Riemann integral

For real continuous functions $f,g$ on a compact interval with $g\ge 0$, some $\alpha$ in the interval satisfies $\int fg=f(\alpha )\int g$. If $\int g>0$, the weighted average lies between the minimum and maximum of $f$, and the intermediate value theorem supplies $\alpha$. If $\int g=0$, the bound $|\int fg|\le \sup |f|\int g$ makes both sides zero. Nonnegativity of the weight is essential to this argument.

A function $g$ on $[0,\infty )$ is directly Riemann integrable when its upper and lower sums on an equal-width mesh are absolutely finite and converge to the same finite integral as the mesh width tends to zero. A locally absolutely continuous integrable function with integrable derivative has this property: the upper-minus-lower sum is bounded by mesh width times its total variation. The derivative bound therefore gives bounded variation as well as tail control. This stronger form of integrability is a hypothesis of the key renewal theorem.

An improper integral is defined by taking a limit when an integration endpoint is infinite or the integrand is unbounded. Its convergence must be established before ordinary integral manipulations are applied.

A logarithmic divergence grows like a logarithm as a cutoff is removed. For example ${\int }_{ϵ}^{1}dx/x=\log (1/ϵ)\to \infty$ as $ϵ\to 0^{+}$. An integrand $f(x)=C/x+O(1)$ near zero gives $C\log (1/ϵ)+O(1)$. This behavior distinguishes logarithmic sensitivity to an infrared or ultraviolet cutoff from power-law divergence.

The improper integral ${\int }_1^{\infty }{u}^{-p}du$ converges exactly when $p>1$, in which case its value is $1/(p-1)$.

If Riemann-integrable functions satisfy $f(x)\le g(x)$ throughout $[a,b]$, thenThis follows directly by comparing their lower and upper Darboux sums, or their Riemann sums on a common sequence of refining partitions.