Riemann sum

For a tagged partition $a=x_0<\cdots <x_n=b$ with $t_i\in [x_{i-1},x_i]$, the corresponding Riemann sum isIts limits over increasingly fine partitions define the Riemann integral.

A right Riemann sum on a partition $a=x_0<\cdots <x_N=b$ samples each subinterval at its right endpoint. For a continuous function its limit as the mesh tends to zero is the integral, by the Riemann sum convergence theorem. On a uniform partition of width $h$, comparison of endpoint weights in the composite trapezoidal rule gives $R_N-T_N=h(f(b)-f(a))/2$. Thus $N^{-5}{\sum }_{j=1}^N{j}^4$ is a right Riemann sum for ${\int }_0^1{x}^{4}dx$, and its limit is $1/5$.

For a bounded function on a fixed closed interval, Riemann integrability with Riemann integral $I$ is equivalent to convergence of all tagged Riemann sums to $I$ as the mesh of a partition tends to zero. Precisely, for every $\epsilon >0$ there is $\delta >0$ such that every partition of an interval of mesh less than $\delta$, with any tags in the corresponding closed subintervals, has its Riemann sum within $\epsilon$ of $I$.

On the geometric progression $x_i=(1+ϵ{)}^i$, with $ϵ>0$, the interval widths are $ϵ(1+ϵ{)}^i$. The left-endpoint Riemann sum for the decreasing integrand $x^{-1-\alpha }$ is the geometric seriesThe right-endpoint sum is $L_{ϵ}=(1+ϵ{)}^{-1-\alpha }{U}_{ϵ}$. Integral comparison on each interval gives $L_{ϵ}\le I\le U_{ϵ}$. Both bounds tend to $1/\alpha$ as $ϵ\to 0^{+}$, proving the improper integral by the squeeze theorem. Although the largest interval width is unbounded on the infinite domain, these explicit upper and lower bounds justify the limit without incorrectly applying a finite-interval mesh argument to the entire half-line.