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In A-Level Further Mathematics, you will meet integrals that push beyond the familiar territory of standard definite integrals. An improper integral is one where either the interval of integration is infinite, or the integrand has a discontinuity (an asymptote) within the interval. This lesson covers both types, how to evaluate them, and the crucial concept of convergence versus divergence.
A standard definite integral requires two things:
When either condition fails, the integral is called improper. We cannot simply apply the Fundamental Theorem of Calculus without extra care.
If we want to integrate f(x) from a to infinity, we define:
integral from a to infinity of f(x) dx = lim (t -> infinity) of integral from a to t of f(x) dx
If this limit exists and is finite, the integral converges. Otherwise it diverges.
Similarly:
integral from -infinity to b of f(x) dx = lim (t -> -infinity) of integral from t to b of f(x) dx
And for a doubly infinite integral:
integral from -infinity to infinity of f(x) dx = integral from -infinity to c of f(x) dx + integral from c to infinity of f(x) dx
where c is any convenient real number (the result does not depend on which c you choose, provided both integrals converge).
Evaluate the integral from 1 to infinity of 1/x² dx.
Solution:
Step 1: Replace the infinite limit with t.
integral from 1 to t of 1/x² dx = integral from 1 to t of x^(-2) dx
Step 2: Integrate.
= [-x^(-1)] from 1 to t = -1/t - (-1/1) = 1 - 1/t
Step 3: Take the limit as t -> infinity.
lim (t -> infinity) (1 - 1/t) = 1 - 0 = 1
The integral converges to 1.
Evaluate the integral from 1 to infinity of 1/x dx.
Solution:
integral from 1 to t of 1/x dx = [ln|x|] from 1 to t = ln t - ln 1 = ln t
As t -> infinity, ln t -> infinity, so the integral diverges.
Key Insight: 1/x² converges but 1/x diverges. The rate at which the function decreases determines convergence.
This is one of the most important results:
integral from 1 to infinity of 1/x^p dx converges if and only if p > 1
When p > 1, the value of the integral is 1/(p - 1).
When p = 1, the integral is ln t -> infinity (diverges).
When p < 1, the integral also diverges.
| p value | Convergence | Value (if convergent) |
|---|---|---|
| p > 1 | Converges | 1/(p - 1) |
| p = 1 | Diverges | -- |
| p < 1 | Diverges | -- |
Evaluate the integral from 0 to infinity of e^(-2x) dx.
Solution:
integral from 0 to t of e^(-2x) dx = [-1/2 e^(-2x)] from 0 to t = -1/2 e^(-2t) + 1/2
As t -> infinity, e^(-2t) -> 0, so the integral converges to 1/2.
Exam Tip: Exponential decay functions like e^(-kx) (with k > 0) always produce convergent integrals over [0, infinity).
Evaluate the integral from 0 to infinity of x e^(-x) dx.
Solution:
Use integration by parts with u = x and dv = e^(-x) dx.
Then du = dx and v = -e^(-x).
integral from 0 to t of x e^(-x) dx = [-x e^(-x)] from 0 to t + integral from 0 to t of e^(-x) dx
= -t e^(-t) + 0 + [-e^(-x)] from 0 to t
= -t e^(-t) - e^(-t) + 1
As t -> infinity, both t e^(-t) -> 0 and e^(-t) -> 0 (the exponential dominates the polynomial).
So the integral converges to 1.
If f(x) has a vertical asymptote at x = c where a <= c <= b, the integral is improper. We evaluate it as a limit:
If the asymptote is at the left endpoint x = a:
integral from a to b of f(x) dx = lim (t -> a+) of integral from t to b of f(x) dx
If the asymptote is at the right endpoint x = b:
integral from a to b of f(x) dx = lim (t -> b-) of integral from a to t of f(x) dx
If the asymptote is at an interior point x = c (where a < c < b), split the integral:
integral from a to b of f(x) dx = integral from a to c of f(x) dx + integral from c to b of f(x) dx
and evaluate each as a separate limit.
Evaluate the integral from 0 to 1 of 1/sqrt(x) dx.
Note that 1/sqrt(x) -> infinity as x -> 0+, so the integrand has an asymptote at x = 0.
