MATH 141 Calculus 2 • McGill University, Montreal

Revision sheet: improper integrals (MATH 141)

Q: How do I know if an integral is improper when both bounds are finite numbers?

Look at the integrand at each bound and at every point between them. If it is undefined and blows up anywhere on the closed interval, for instance where a denominator vanishes or where tangent or secant has an asymptote, the integral is improper there. Split at that point and treat each piece as a limit. The endpoints looking harmless proves nothing.

Q: Why can't the positive and negative infinite parts cancel out?

Because an integral over an interval with a problem point is defined piece by piece, and each piece must be a finite number on its own. Infinity minus infinity is not a number: letting the two ends approach at different speeds produces any value you like. So as soon as one piece diverges, the whole integral diverges, whatever the other piece does.

Q: When does the integral of one over x to the p converge?

It depends on where the problem is. From one to infinity it converges exactly when p is greater than one, because the function must die fast enough. From zero to one it converges exactly when p is less than one, because the function must not explode too fast. At p equal to one it diverges on both, and from zero to infinity it diverges for every p.

Q: How do I use the comparison test for improper integrals without getting it backwards?

Only two situations conclude. If your integrand is smaller than a positive function whose integral converges, yours converges. If it is larger than a positive function whose integral diverges, yours diverges. Being smaller than something divergent, or larger than something convergent, proves nothing, so choose the bound in the direction you want to prove.

This sheet is not a summary of section 7.8 of Stewart: you already have the course notes. It answers one question only, what makes students lose marks on improper integrals in MATH 141 at McGill University, and which precise gesture avoids each loss.

Every number below is done by hand, as on the exam, and every value quoted has been checked by an independent computation. The rule that runs through the whole sheet: locate every problem point first, then write one limit per piece.

The thread of the chapter

An improper integral is a LIMIT, one limit per problem point: find every infinite bound, every endpoint where the integrand blows up and every asymptote hidden inside the interval BEFORE integrating, and the integral converges only if each piece converges on its own. Infinities never cancel.

This chapter is part of MATH 141, Calculus 2 (McGill)

What is on this sheet

1. The essentials
2. The rules in table form
3. The mistakes that cost marks
4. Which method to choose
5. How the answer is expected to be written
6. Check before you hand in
7. The typical problem, taken apart
8. Learn by heart
9. Frequently asked questions

Topics covered

Improper integral of type 1
Improper integral of type 2
Convergence and divergence
Power integrals (p-integrals)
Comparison theorem for integrals
Interior vertical asymptote
Principal value
L'Hôpital's rule
Fundamental Theorem of Calculus
Indeterminate forms

The essentials

Where is it improper? Then one limit per piece

• Type 1, an infinite bound: $\int_a^\infty f(x)\,dx = \lim_{t\to\infty}\int_a^t f(x)\,dx$ . Type 2, an integrand unbounded near $a$ : $\int_a^b f(x)\,dx = \lim_{t\to a^+}\int_t^b f(x)\,dx$ .
• The integral CONVERGES when the limit exists and is finite. An infinite limit, or no limit at all, means it DIVERGES.
• Before integrating, check three places: the bounds, the integrand AT each bound, and every point INSIDE $[a, b]$ where the integrand is undefined.
• Several problem points: split, $\int_a^b = \int_a^c + \int_c^b$ , with independent limits and different letters. The integral converges only if EVERY piece converges.
• Two infinite bounds: $\int_{-\infty}^\infty f = \int_{-\infty}^c f + \int_c^\infty f$ , never $\lim_{t\to\infty}\int_{-t}^t f$ .

Left of the dashed line

\frac{1}{x^2}

is ABOVE

\frac{1}{\sqrt{x}}

, right of it BELOW: the same exponent helps at infinity and hurts near

0

, which is why the two

p

-rules point in opposite directions.

The first line of every answer names the problem points. A copy that starts with an antiderivative has already skipped the step that decides whether the integral exists.

