Riemann integral

A proper Riemann integral on a compact interval requires a bounded integrand, but a derivative can be unbounded. For $f(x)=x^2\sin (1/x^2)$ away from zero and $f(0)=0$, the difference quotient gives $f^{\prime }(0)=0$, whereas $f^{\prime }(x)=2x\sin (1/x^2)-2\cos (1/x^2)/x$ for $x\ne 0$. At $x_n=(2\pi n{)}^{-1/2}$, this derivative is $-2/x_n$ and is unbounded. The fundamental theorem of calculus version that integrates a derivative therefore needs an explicit integrability hypothesis.

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.

If $f,g:[a,b]\to \mathbb{R}$ are continuous functions, $g\ge 0$, and $f$ is a nondecreasing monotonic function, there is $d\in [a,b]$ satisfying the identity in the title. It expresses the weighted integral as a mixture of the two endpoint values of $f$, with the split occurring in the accumulated weight.

Put $G={\int }_a^{b}g$ and $I={\int }_a^{b}fg$. The monotonicity of the Riemann integral gives $f(a)G\le I\le f(b)G$. The function $H(x)={\int }_a^{x}g$ is continuous; indeed, $|H(y)-H(x)|\le {\max }_{[a,b]}|g||y-x|$ by the extreme value theorem and the triangle inequality for the Riemann integral. Thereforeis continuous, with $K(a)=f(b)G$ and $K(b)=f(a)G$. The intermediate value theorem gives $K(d)=I$. If the endpoint values coincide, the bounds force $I$ to have that common value and an endpoint works. The proof covers zero total weight and a degenerate interval without division.

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.

Let $h$ be measurable, integrable over a positive period $L$, and have mean zero there. Its positive and negative parts have equal integrals $H$. On the $k$-th period the signed integral of $h(x)/x$ has absolute value at mostThis follows by bounding the positive and negative weighted integrals between $H/((k+1)L)$ and $H/(kL)$. The bounds are summable, so the integrals at period endpoints converge. A leftover partial period has absolute integral at most $\Vert h{\Vert }_{L^1(0,L)}/(kL)$, tending to zero, which proves convergence for arbitrary upper endpoints too.

In contrast, each full-period integral of $|h(x)|/x$ is at least $2H/((k+1)L)$. It diverges in total unless $H=0$, equivalently $h=0$ almost everywhere. Thus a nonzero mean-zero periodic function gives a convergent improper integral here but no Lebesgue integrability.

For $p>0$, the function $x\mapsto \sin x/x^p$ is Lebesgue integrable on $(0,\infty )$ exactly when $1<p<2$. Near zero, $\sin x/x\to 1$, so absolute integrability is equivalent to that of $x^{1-p}$, which requires $p<2$. At infinity, $p>1$ gives the upper bound $x^{-p}$. For $0<p\le 1$, on each interval $[k\pi ,(k+1)\pi ]$,and the corresponding P-series diverges.

The signed improper integral nevertheless converges for every $0<p<2$: near zero the same local comparison applies, and at infinity the Dirichlet test for improper integrals uses the bounded antiderivative of sine and the decreasing factor $x^{-p}\to 0$. Thus $0<p\le 1$ gives conditional convergence, while $1<p<2$ gives absolute integrability. In particular, substituting $t=1/x$ turns ${\int }_0^1|x^{-1}\sin (1/x)|dx$ into the divergent tail ${\int }_1^{\infty }|\sin t|/tdt$.

For a locally integrable real or complex function, the displayed tail condition is equivalent to ordinary convergence of its improper integral at infinity. Apply the Cauchy criterion for a function limit to its truncated integrals, using completeness of the real or complex numbers. The necessity follows by subtracting two nearby-to-limit truncations; sufficiency follows by first taking a convergent sequence of truncations and then bounding all other late truncations against it. The analogous criterion applies at a finite singular endpoint. Bounding the modulus of every tail by its absolute-value integral proves that an absolutely convergent improper integral converges ordinarily.

An improper integral whose absolute-value integral is finite. The triangle inequality for an integral bounds each truncated tail by the corresponding tail of $|f|$, so the Cauchy criterion for an improper integral gives ordinary convergence. Absolute convergence is stronger than a Cauchy principal value, which may exist only because two divergent sides cancel. The continuous sine-product integrand $\sin (ax)\sin (bx)/x^2$ has a removable value $ab$ at zero and an integrable $1/x^2$ tail; its auxiliary complex contour integrand can nevertheless require a principal value.

If $F(t)={\int }_a^{t}f(x)dx$ is bounded and $g$ is a nonnegative differentiable decreasing function tending to zero, then ${\int }_a^{\infty }f(x)g(x)dx$ converges as an improper integral. Integration by parts bounds the tail ${\int }_A^{B}fg$ by a constant multiple of $g(A)$, because $F$ is bounded and ${\int }_A^B|g^{\prime }|=g(A)-g(B)$. In particular ${\int }_1^{\infty }{x}^{-\alpha }\sin xdx$ converges for every $\alpha >0$, even though absolute integrability of that tail requires $\alpha >1$.

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 $1/(p-1)$.

A function has an integrable singularity at a point if it becomes unbounded there but its absolute value has a finite improper integral over a sufficiently small surrounding interval, one-sided at an endpoint. The endpoint behavior of the arcsine distribution's probability density function is an example.

An endpoint integrable singularity with $f(x)\sim C/\sqrt{x-x_0}$ as $x\downarrow x_0$, for finite nonzero $C$. It is an integrable singularity because ${\int }_{x_0}^{x_0+ϵ}(x-x_0{)}^{-1/2}dx=2\sqrt{ϵ}$. One common origin is a stationary change of variables: if $y-y_0=b{z}^2+O(z^4)$ with $b>0$ and a smooth nonzero probability density for $z$ at zero, the two inverse branches give a probability density for $y$ proportional to $(y-y_0{)}^{-1/2}$.

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.