Improper Integrals

Item	Statement
Type I (infinite interval)	${\displaystyle {\int }_a^{\mathrm{\infty }}f(x)dx=\underset{b\to \mathrm{\infty }}{lim}{\int }_a^{b}f(x)dx}$
Type II (unbounded integrand)	${\displaystyle {\int }_a^{b}f(x)dx=\underset{ϵ\to 0^{+}}{lim}{\int }_{a+ϵ}^{b}f(x)dx}$
Converges	the limit exists and is finite
Diverges	the limit is infinite or fails to exist
Two bad points	split the integral, and require each piece to converge on its own
$p$-test at infinity	${\displaystyle {\int }_1^{\mathrm{\infty }}\frac{dx}{x^p}}$ converges
$p$-test at a singularity	${\displaystyle {\int }_0^1\frac{dx}{x^p}}$ converges
Direct Comparison Test	$0\le f\le g$: $\int g$ converges $\Rightarrow \int f$ converges; $\int f$ diverges $\Rightarrow \int g$ diverges
Limit Comparison Test	${\displaystyle lim\frac{f}{g}=L}$ with : both converge or both diverge
Cauchy principal value	the symmetric limit ${\displaystyle \underset{\epsilon \to 0^{+}}{lim}[{\int }_a^{c-\epsilon }+{\int }_{c+\epsilon }^b]}$
Absolute convergence	\int f$ converges but $\int
The classic trap	applying the Fundamental Theorem across an interior infinite discontinuity
Reference values	${\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{dx}{1+x^2}=\pi }$, ${\displaystyle {\int }_0^{\mathrm{\infty }}{x}^n{e}^{-x}dx=n!}$, ${\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-x^2}dx=\sqrt{\pi }}$

What Makes an Integral Improper

The definite integral ${\int }_a^{b}f(x)dx$ was built on a Riemann sum over a finite interval $[a,b]$, using a function that is continuous, and therefore bounded, on that interval. Both conditions were essential. A finite interval is what lets us chop $[a,b]$ into finitely many subintervals of shrinking width, and boundedness is what keeps each rectangle's area finite.

Two familiar-looking integrals break the rules:

$${\int }_1^{\mathrm{\infty }}{e}^{-x}dx\text{and}{\int }_0^1\frac{dx}{\sqrt{x}}.$$

The first extends over an interval of infinite length. The second has an integrand that runs off to $+\mathrm{\infty }$ as $x\to 0^{+}$. Neither is a Riemann integral at all, so strictly speaking neither has been defined yet. This chapter defines them.

The definitions are not exotic. In each case we replace the offending endpoint with a movable one, compute an honest definite integral, and then let the movable endpoint approach the trouble spot:

finite integral first, limit second.

If that limit exists and is finite, the improper integral converges and the limit is its value. Otherwise it diverges. That single sentence is the whole chapter in miniature; everything else is machinery for deciding which of the two happens when the limit is hard to compute.

An Unbounded Region With Finite Area

The idea that catches most students by surprise is that a region of infinite extent can have finite area. Compare two curves that look nearly identical near $x=1$:

Both curves pass through $(1,1)$, both decrease, both approach the $x$-axis. Yet the region under $y=1/x^2$ to the right of $x=1$ has area exactly $1$, while the region under $y=1/x$ has infinite area. The only difference is the rate of decay, and the threshold that separates the two cases turns out to be sharp: the exponent $p=1$. Pinning that threshold down is the job of the $p$-test.

The same surprise happens vertically. The region under $y=1/\sqrt{x}$ over $(0,1]$ is unbounded in height but has area $2$, while the region under $y=1/x$ over the same interval has infinite area. Again the threshold is $p=1$, but now the inequality points the other way.

