Riemann-like sum for infinite integrals Under what conditions does $\displaystyle\lim_{n \to \infty} \frac{1}{n} \sum_{k=0}^{\infty} f \left(\frac{k}{n} \right) = \int_{0}^{\infty} f(x) \ dx ? $
I think I read somewhere that it's only true for monotonic functions (assuming that the integral converges, of course). But it seems to be true for other functions as well. And do we refer to this as a Riemann sum?
 A: This may not be too useful but it is something. Assume that $f$ is continuous. Define $g_n(m) = \frac{1}{n}\sum_{k=0}^{n-1} f(\frac{k}{n} + m)$. We have the following equalities:
$
\int_0^\infty f = \sum_{m=0}^\infty \int_m^{m+1}f = \sum_{m=0}^\infty\left(\lim_{n \to \infty} \frac{1}{n} \sum_{k=0}^{n-1} f(\frac{k}{n} + m)\right) = \sum_{m = 0}^\infty \lim_{n \to \infty} g_n(m)
$
If we could exchange the limit and sum we would have
$
\int_0^\infty f = \sum_{m = 0}^\infty \lim_{n \to \infty} g_n(m) = \lim_{n \to \infty} \sum_{m = 0}^\infty g_n(m) = \lim_{n \to \infty} \frac{1}{n} \sum_{m = 0}^\infty \sum_{k=0}^{n-1} f(\frac{k}{n} + m) = \lim_{n \to \infty} \frac{1}{n} \sum_{k=0}^{\infty} f(\frac{k}{n})
$
We wish to know under what circumstances we can exchange the limit and the sum. By Lebesgue's Dominated Convergence Theorem this is possible when $|g_n| \leq h$ for all $n$ for some integrable function $h$, i.e. when we can find $h(m)$ with $\int_0^\infty |h| < \infty$ and $h (m) \geq \frac{1}{n} \sum_{k=0}^{n-1} f(\frac{k}{n} + m)$ for all $n$.
