Unit 12 · Term 2

Improper Integrals

Integrals over an infinite interval, or of a function that blows up, are defined as limits of ordinary integrals. The question is always the same: does that limit exist?

By the end you can

Evaluate type I (infinite interval) and type II (unbounded integrand) improper integrals.
Apply the $p$-test.
Use the comparison test for convergence.

Why it matters & where it comes from

Why it exists

Many of the most important integrals run to infinity or blow up somewhere — distributions over all of space, decaying signals, probabilities. Improper integrals define them carefully as limits.

In the real world

The Gaussian integral $\int_{-\infty}^{\infty} e^{-x^2}\,dx=\sqrt{\pi}$ is what normalizes the bell curve. Laplace and Fourier transforms — over infinite domains — are the backbone of signal processing and control engineering. Expected values in probability are improper integrals.

Story & open questions

Torricelli's "Gabriel's Horn" (1640s) has finite volume but infinite surface area — you could fill it with paint but never paint it. The paradox shocked mathematicians and is a perfect demonstration that infinity does not behave the way intuition expects.

Improper integral workflow

An improper integral is not an ordinary integral with a strange endpoint. It is a limit. The first step is always to identify every place where ordinary Riemann integration fails.

Checklist

Is an endpoint infinite? Replace it with a finite variable and take a limit.

Does the integrand blow up at an endpoint? Approach that endpoint from inside the interval.

Does the integrand blow up in the middle? Split the integral there, and require both pieces to converge.

Only after the correct limits are written should you compute or compare.

Comparison strategy

When an antiderivative is hard or impossible, compare the integrand to a known benchmark. The benchmark is usually a $p$-integral, an exponential, or a simpler rational function.

To prove convergence of $f$, find a larger convergent $g$ with $0\le f\le g$.
To prove divergence of $g$, find a smaller divergent $f$ with $0\le f\le g$.
For large $x$, compare leading powers; near $0$, compare the most singular power.

Type I — infinite intervals

Definition

$\displaystyle\int_a^\infty f=\lim_{t\to\infty}\int_a^t f$, when the limit exists (then the integral converges). Integrals over $(-\infty,\infty)$ must be split at a finite point; both pieces must converge.

Example

$\displaystyle\int_1^\infty\frac{dx}{x^2}=\lim_{t\to\infty}\Big[-\frac1x \Big]_1^t=\lim_{t\to\infty}\Big(1-\frac1t\Big)=1.$

Type II — unbounded integrand

Definition

If $f$ blows up at $b$, $\displaystyle\int_a^b f=\lim_{t\to b^-}\int_a^t f$; if it blows up at $a$, use $\lim_{t\to a^+}\int_t^b f$. At an interior singularity, split the integral there; every piece must converge.

Example

$\displaystyle\int_0^1\frac{dx}{\sqrt{x}}=\lim_{t\to0^+}\big[2\sqrt x\big]_t^1=2.$ Converges.

The $p$-test

Theorem

$\displaystyle\int_1^\infty\frac{dx}{x^p}$ converges iff $p>1$. $\displaystyle\int_0^1\frac{dx}{x^p}$ converges iff $p<1$.

Comparison test

Theorem

If $0\le f\le g$ near the same improper endpoint and $\int g$ converges, so does $\int f$. If $\int f$ diverges, so does $\int g$.

Example

$\displaystyle\int_1^\infty\frac{dx}{x^2+1}$ converges, since $\dfrac{1}{x^2+1}\le\dfrac{1}{x^2}$ and $\int_1^\infty x^{-2}$ converges.

Common pitfalls

• Always rewrite as a limit; don't plug $\infty$ in directly.
• Watch for hidden type-II behaviour (a singularity inside the interval).
• For the comparison test the functions must be nonnegative.

Practice

Problems adapted from the MAT137 question bank by Alfonso Gracia-Saz, used under CC BY-SA 4.0. Write full solutions before opening the answer.

Practice 1 · Evaluate

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

Answer

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

Practice 2 · Divergence

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

Answer

No: $p=1$, and $\int_1^t x^{-1}\,dx=\ln t\to\infty$.