Sections in This Chapter

Section	Topic	What it answers
9.1	Type I: Infinite Limits of Integration	How do we integrate over $[a,\mathrm{\infty })$, $(-\mathrm{\infty },b]$, or all of $ℝ$?
9.2	Type II: Discontinuous Integrands	What if the integrand blows up at an endpoint, or worse, inside the interval?
9.3	The Cauchy Principal Value	Can a divergent integral still be assigned a sensible finite number?
9.4	The $p$-Integrals	Exactly where is the borderline between convergence and divergence?
9.5	Comparison Tests	How do we decide convergence when the antiderivative is out of reach?
9.6	Absolute and Conditional Convergence	What changes when the integrand keeps changing sign?

Read them in order. Sections 9.1 and 9.2 give the two definitions and are the only sections that are strictly required; everything after them is a tool for using those definitions well. Section 9.3 is a short aside on a notion that looks like convergence but is not. Section 9.4 supplies the reference integrals, and Section 9.5 is where the reference integrals earn their keep, since most improper integrals worth testing have no elementary antiderivative. Section 9.6 removes the last restriction, the assumption that the integrand keeps one sign.

A Strategy for Any Improper Integral

Almost every problem in this chapter yields to the same four steps.

1. Find every bad point. Scan the interval of integration for infinite endpoints and for points where $|f(x)|\to \mathrm{\infty }$, including points strictly inside the interval. Interior singularities are the ones people miss, and missing one produces answers that are not merely wrong but impossible, such as a negative value for the integral of a positive function.
2. Split so that each piece has exactly one bad point, and that bad point is an endpoint of the piece. An integral like ${\int }_0^{\mathrm{\infty }}\frac{dx}{\sqrt{x}(1+x)}$ has two, a singularity at $0$ and an infinite upper limit, so it splits into two pieces at any convenient interior point.
3. Test each piece. If an antiderivative is available, use the definition directly. If not, compare the piece with a $p$-integral or an exponential using the Direct or Limit Comparison Test. The whole integral converges only if every piece converges.
4. Evaluate, if a value is wanted. Convergence and value are separate questions, and the comparison tests answer only the first.

Where Improper Integrals Are Used

• Probability and statistics. A continuous probability density $f$ must satisfy ${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f(x)dx=1$, an improper integral by definition. The normal distribution rests on ${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-x^2}dx=\sqrt{\pi }$, and the exponential and gamma distributions on ${\int }_0^{\mathrm{\infty }}{x}^n{e}^{-x}dx=n!$. Means, variances, and moments are all improper integrals.
• The Laplace transform. ${\displaystyle F(s)={\int }_0^{\mathrm{\infty }}{e}^{-st}f(t)dt}$ converts differential equations into algebra, and its entire theory is a study of when this improper integral converges. Engineers use it daily for circuits, control systems, and mechanical vibrations.
• Fourier analysis and signal processing. The Fourier transform integrates over the whole real line, and the Dirichlet integral ${\int }_0^{\mathrm{\infty }}\frac{\sin x}{x}dx=\frac{\pi }{2}$ from Section 9.5 is the prototype of a conditionally convergent transform integral.
• Gravitation and electrostatics. The work required to move a mass from the surface of a planet to infinity, or the potential energy of a charge distribution, is ${\int }_R^{\mathrm{\infty }}\frac{k}{x^2}dx$, which converges precisely because . Escape velocity is a $p$-test in disguise.
• Geometry of unbounded solids. Gabriel's Horn, the solid obtained by revolving $y=1/x$ on $[1,\mathrm{\infty })$ about the $x$-axis, has volume $\pi$ but infinite surface area, since the volume integral has $p=2$ and the surface integral has $p=1$.
• Quantum mechanics. Normalising a wave function means requiring ${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}|\psi {|}^{2}dx=1$, and expectation values are improper integrals over all space.
• Infinite series. The integral test in the next chapter converts a series $\sum a_n$ into an improper integral ${\int }_1^{\mathrm{\infty }}f(x)dx$. The reason $\sum 1/n$ diverges while $\sum 1/n^2$ converges is exactly the $p$-test of Section 9.4.