \to \infty}\frac{1}{N} \sum_{k=0}^{M} f(k/N) = \int_0^\infty f(x) \, dx.$$

Can this still hold if $f$ is not monotonic?

For example, consider $f(x) = \sin x /x$, where

$$\int_0^\infty \frac{\sin x}{x} \, dx = \frac{\pi}{2}.$$

Examining the corresponding series (WLOG starting with $k=1$) we find

$$\frac{1}{N}\sum_{k = 1}^{\infty} \frac{\sin (k/N)}{k/N} = \sum_{k = 1}^{\infty} \frac{\sin (k/N)}{k} \\ = \frac{\pi}{2}-\frac{1}{2N} \\ \longrightarrow_{N \to \infty} \frac{\pi}{2}.$$

I have not yet found a counterexample for a non-monotone function. As the integral test can be generalized to $C^1$ functions of bounded variation, I suspect this may characterize a wider class of functions for which this result holds.

This could be shown by considering

$$\left|\int_{k/N}^{(k+1)/N} f(x) \, dx - \frac{f(k/N)}{N} \right| \leqslant \int_{k/N}^{(k+1)/N} |f(x) - f(k/N)| \, dx $$

and then summing over $k$, applying the mean value theorem when $f$ is differentiable, and using $\int_0^\infty |f'(x)|\, dx < \infty$ to show that the sum converges to the integral as $N \to \infty.$

$\endgroup$
3
  • $\begingroup$ Very nice answer, thank you. If I may ask, does the order you specify the limits in matter? First M-->\infty, than N-->\infty ? The question relates to k/N. Becase k-->M, and k/N-->\infty (because the count goes to infinity, hence the upper limit of integration),this implies that k "overgrows" N, hence M does too. I hope my wording is understandable. $\endgroup$ Feb 5, 2017 at 9:13
  • $\begingroup$ You're welcome. The order does matter. Here we take the limit as $M \to \infty$ first followed by the limit as $N \to \infty$. If we switch, then with $M$ fixed we have $\frac{1}{N} \sum_{k=0}^M f(k/N) \to 0$ as $N \to \infty$ if for example $f$ is continuous and $f(0)$ is finite. $\endgroup$
    – RRL
    Feb 5, 2017 at 9:26
  • $\begingroup$ This was my hypothesis, and the original reason for thinking about this question. Thank you for helping me understand. $\endgroup$ Feb 5, 2017 at 9:31

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .