We know that Riemann sum gives us the following formula for a function $f\in C^1$:
$$\lim_{n\to \infty}\frac 1n\sum_{k=0}^n f\left(\frac kn\right)=\int_0^1f(x) dx.$$
I am looking for an example where the exact calculation of $\int f$ would be interesting with a Riemann sum.
We usually use integrals to calculate a Riemann sum, but I am interesting in the other direction.
Edit.
I actually found an example of my own today. You can compute
$$I(\rho)=\int_0^\pi \log(1-2\rho \cos \theta+\rho^2)\mathrm d \theta$$
using Riemann sums.
Best Answer
When definite integrals are amenable to exact valuation, it is typically the case that the more expedient approach involves an anti-derivative rather than the limit of a Riemann sum. Often computation of the limit may be straightforward or even trivial, but somewhat tedious, as is the case for integrals of $f: x \mapsto x$ or $f: x \mapsto x^2$.
On the other hand, integrals with simple integrands and easily recognized anti-derivatives such as $f: x\mapsto x^{-2}$ are more challenging with regard to the limit of Riemann sum -- and in that sense the Riemann sum may be "interesting."
To make this more explicit, consider computing the integral
$$\int_a^b x^{-2} \, dx = \lim_{n \to \infty}S_n$$
where
$$S_n =\frac{b-a}{n}\sum_{k=1}^n \left(a + \frac{b-a}{n}k\right)^{-2}.$$
We have
$$\frac{b-a}{n}\sum_{k=1}^n \left(a + \frac{b-a}{n}k\right)^{-1}\left(a + \frac{b-a}{n}(k+1)\right)^{-1} \leqslant S_n \\ \leqslant \frac{b-a}{n}\sum_{k=1}^n \left(a + \frac{b-a}{n}k\right)^{-1}\left(a + \frac{b-a}{n}(k-1)\right)^{-1},$$
and decomposing into partial fractions,
$$\sum_{k=1}^n \left\{\left(a + \frac{b-a}{n}k\right)^{-1}-\left(a + \frac{b-a}{n}(k+1)\right)^{-1}\right\} \leqslant S_n \\\leqslant \sum_{k=1}^n \left\{\left(a + \frac{b-a}{n}(k-1)\right)^{-1}-\left(a + \frac{b-a}{n}k\right)^{-1}\right\}. $$
Since the sums are telescoping, we have
$$\left(a + \frac{b-a}{n}\right)^{-1}-\left(a + \frac{b-a}{n}(n+1)\right)^{-1} \leqslant S_n \leqslant a^{-1} - b^{-1}.$$
By the squeeze theorem, we get the value of the integral as
$$\lim_{n \to \infty}S_n = a^{-1} - b^{-1}.$$
An example where I found the Riemann sum an interesting and, perhaps, most expedient approach is:
Bronstein Integral 21.42