Unit 02 · Analysis 1B

Improper Integrals

Integrals over infinite intervals, or of functions that blow up, defined carefully as limits. The recurring question is always the same: does the limit exist? The tests here are a direct rehearsal for the series tests two units from now.

Why it matters & where it comes from

Why it exists

Many vital integrals run to infinity or have a singularity — probability distributions, decaying signals, gravitational potentials. Improper integrals give them a rigorous meaning by taking a limit of ordinary, proper integrals.

In the real world

The Gaussian integral $\int_{-\infty}^{\infty}e^{-x^2}\,dx=\sqrt{\pi}$ normalizes the bell curve; Laplace and Fourier transforms over infinite domains power signal processing and control engineering; the exponential waiting-time distribution in queueing theory is defined by one.

Story & open questions

Torricelli's "Gabriel's Horn" (1640s) has finite volume but infinite surface area — fillable with a finite amount of paint, yet impossible to paint. A vivid early warning that infinity defies intuition, and a direct ancestor of the $p$-test below.

By the end you can

Evaluate type I (infinite interval) and type II (singular) improper integrals as limits.
Apply the $p$-test to both.
Use direct and limit comparison to decide convergence without evaluating.
Explain the difference between convergence and absolute convergence for improper integrals.

Definitions and the $p$-test

Definition

$\displaystyle\int_a^\infty f=\lim_{t\to\infty}\int_a^t f$. If $f$ is unbounded near $b$, $\displaystyle\int_a^b f=\lim_{t\to b^-}\int_a^t f$. The integral converges exactly when this limit exists (as a finite number); otherwise it diverges. An integral improper at both ends is split at any interior point, and both pieces must converge.

$p$-test

$\displaystyle\int_1^\infty x^{-p}\,dx$ converges iff $p>1$; $\displaystyle\int_0^1 x^{-p}\,dx$ converges iff $p<1$. The two cases are mirror images: near infinity, large $p$ decays fast enough; near $0$, only small $p$ keeps the blow-up mild enough.

Proof of the type I case

For $p\ne1$: $\displaystyle\int_1^t x^{-p}dx= \dfrac{t^{1-p}-1}{1-p}$. As $t\to\infty$, $t^{1-p}\to0$ iff $1-p<0$, i.e. $p>1$.

For $p=1$: $\int_1^t x^{-1}dx=\ln t\to\infty$, so it diverges — the boundary case fails.

Comparison tests

Direct comparison

If $0\le f\le g$ on $[a,\infty)$ and $\int_a^\infty g$ converges, so does $\int_a^\infty f$; if $\int_a^\infty f$ diverges, so does $\int_a^\infty g$. (Squeeze a nonnegative function between $0$ and something integrable, or below something that blows up.)

Limit comparison

If $f,g\ge0$ and $\displaystyle\lim_{x\to\infty}\dfrac{f(x)}{g(x)}=L$ with $0<L<\infty$, then $\int_a^\infty f$ and $\int_a^\infty g$ either both converge or both diverge. Useful when direct comparison is awkward to set up.

Example Direct comparison

$\int_1^\infty\dfrac{dx}{x^2+1}$ converges: $0\le\dfrac1{x^2+1}\le \dfrac1{x^2}$ and $\int_1^\infty x^{-2}dx$ converges ($p=2>1$).

Example Limit comparison

$\int_1^\infty\dfrac{x}{\sqrt{x^5+1}}\,dx$: compare with $g(x)=x^{-3/2}$.

$\dfrac{f(x)}{g(x)}=\dfrac{x\cdot x^{3/2}}{\sqrt{x^5+1}}= \dfrac{x^{5/2}}{\sqrt{x^5+1}}\to1$ as $x\to\infty$.

Since $\int_1^\infty x^{-3/2}dx$ converges ($p=3/2>1$), so does the original.

Absolute convergence If $\int_a^\infty|f|$ converges, then $\int_a^\infty f$ converges too — this lets comparison tests (which need $f\ge0$) apply indirectly to signed functions, exactly as with series later. The converse can fail: an integral can converge only conditionally.

Common pitfalls

• Always rewrite as a limit; never substitute $\infty$ or the singular point directly into the antiderivative.
• Watch for a hidden singularity inside the interval, not just at the endpoints — split the integral there.
• Comparison (direct or limit) needs eventually nonnegative functions.
• Limit comparison requires the limit to be a positive finite number — $L=0$ or $L=\infty$ only give one-directional information.

