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Calculus BC · Unit 6 · 6.13
AP Calculus BC · Integration and Accumulation of Change
Evaluating Improper Integrals
Replace the bad endpoint with a variable, integrate, then take the limit — in that order.
key idea$\int_a^{\infty} f=\lim_{b\to\infty}\int_a^{b} f\quad\text{— replace the bad end with a limit}$
The procedure
An integral is improper when the interval is unbounded or the integrand blows up somewhere on it. The Fundamental Theorem does not apply directly, so you replace the offending endpoint with a variable, integrate, and then take a limit.
1
Identify why it is improper
Type 1: an infinite limit of integration. Type 2: the integrand is unbounded at an endpoint or inside the interval. An interior blow-up is the one people miss.
in the integral out the type
2
Replace the bad endpoint with a variable
$\displaystyle\int_a^{\infty}f\to\lim_{b\to\infty}\int_a^{b}f$, or $\displaystyle\int_a^{c}f\to\lim_{t\to c^{-}}\int_a^{t}f$ when the trouble is at $c$.
out a proper integral inside a limit
3
Split if there are two problems
An integral with both ends infinite, or a blow-up in the middle, must be split so each piece has exactly one problem. Every piece must converge.
out one problem per integral
4
Integrate, then take the limit
Evaluate the proper integral first, giving an expression in the variable endpoint. Only then let it approach the bad value.
out a number, or divergence
5
Declare convergence or divergence
A finite limit means the integral converges to that value. An infinite or non-existent limit means it diverges — and ‘diverges’ is the complete answer.
out the verdict
| Integral | Why improper | Converges? |
|---|
| $\displaystyle\int_{1}^{\infty}\dfrac{dx}{x^{2}}$ | infinite upper limit | yes, to $1$ |
| $\displaystyle\int_{1}^{\infty}\dfrac{dx}{x}$ | infinite upper limit | no — diverges |
| $\displaystyle\int_{0}^{1}\dfrac{dx}{\sqrt{x}}$ | unbounded at $x=0$ | yes, to $2$ |
| $\displaystyle\int_{0}^{1}\dfrac{dx}{x}$ | unbounded at $x=0$ | no — diverges |
| $\displaystyle\int_{-1}^{1}\dfrac{dx}{x^{2}}$ | unbounded inside | no — must be split first |
For $\displaystyle\int_{1}^{\infty}x^{-p}dx$ convergence needs $p\gt 1$; for $\displaystyle\int_{0}^{1}x^{-p}dx$ it needs $p\lt 1$. The opposite condition at each end — and $p=1$ diverges at both.
the p-test
- A finite limit means converges; an infinite or non-existent limit means diverges.
- $\displaystyle\int_{1}^{\infty}\dfrac{dx}{x^{p}}$ converges exactly when $p\gt 1$ — the tail must decay faster than $1/x$.
- $\displaystyle\int_{0}^{1}\dfrac{dx}{x^{p}}$ converges exactly when $p\lt 1$ — the spike must be milder than $1/x$.
- $p=1$ is the borderline and diverges at both ends. $\ln$ is the reason.
$\displaystyle\int_1^{\infty}\frac{dx}{x^{p}}$
converges when $p\gt 1$ · diverges when $p\le 1$
$\displaystyle\int_0^{1}\frac{dx}{x^{p}}$
the opposite: converges when $p\lt 1$ · diverges when $p\ge 1$
$p=1$ either way
diverges — $\ln$ grows without bound, however slowly
why the two disagree
far from the origin a large $p$ makes the tail small; near $0$ a large $p$ makes the spike tall
Worked examples
EX 1
Infinite upper limit, convergent
$\int_1^{\infty}x^{-2}dx$Evaluate $\displaystyle\int_{1}^{\infty}\dfrac{dx}{x^{2}}$.
1
Name the impropriety
Type 1 — the upper limit is infinite.
2
Replace it with a variable
$\displaystyle\lim_{b\to\infty}\int_{1}^{b}x^{-2}dx$.
3
Integrate the proper integral
$\left[-\dfrac1x\right]_{1}^{b}=-\dfrac1b+1$.
4
Take the limit
$\displaystyle\lim_{b\to\infty}\left(1-\dfrac1b\right)=1$.
5
State the verdict
The integral converges to $1$. An unbounded region with finite area — the tail decays fast enough.
EX 2
The borderline case
$\int_1^{\infty}x^{-1}dx$Evaluate $\displaystyle\int_{1}^{\infty}\dfrac{dx}{x}$.
1
Set up the limit
$\displaystyle\lim_{b\to\infty}\int_{1}^{b}\dfrac{dx}{x}$.
2
Integrate
$\left[\ln x\right]_{1}^{b}=\ln b-\ln1=\ln b$.
3
Take the limit
$\displaystyle\lim_{b\to\infty}\ln b=\infty$.
4
State the verdict
The integral diverges.
5
Contrast with Example 1
$x^{-2}$ converges, $x^{-1}$ diverges. The boundary sits exactly at $p=1$, and $\ln$ growing without bound — however slowly — is what decides it.
EX 3
Unbounded at an endpoint
$\int_0^1 x^{-1/2}dx$Evaluate $\displaystyle\int_{0}^{1}\dfrac{dx}{\sqrt{x}}$.
1
Name the impropriety
Type 2 — the integrand blows up at the lower endpoint $x=0$.
2
Replace that endpoint
$\displaystyle\lim_{t\to0^{+}}\int_{t}^{1}x^{-1/2}dx$.
3
Integrate
$\left[2\sqrt{x}\right]_{t}^{1}=2-2\sqrt{t}$.
4
Take the limit
$\displaystyle\lim_{t\to0^{+}}\left(2-2\sqrt{t}\right)=2$.
5
State the verdict
Converges to $2$. An infinitely tall region can still enclose finite area, provided it is thin enough fast enough.
EX 4
A blow-up in the middle
must splitEvaluate $\displaystyle\int_{-1}^{1}\dfrac{dx}{x^{2}}$.
1
Spot the interior problem
$\dfrac{1}{x^{2}}$ is unbounded at $x=0$, which lies inside $[-1,1]$.
2
Split at the trouble point
$\displaystyle\int_{-1}^{0}+\int_{0}^{1}$, each now improper at one end only.
3
Test the second piece
$\displaystyle\lim_{t\to0^{+}}\int_{t}^{1}x^{-2}dx=\lim_{t\to0^{+}}\left(-1+\dfrac1t\right)=\infty$.
4
Conclude immediately
One piece diverges, so the whole integral diverges. The other piece need not be checked.
5
The trap
Applying the FTC blindly gives $\left[-\tfrac1x\right]_{-1}^{1}=-2$ — a negative answer for a positive integrand. Always scan the interval for interior blow-ups first.
Where it goes wrong
Scan the whole interval, not just the ends
An interior asymptote makes the integral improper even when both limits are finite. Example 4 produces a negative answer for a positive function if you miss it.
Integrate first, take the limit second
Substituting $\infty$ into an antiderivative is not valid notation. Evaluate to an expression in $b$, then let $b\to\infty$.
Every piece must converge
A split integral converges only if all pieces do. One divergent piece settles it — the others are irrelevant.
‘Diverges’ is a complete answer
You are not expected to produce a number. Showing the limit is infinite or non-existent finishes the problem.