The yardsticks: p-integrals and comparison

• $\int_1^\infty\frac{dx}{x^p}$ converges iff $p > 1$ , to $\frac{1}{p - 1}$ . $\int_0^1\frac{dx}{x^p}$ converges iff $p < 1$ , to $\frac{1}{1 - p}$ . At $p = 1$ both diverge.
• No power converges on $(0, \infty)$ : split at $1$ , and the two pieces ask for $p < 1$ and $p > 1$ at once.
• Comparison, with $0 \le g \le f$ : $\int f$ converges gives $\int g$ converges; $\int g$ diverges gives $\int f$ diverges. The two other directions prove nothing.
• A constant factor never changes convergence: $\frac{1}{\sqrt{x} + 1} \ge \frac{1}{2\sqrt{x}}$ on $[1, \infty)$ is enough to conclude divergence.
• $f(x) \to 0$ does NOT imply convergence: $\frac{1}{\sqrt{x}} \to 0$ and $\int_1^\infty\frac{dx}{\sqrt{x}}$ diverges. What decides is how fast.

The rules in table form

Each row reads from left to right: the assumptions, then the result. A red cell is not an answer, it is the finding that the form settles nothing and the instruction to rewrite it. Every case is followed by a worked example.

The final limit: what closes each computation

Almost every improper integral ends on one of these limits. The left column is the limit as it appears after the antiderivative is evaluated; the red line is the one that students turn into a value.

Limit	Typical source	Result
$e^{-at}$ , $t \to \infty$ , $a > 0$	$\int_0^\infty e^{-ax}\,dx$	$0$
Example: $\int_0^\infty e^{-3x}\,dx = \lim_{t\to\infty}\frac{1 - e^{-3t}}{3} = \frac{1}{3}$ .
$\arctan t$ , $t \to \pm\infty$	$\int\frac{dx}{x^2 + a^2}$	$\pm\frac{\pi}{2}$
Example: $\int_{-\infty}^\infty\frac{dx}{x^2 + 2x + 5} = \frac{1}{2}\left(\frac{\pi}{2} + \frac{\pi}{2}\right) = \frac{\pi}{2}$ .
$\ln t$ , $t \to \infty$ or $t \to 0^+$	$\int\frac{dx}{x}$ , $\int\frac{dx}{x - c}$	$\pm\infty$ : diverges
Example: $\int_1^\infty\frac{dx}{x} = \lim_{t\to\infty}\ln t = \infty$ , the boundary case $p = 1$ .
$t^n e^{-t}$ , $\frac{\ln t}{t}$ , $t \to \infty$	parts on $x^n e^{-x}$ or $\frac{\ln x}{x^2}$	$0$ , by L'Hôpital
Example: $\int_0^\infty x e^{-x}\,dx = \lim_{t\to\infty}\left(1 - t e^{-t} - e^{-t}\right) = 1$ .
$t\ln t$ , $t \to 0^+$	parts on $\ln x$ near $0$	$0$ , by L'Hôpital on $\frac{\ln t}{1/t}$
Example: $\int_0^1\ln x\,dx = \lim_{t\to0^+}\left(-1 - t\ln t + t\right) = -1$ .
$\sin t$ or $\cos t$ , $t \to \infty$	$\int_0^\infty\cos x\,dx$	no limit: diverges no limit, no value
Example: $\int_0^t\cos x\,dx = \sin t$ takes the values $1$ at $t = \frac{\pi}{2} + 2k\pi$ and $-1$ at $t = \frac{3\pi}{2} + 2k\pi$ . What to do: Write diverges: a bounded oscillation is not a limit, and no averaging is allowed. If the integrand also decays, as $\frac{\cos x}{x^2}$ , compare its absolute value instead.

The red line is the only one that is not a number. An indeterminate form such as $\infty \cdot 0$ or $\frac{\infty}{\infty}$ is never read off: it is rewritten as a quotient and handed to L'Hôpital's rule.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Substituting infinity into the antiderivative

1 to 2 marks, the method mark of the question

What not to write

“ $\int_1^\infty\frac{dx}{x^3} = \left[-\frac{1}{2x^2}\right]_1^\infty = -\frac{1}{2\infty^2} + \frac{1}{2} = \frac{1}{2}$ .”

What to write

“ $\int_1^\infty\frac{dx}{x^3} = \lim_{t\to\infty}\int_1^t x^{-3}\,dx = \lim_{t\to\infty}\left(\frac{1}{2} - \frac{1}{2t^2}\right) = \frac{1}{2}$ , so the integral converges.”

Why: $\infty$ is not a number and the improper integral is DEFINED as the limit. Markers look for the limit on the first line; the right number without it is read as a guess.

2. Using the Fundamental Theorem across a hidden asymptote

the whole question

What not to write

“ $\int_0^3\frac{dx}{(x - 2)^2} = \left[-\frac{1}{x - 2}\right]_0^3 = -1 - \frac{1}{2} = -\frac{3}{2}$ .”

What to write

“ $\frac{1}{(x - 2)^2}$ is unbounded at $2 \in [0, 3]$ . $\int_2^3\frac{dx}{(x - 2)^2} = \lim_{s\to2^+}\left(-1 + \frac{1}{s - 2}\right) = \infty$ , so the integral diverges.”

The curve is positive everywhere and climbs without bound on both sides of

x = 2

: its integral over

[0, 3]

cannot be

-\frac{3}{2}

, and in fact it is infinite.

Why: The FTC needs $f$ continuous on the whole of $[a, b]$ . The endpoints are harmless here, so nothing warns you, except the result: a NEGATIVE value for a positive integrand is impossible, and it is the alarm to go back and look inside the interval.

3. Adding two infinities into a finite number

the whole question

What not to write

“ $\int_0^3\frac{dx}{x - 1}$ : the left part is $-\infty$ and the right part $+\infty$ , they cancel and the answer is $\ln 2$ .”

What to write

“ $\int_1^3\frac{dx}{x - 1} = \lim_{s\to1^+}\left(\ln 2 - \ln(s - 1)\right) = \infty$ : one piece diverges, so the integral diverges.”

Why: $\infty - \infty$ is not a value. The integral exists only if EACH piece is a finite number on its own; the moment one piece diverges, the verdict is in and the other piece need not be computed.

4. Taking the symmetric limit for the integral

2 to 3 marks

What not to write

“ $\int_{-\infty}^\infty\frac{x}{1 + x^2}\,dx = \lim_{t\to\infty}\int_{-t}^t\frac{x\,dx}{1 + x^2} = 0$ , the integrand is odd.”

What to write

“ $\int_0^\infty\frac{x\,dx}{1 + x^2} = \lim_{s\to\infty}\frac{1}{2}\ln(1 + s^2) = \infty$ , so the integral diverges.”

Why: The symmetric limit is the principal value, not the integral. Moving the bounds as $-t$ and $2t$ gives $\ln 2$ instead of $0$ : a value that depends on how the bounds move is not a value. Symmetry gives $0$ only AFTER one half is shown to converge, as for $x e^{-x^2}$ .

5. Applying the p-rule of infinity near zero

the whole question

What not to write

“ $\int_0^1\frac{dx}{x^2}$ converges because $p = 2 > 1$ .”

What to write

“On $(0, 1]$ the rule is $p < 1$ . Here $p = 2 \ge 1$ : $\int_t^1 x^{-2}\,dx = \frac{1}{t} - 1 \to \infty$ , the integral diverges.”

Why: Near $0$ a larger $p$ makes the function EXPLODE faster; at infinity it makes it DIE faster. Always state the rule with its interval, and check it on the figure of the essentials: left of $1$ the order of the curves is reversed.

6. Comparing in the direction that proves nothing

the whole question, since the verdict itself is wrong

What not to write

“ $\frac{1}{x^2 + x} \le \frac{1}{x}$ and $\int_1^\infty\frac{dx}{x}$ diverges, so $\int_1^\infty\frac{dx}{x^2 + x}$ diverges.”

What to write

“ $0 \le \frac{1}{x^2 + x} \le \frac{1}{x^2}$ on $[1, \infty)$ and $\int_1^\infty\frac{dx}{x^2}$ converges ( $p = 2$ ), so $\int_1^\infty\frac{dx}{x^2 + x}$ converges.”

The shaded region under

\frac{1}{x^2}

has area

1

and lies entirely under

\frac{1}{x}

, whose area is infinite: being below a divergent integral decides nothing.

Why: Smaller than a divergent integral, or larger than a convergent one, says nothing. The two usable directions are smaller than convergent and larger than divergent. Here the exact value is $\ln 2$ , so the faux line even reached the wrong verdict.

7. Concluding from the integrand tending to zero

the whole question

What not to write

“ $\frac{1}{\sqrt{x}} \to 0$ as $x \to \infty$ , so $\int_1^\infty\frac{dx}{\sqrt{x}}$ converges.”

What to write

“ $\int_1^t\frac{dx}{\sqrt{x}} = 2\sqrt{t} - 2 \to \infty$ : it diverges. A limit of $0$ is necessary for nothing to go wrong, and never sufficient.”

Why: What decides is HOW FAST the integrand tends to $0$ , measured against $\frac{1}{x^p}$ : $p = \frac{1}{2}$ is too slow. The only safe conclusion from a limit is the other one: an integrand tending to a positive number has a divergent integral.

8. Writing infinity times zero as zero

1 mark, even with the right final value

What not to write

“ $\lim_{t\to\infty}t e^{-t} = \infty \cdot 0 = 0$ .”

What to write

“ $t e^{-t} = \frac{t}{e^t}$ , of the form $\frac{\infty}{\infty}$ ; by L'Hôpital's rule the limit is $\lim_{t\to\infty}\frac{1}{e^t} = 0$ .”

Why: $\infty \cdot 0$ is indeterminate: $t \cdot \frac{1}{t} \to 1$ and $t \cdot \frac{1}{\sqrt{t}} \to \infty$ have the same form. Rewrite as a quotient, name the form, name the rule.

Which method to choose

Which tool, by the FORM of the integral

Before any antiderivative, list the problem points, then read the form of the integrand near each of them

If one problem point and an antiderivative you can find → write the limit, integrate on the finite piece, take the limit

Example: $\int_0^4\frac{dx}{\sqrt{x}} = \lim_{t\to0^+}\left(4 - 2\sqrt{t}\right) = 4$
If a point inside the interval where the integrand is undefined → split there; stop at the first piece that diverges

Example: $\int_0^\pi\sec^2 x\,dx$ : split at $\frac{\pi}{2}$ , $\int_0^{\pi/2}$ diverges
If both bounds infinite → split at a convenient $c$ , two independent limits, never the symmetric one

Example: $\int_{-\infty}^\infty\frac{dx}{x^2 + 2x + 5}$ : split at $-1$ , two halves of $\frac{\pi}{4}$
If the integrand is a power $\frac{1}{x^p}$ , or behaves like one → apply the $p$ -rule of the right interval: $p > 1$ at infinity, $p < 1$ near $0$

Example: $\int_0^1\frac{dx}{\sqrt[3]{x^2}}$ : $p = \frac{2}{3} < 1$ , converges to $3$
If no antiderivative, or only the verdict is asked → compare with a $p$ -integral or with $e^{-x}$ , in a direction that concludes

Example: $e^{-x^2} \le e^{-x}$ for $x \ge 1$ : $\int_1^\infty e^{-x^2}\,dx$ converges, below $\frac{1}{e}$
If improper at $0$ AND at infinity → split at 1 and use one yardstick per piece

Example: $\frac{1}{\sqrt{x} + x^2} \le \frac{1}{\sqrt{x}}$ on $(0, 1]$ , $\le \frac{1}{x^2}$ on $[1, \infty)$ : converges

If the integrand is undefined at a point but has a finite limit there, as $\frac{\sin^2 x}{x^2}$ at $0$ , it is bounded and that piece needs no $p$ -rule: only UNBOUNDED behaviour makes an integral improper in any serious way.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Writing an improper integral computation

When to use it: Any question that says evaluate or show that it diverges, on an interval with an infinite bound or an integrand that blows up

1 Name the problem points: the infinite bounds, and the points of $[a, b]$ where $f$ is unbounded, interior ones included.
2 If there is more than one, split the integral at them, and write each piece as its own limit with its own letter.
3 Find the antiderivative, and check it by differentiating before using it.
4 Evaluate on the finite interval, then take each limit, naming L'Hôpital's rule when an indeterminate form appears.
5 Conclude with a sentence: converges to the value, or diverges, and say which piece decided.

Concluding sentence

“ $\int_0^\infty\frac{dx}{\sqrt{x}(1 + x)} = \lim_{t\to0^+}\int_t^1 + \lim_{s\to\infty}\int_1^s = \frac{\pi}{2} + \frac{\pi}{2}$ : both pieces converge, so the integral converges to $\pi$ .”

The trap: Writing one limit for two problems, such as $\lim_{t\to\infty}\int_{1/t}^{t}$ , which ties the two ends together exactly as the principal value does.