Practice

Decide convergence before you try to evaluate — often you don't need the antiderivative at all.

Practice 1

$\int_0^\infty e^{-x}\,dx$.

Answer

$[-e^{-x}]_0^\infty=1$.

Practice 2

Does $\int_1^\infty\dfrac{dx}{x}$ converge?

Answer

No; $\ln t\to\infty$ ($p=1$, the boundary case).

Practice 3

Converge or diverge: $\int_1^\infty\dfrac{2+\sin x}{x^2}\,dx$?

Answer

$0\le \tfrac{2+\sin x}{x^2}\le\tfrac{3}{x^2}$, and $\int x^{-2}dx$ converges, so it converges by direct comparison.

Practice 4 · Type II

Does $\int_0^1\dfrac{dx}{\sqrt x}$ converge? Evaluate it.

Solution

$p=1/2<1$, converges. $\lim_{t\to0^+}\int_t^1 x^{-1/2}dx=\lim_{t\to0^+}(2-2\sqrt t)=2$.

Practice 5 · Both ends

Does $\int_0^\infty\dfrac{dx}{\sqrt x(1+x)}$ converge?

Solution

Split at $1$. Near $0$, behaves like $x^{-1/2}$ ($p=1/2<1$, converges). Near $\infty$, behaves like $x^{-3/2}$ ($p=3/2>1$, converges). Both pieces converge, so the whole integral converges.

Practice 6 · Limit comparison

Does $\int_2^\infty\dfrac{\ln x}{x^2}\,dx$ converge?

Hint

For any $\varepsilon>0$, eventually $\ln x\le x^\varepsilon$, so compare with $x^{-2+\varepsilon}$ for small $\varepsilon$; it converges (and can also be evaluated directly by parts).

Practice 7 · Absolute convergence

Show $\int_1^\infty\dfrac{\cos x}{x^2}\,dx$ converges.

Answer

$\left|\dfrac{\cos x} {x^2}\right|\le\dfrac{1}{x^2}$, so the integral converges absolutely, hence converges.

Practice 8 · Gabriel's Horn

The horn formed by rotating $y=1/x$ for $x\ge1$ has volume $\int_1^\infty\pi/x^2\,dx$ and surface area at least $\int_1^\infty2\pi/x\,dx$. Show the volume is finite but the surface area is infinite.

Answer

Volume: $p=2>1$, converges to $\pi$. Surface area: bounded below by $\int 2\pi/x\,dx$, which is the $p=1$ boundary case and diverges — so the true surface area is infinite.

Expanded topic checklist

Course coverage. This unit is part of the complete MAT159H1 topic map. It should cover the following items before you move on:

Infinite intervals. Rewrite integrals over unbounded intervals as limits.
Singular endpoints. Split intervals at vertical asymptotes and take one-sided limits.
Comparison. Use direct and limit comparison with p-integrals, exponentials, and rational functions.
Absolute convergence. Compare $\int f$ and $\int |f|$ for sign-changing functions.

Textbook/source alignment

Use the current UofT MAT159H1 calendar topics as the coverage spine, and pair lecture notes with a Spivak-style proof-calculus text for extra reading. The exact assigned textbook should still be checked against the instructor syllabus.

Core theorem map

The improper-integral spine is: locate every improper point, split the integral there, and require every resulting limit to converge.

Study routine. Most mistakes come from evaluating an antiderivative across a singularity without splitting; make the split visible in your work.

Deep practice extensions

Use these after the main practice set. They are designed to force definition-level reasoning, theorem selection, and mixed application rather than routine calculation only.

Extension 1 · p-test

Decide convergence of $\int_2^\infty dx/(x(\ln x)^2)$.

Guide

Use $u=\ln x$; the integral becomes $\int_{\ln2}^\infty u^{-2}du$, which converges.

Extension 2 · Limit comparison

Determine convergence of $\int_1^\infty (3x^2+1)/(x^5+x)dx$.

Guide

The integrand behaves like $3/x^3$, and $\int_1^\infty x^{-3}dx$ converges, so limit comparison gives convergence.

Extension 3 · Singularity

Analyze $\int_0^1 dx/x$.

Guide

It equals $\lim_{a\to0^+}\int_a^1 dx/x=\lim_{a\to0^+}-\ln a=\infty$, so it diverges.

Extension 4 · Absolute value

If $\int |f|$ converges, what can you conclude about $\int f$?

Guide

It converges absolutely, hence converges. Proof uses Cauchy criterion or comparison of positive and negative parts.

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