Solution:
integral from t to 1 of x^(-1/2) dx = [2x^(1/2)] from t to 1 = 2(1) - 2 sqrt(t) = 2 - 2 sqrt(t)
As t -> 0+, 2 sqrt(t) -> 0, so the integral converges to 2.
Evaluate the integral from 0 to 1 of 1/x dx.
Here 1/x -> infinity as x -> 0+.
Solution:
integral from t to 1 of 1/x dx = [ln|x|] from t to 1 = ln 1 - ln t = -ln t
As t -> 0+, -ln t -> infinity. The integral diverges.
integral from 0 to 1 of 1/x^p dx converges if and only if p < 1
Notice this is the opposite condition to the p-test at infinity. The two tests complement each other.
| p value | Near 0 (integral from 0 to 1) | At infinity (integral from 1 to infinity) |
|---|---|---|
| p < 1 | Converges | Diverges |
| p = 1 | Diverges | Diverges |
| p > 1 | Diverges | Converges |
Evaluate the integral from 0 to 4 of 1/(4 - x)^(2/3) dx.
The integrand has an asymptote at x = 4.
Solution:
integral from 0 to t of (4 - x)^(-2/3) dx
Let u = 4 - x, du = -dx.
= -integral of u^(-2/3) du = -[3u^(1/3)] = -3(4 - x)^(1/3)
Evaluating from 0 to t:
= -3(4 - t)^(1/3) + 3(4)^(1/3) = 3 * 4^(1/3) - 3(4 - t)^(1/3)
As t -> 4-, (4 - t)^(1/3) -> 0, so the integral converges to 3 * 4^(1/3) = 3 * cube root of 4.
When you cannot find an antiderivative, you can still determine convergence by comparing with a known integral.
Theorem (Comparison Test):
Suppose 0 <= f(x) <= g(x) for all x >= a.
Does the integral from 1 to infinity of 1/(x^2 + 1) dx converge?
Solution:
For x >= 1, we have x^2 + 1 > x^2, so 1/(x^2 + 1) < 1/x^2.
Since integral from 1 to infinity of 1/x^2 dx converges (p-test with p = 2 > 1), by the comparison test, integral from 1 to infinity of 1/(x^2 + 1) dx also converges.
In fact, this integral equals arctan(x) evaluated from 1 to infinity = pi/2 - pi/4 = pi/4.
Forgetting to check for discontinuities. Always inspect the integrand for asymptotes inside the interval before integrating.
Ignoring the limit process. You must formally write the limit; you cannot just "plug in infinity."
Misapplying the Fundamental Theorem. If you integrate 1/x^2 from -1 to 1 directly as [-1/x] from -1 to 1 = -1 - 1 = -2, you get a negative answer for a positive function! The issue is that 1/x^2 has an asymptote at x = 0, so you must split at 0 — and both halves diverge.
Confusing the two p-tests. The integral from 1 to infinity converges for p > 1; the integral from 0 to 1 converges for p < 1.
Evaluate the integral from 0 to infinity of e^(-x) dx.
This has an infinite upper limit (Type 1) and the lower limit is fine (e^(-0) = 1 is finite).
integral from 0 to t of e^(-x) dx = [-e^(-x)] from 0 to t = -e^(-t) + 1
As t -> infinity, the integral converges to 1.
Evaluate the integral from 0 to infinity of 1/(1 + x^2) dx.
Solution:
integral from 0 to t of 1/(1 + x^2) dx = [arctan x] from 0 to t = arctan t - 0
As t -> infinity, arctan t -> pi/2. The integral converges to pi/2.
Improper integrals appear throughout mathematics and science:
| Type | Feature | Method |
|---|---|---|
| Type 1 | Infinite limit(s) | Replace with t, take limit |
| Type 2 | Discontinuous integrand | Approach the asymptote as a limit |
| Both | Infinite limit AND asymptote | Split and handle each |
Key results to memorise:
AQA Exam Tip: Questions on improper integrals typically ask you to (a) explain why the integral is improper, (b) evaluate it using limits, and (c) state whether it converges or diverges. Always show the limit notation clearly.