Marking: Typically 1 mark for identifying the problem points and writing the limits, 1 to 2 for the antiderivative, 1 for the limits with their justification, and the conclusion sentence is what makes the answer count.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

The sign test

A positive integrand has a positive integral, or a divergent one. A negative or zero value for a positive integrand means the Fundamental Theorem was used across an asymptote.

$\int_0^\pi\sec^2 x\,dx = 0$ and $\int_0^3\frac{dx}{(x - 2)^2} = -\frac{3}{2}$ are both impossible, and both integrals diverge.
The yardstick test

Before computing, decide convergence by comparing with $\frac{1}{x^p}$ near each problem point. The computed verdict must agree with it.

$\frac{\ln x}{x^2} \le \frac{1}{x^{3/2}}$ predicts convergence with a value at most $2$ ; the exact value $1$ agrees.
The derivative test

Differentiate the antiderivative before taking any limit. Sign slips from the chain rule are the most common source of an impossible value.

$\frac{d}{dx}\left(-2\sqrt{3 - x}\right) = \frac{1}{\sqrt{3 - x}}$ : without the minus sign, $\int_2^3\frac{dx}{\sqrt{3 - x}}$ would come out as $-2$ instead of $2$ .
The one-half test

For an odd integrand on a symmetric interval, compute ONE half first. If it diverges, so does the integral, whatever symmetry suggests.

$\int_0^\infty x e^{-x^2}\,dx = \frac{1}{2}$ converges, so $\int_{-\infty}^\infty x e^{-x^2}\,dx = 0$ is legitimate.

The typical problem, taken apart

An infinite bound, parts, and L'Hôpital for the last step

Evaluate $\int_1^\infty\frac{\ln x}{x^3}\,dx$ , or show that it diverges.

No calculator. Every step must be justified as on a MATH 141 final.

Step 1

Problem points: only the upper bound. $\frac{\ln x}{x^3}$ is continuous on $[1, \infty)$ . So $\int_1^\infty\frac{\ln x}{x^3}\,dx = \lim_{t\to\infty}\int_1^t\frac{\ln x}{x^3}\,dx$ .

Why

Naming the problem points and writing the limit is the first mark. At $x = 1$ the integrand is $0$ , nothing to split.

Step 2

Predict: $\ln x \le x$ for $x \ge 1$ , so $0 \le \frac{\ln x}{x^3} \le \frac{1}{x^2}$ , and $\int_1^\infty\frac{dx}{x^2} = 1$ . The integral converges, to a value between $0$ and $1$ .

Why

The comparison costs one line and tells you what the computation must produce. It also justifies the limit in the last step before you reach it.

Step 3

Parts: $u = \ln x$ , $dv = x^{-3}\,dx$ , $du = \frac{dx}{x}$ , $v = -\frac{1}{2x^2}$ . $\int_1^t\frac{\ln x}{x^3}\,dx = \left[-\frac{\ln x}{2x^2}\right]_1^t + \frac{1}{2}\int_1^t x^{-3}\,dx = -\frac{\ln t}{2t^2} + \frac{1}{4} - \frac{1}{4t^2}$ .

Why

$u = \ln x$ because differentiating it removes the logarithm; the other choice would ask for an antiderivative of $\ln x$ and make things worse.

Step 4

Limit: $\frac{\ln t}{2t^2}$ is of the form $\frac{\infty}{\infty}$ ; by L'Hôpital's rule, $\lim_{t\to\infty}\frac{1/t}{4t} = \lim_{t\to\infty}\frac{1}{4t^2} = 0$ . So the integral equals $0 + \frac{1}{4} - 0 = \frac{1}{4}$ .

Why

The indeterminate form is named and the rule is named. Writing $\frac{\infty}{\infty} = 0$ without the rule is where the marks go.

Step 5

Check: $\frac{1}{4}$ is positive and below the bound $1$ of step 2. Derivative of the antiderivative: $\frac{d}{dx}\left(-\frac{\ln x}{2x^2} - \frac{1}{4x^2}\right) = -\frac{1}{2x^3} + \frac{\ln x}{x^3} + \frac{1}{2x^3} = \frac{\ln x}{x^3}$ .

Why

Two checks, ten seconds: the sign and size test against the comparison, and the derivative test against a slip in the parts.

The conclusion, written out

“ $\int_1^\infty\frac{\ln x}{x^3}\,dx = \lim_{t\to\infty}\left(\frac{1}{4} - \frac{\ln t}{2t^2} - \frac{1}{4t^2}\right) = \frac{1}{4}$ : the integral converges to $\frac{1}{4}$ .”

The classic mistake on this problem: Writing $\left[-\frac{\ln x}{2x^2} - \frac{1}{4x^2}\right]_1^\infty$ and declaring the value at infinity $0$ without a limit, or computing $v = -\frac{1}{3x^2}$ from a wrong power rule, which gives $\frac{1}{9}$ and still passes the bound test: only the derivative test catches it.

Learn by heart

• An improper integral is a LIMIT. Never substitute $\infty$ ; never cross an asymptote with the FTC.
• Problem points: infinite bounds, bad endpoints, bad INTERIOR points. One limit per piece, all pieces must converge.
• $\int_1^\infty\frac{dx}{x^p}$ : converges iff $p > 1$ . $\int_0^1\frac{dx}{x^p}$ : converges iff $p < 1$ . $p = 1$ diverges on both.
• $\int_{-\infty}^\infty f = \int_{-\infty}^c f + \int_c^\infty f$ , never the symmetric limit.
• Comparison concludes only in two directions: under convergent, above divergent.
• $\infty - \infty$ , $\infty \cdot 0$ , $\frac{\infty}{\infty}$ are not values: L'Hôpital, or diverges.
• $f(x) \to 0$ is not enough. $\sin t$ and $\cos t$ have no limit at infinity.

Frequently asked questions

How do I know if an integral is improper when both bounds are finite numbers?

Look at the integrand at each bound and at every point between them. If it is undefined and blows up anywhere on the closed interval, for instance where a denominator vanishes or where tangent or secant has an asymptote, the integral is improper there. Split at that point and treat each piece as a limit. The endpoints looking harmless proves nothing.

Why can't the positive and negative infinite parts cancel out?

Because an integral over an interval with a problem point is defined piece by piece, and each piece must be a finite number on its own. Infinity minus infinity is not a number: letting the two ends approach at different speeds produces any value you like. So as soon as one piece diverges, the whole integral diverges, whatever the other piece does.

When does the integral of one over x to the p converge?

It depends on where the problem is. From one to infinity it converges exactly when p is greater than one, because the function must die fast enough. From zero to one it converges exactly when p is less than one, because the function must not explode too fast. At p equal to one it diverges on both, and from zero to infinity it diverges for every p.

How do I use the comparison test for improper integrals without getting it backwards?

Only two situations conclude. If your integrand is smaller than a positive function whose integral converges, yours converges. If it is larger than a positive function whose integral diverges, yours diverges. Being smaller than something divergent, or larger than something convergent, proves nothing, so choose the bound in the direction you want to prove.

Practise it

Corrected exercises: Improper integrals, MATH 141 at McGill

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

10 corrected exercises
100 points
150 minutes

Do the exercises

Previous sheet Strategy for integration Next sheet Areas between curves and average value

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Get in touch for a first session. Improper integrals are where the techniques of integration meet the limits of MATH 140, and where the method marks are the limits themselves.

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Revision sheet: improper integrals (MATH 141)

The essentials

Where is it improper? Then one limit per piece

The yardsticks: p-integrals and comparison

The rules in table form

The final limit: what closes each computation

The mistakes that cost marks

1. Substituting infinity into the antiderivative

2. Using the Fundamental Theorem across a hidden asymptote

3. Adding two infinities into a finite number

4. Taking the symmetric limit for the integral

5. Applying the p-rule of infinity near zero

6. Comparing in the direction that proves nothing

7. Concluding from the integrand tending to zero

8. Writing infinity times zero as zero

Which method to choose

Which tool, by the FORM of the integral

How the answer is expected to be written

Writing an improper integral computation

Check before you hand in

The typical problem, taken apart

An infinite bound, parts, and L'Hôpital for the last step

Learn by heart

Frequently asked questions

How do I know if an integral is improper when both bounds are finite numbers?

Why can't the positive and negative infinite parts cancel out?

When does the integral of one over x to the p converge?

How do I use the comparison test for improper integrals without getting it backwards?

Corrected exercises: Improper integrals, MATH 141 at McGill

See also

Looking for a MATH 141 tutor in Montreal?