Let be periodic (say, 1-periodic) and smooth. Our goal is to approximate . Here are the first two ways that come to mind:

• Use a left (or right) Riemann sum:
  
• Use a trapezoidal Riemann sum:
  

How good is each approximation as ? In other words, if is the approximation and is the integral, how does decay? What is the largest exponent such that for some constant (that is allowed to depend on )? Are there any schemes that induce an even faster convergence?

Try to guess the answer before moving on.

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Ok time's up.

Theorem: Let be smooth and 1-periodic. Then
for every !


So you're not going to do better than the classic Riemann sum. I leave as an exercise to show that "any scheme" will also achieve this extremely rapid decay.

(By the way, did you notice that left, right, and trapezoidal Riemann sums are all completely identical for periodic functions?)

Proof. By periodicity we may consider the Fourier series
By smoothness the above relationship is a "true" pointwise equality, holding for every . Now take the left Riemann sum of both sides to find
where the interchange is justified (why?). But observe that
hence
Since is smooth, the Fourier coefficients decay rapidly; say, for as large as we wish. Then the error is bounded as
for as large as we wish. .

We can also try this trick for not necessarily periodic. If is integrable and continuous then we may safely apply Poisson summation to the function to find
(Recall that if then .) The LHS is a Riemann sum using rectangles of width and the error is then given by , which is smaller the more regular is. In particular if then we can expect the error to be of order .