Practice 3 · Comparison

Determine convergence of $\displaystyle\int_1^\infty\frac{2+\sin x}{x^2} \,dx$.

Hint

$0\le\dfrac{2+\sin x}{x^2} \le\dfrac{3}{x^2}$; converges by comparison.

Practice 4 · Evaluate

Evaluate $\displaystyle\int_1^\infty\frac{dx}{x^2+x}$.

Solution

$\dfrac1{x^2+x}=\dfrac1x -\dfrac1{x+1}$, so $\int_1^t=\left[\ln\dfrac{x}{x+1}\right]_1^t\to0-\ln\tfrac12 =\ln2.$

Practice 5 · p-test

For which $p$ do $\displaystyle\int_1^\infty\frac{dx}{x^p}$ and $\displaystyle\int_0^1\frac{dx}{x^p}$ converge?

Answer

$\int_1^\infty$ converges iff $p>1$; $\int_0^1$ converges iff $p<1$. (The trouble is at $\infty$ in the first and at $0$ in the second.)

Practice 6 · A subtle one

What is $\displaystyle\int_{-1}^{1}\frac{dx}{x}$?

Answer

It diverges: it is improper at $0$, and $\int_0^1 dx/x$ already diverges, so the symmetry "$=0$" argument is invalid.

Practice 7 · Comparison

Determine convergence of $\displaystyle\int_1^\infty\frac{dx}{x+e^x}$.

Answer

$0<\dfrac1{x+e^x} <e^{-x}$ and $\int_1^\infty e^{-x}\,dx$ converges, so it converges.

Practice 8 · Compare two

Which converge: $\displaystyle\int_1^\infty\frac{1+\cos^2x}{x^{2/3}}\,dx$ or $\displaystyle\int_1^\infty\frac{1+\cos^2x}{x^{4/3}}\,dx$?

Answer

The numerator lies in $[1,2]$. The first diverges ($p=\tfrac23\le1$, bound below by $1/x^{2/3}$); the second converges ($p=\tfrac43>1$, bound above by $2/x^{4/3}$).

Expanded topic checklist

Course coverage. This unit is part of the complete MAT137 topic map. It should cover the following items before you move on:

Improper intervals. Handle infinite intervals by limits of proper integrals.
Unbounded integrands. Handle vertical asymptotes by cutting the interval at the singular point.
Comparison tests. Use direct and limit comparison to prove convergence or divergence.
Absolute convergence. Separate convergence of $\int f$ from convergence of $\int |f|$ when signs oscillate.

Textbook/source alignment

Use the Alfonso Gracia-Saz MAT137 materials as the primary local spine, with the current UofT single-variable calculus topic split as the completeness check. The added items are original study prompts, not copied textbook exercises.

Core theorem map

The improper-integral spine is: replace the suspect endpoint by a variable endpoint, compute or estimate the proper integral, then take a limit.

Study routine. Always identify the exact source of impropriety before integrating; a single integral may need more than one limiting piece.

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-integral

For which $p$ does $\int_1^\infty x^{-p}dx$ converge?

Guide

It converges exactly when $p>1$; integrating gives a finite limit only in that case.

Extension 2 · Vertical asymptote

Determine whether $\int_0^1 x^{-1/2}dx$ converges.

Guide

Use $\lim_{a\to0^+}\int_a^1 x^{-1/2}dx=\lim_{a\to0^+}2(1-\sqrt a)=2$, so it converges.

Extension 3 · Comparison

Show $\int_1^\infty e^{-x^2}dx$ converges.

Guide

For $x\ge1$, $x^2\ge x$, so $e^{-x^2}\le e^{-x}$. Since $\int_1^\infty e^{-x}dx$ converges, comparison applies.

Extension 4 · Split point

Why must $\int_{-1}^1 1/x\,dx$ not be assigned value $0$ by symmetry?

Guide

It is improper at $0$ and must be split into $\int_{-1}^0$ and $\int_0^1$. Both diverge, so the improper integral diverges even though the principal value is $0$.

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