A definite integral is called improper when one or both limits of integration are infinite, or when the integrand has a singularity — a point where it blows up — inside or at the boundary of the interval. Neither situation is a dead end. The strategy in both cases is the same: replace the problematic point with a parameter and ask whether the integral tends to a finite limit as the parameter approaches the problem.
Improper integrals sit in compulsory Further Pure at all four main boards. Evaluating the limits that arise often requires L'Hôpital's rule — the two topics appear together frequently.
The core idea: replace, integrate, take the limit
Replace the problematic endpoint (∞, −∞, or a singularity) with a finite parameter t. Integrate normally. Then take the limit as t approaches the problem point.
If the limit exists and is finite, the integral converges to that value. If the limit is ±∞ or does not exist, the integral diverges.
Type 1 — infinite limits of integration
A-level Further Maths — Further Pure (compulsory)Edexcel (Core Pure 2)AQA (Further Pure 2)OCR (Pure Core 2)OCR MEI (Core Pure B)
Example 1 — a convergent integral
The integral converges to 1.
Try it yourself
Evaluate ∫1∞ x−3 dx.
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Example 2 — a divergent integral
The limit is infinite, so the integral diverges — even though 1/x becomes arbitrarily small. The function decreases, but not fast enough for the area to remain finite.
Try it yourself
Does ∫1∞ x−1/2 dx converge or diverge? Use the p-integral result to justify your answer.
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This is the p-integral with p = ½. Since p = ½ ≤ 1, the integral diverges.
The p-integral — a benchmark result
The single family of integrals ∫1∞ 1/xp dx neatly separates convergence from divergence:
For p = 1 the antiderivative is ln x (as in Example 2), which diverges separately. For p < 1, x1−p grows without bound as t → ∞. The borderline p = 1 is worth remembering: the harmonic series and its integral both diverge, despite the terms tending to zero.
Type 2 — singularities in the integrand
A-level Further Maths — Further Pure (compulsory)Edexcel (Core Pure 2)AQA (Further Pure 2)OCR (Pure Core 2)OCR MEI (Core Pure B)
When f(x) → ±∞ as x → b (a boundary of the interval), replace b with a parameter approaching from inside the interval:
Example 3 — singularity at the lower limit
Although 1/√x → ∞ as x → 0, the singularity is mild enough that the area remains finite. The integral converges to 2.
Try it yourself
Evaluate ∫01 x−1/3 dx.
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p = 1/3 < 1, so the integral converges — consistent with the p-integral result.
Example 4 — a divergent singularity
The singularity is too strong; the integral diverges.
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Does ∫01 x−3/2 dx converge? Justify using the p-integral result.
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p = 3/2 ≥ 1, so the integral diverges.
The p-integral near zero — the mirror result
Notice the condition flips relative to the infinite-limit case: near zero, mild singularities (p < 1) are integrable; near infinity, functions must decay faster than 1/x (p > 1) to be integrable.
Where L'Hôpital's rule comes in
After integrating, you are left with a limit to evaluate. When the antiderivative involves a product of a polynomial and an exponential or logarithm, the limit as t → ∞ is often an indeterminate form — exactly the situation L'Hôpital's rule is designed for.
Example 5 — integration by parts then L'Hôpital
Evaluate ∫0∞ x e−x dx.
Integrate by parts with u = x, dv/dx = e−x:
Now take the limit as t → ∞. The term −e−t → 0. For −te−t:
Therefore:
Try it yourself
Evaluate ∫0∞ x2e−x dx. (Hint: integrate by parts twice, or use the result from Example 5.)
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Integrate by parts with u = x², dv = e−x dx:
As t → ∞, all terms with e−t vanish (L'Hôpital):
Example 6 — logarithm growth killed by a power
Evaluate ∫1∞ ln(x)/x2 dx.
Integrate by parts with u = ln x, dv/dx = x−2:
As t → ∞, −1/t → 0. For the logarithm term, L'Hôpital gives:
The general pattern: any power of t grows slower than et, and any power of ln t grows slower than any positive power of t. L'Hôpital makes these comparisons precise with a single differentiation.
Try it yourself
Evaluate ∫1∞ ln(x) / x3 dx.
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Integrate by parts with u = ln x, dv = x−3 dx:
As t → ∞, ln(t)/t² → 0 (L'Hôpital) and 1/t² → 0, so:
Integrals with two problems — split first
When an integral has problems at both ends, or at an interior point, it must be split into separate improper integrals at a convenient intermediate point. Both parts must converge independently for the whole integral to converge. It is not valid to let a single limit simultaneously handle both ends.
Integral over all of ℝ
Each half is a separate improper integral. Both must converge.
Example 7 — an integral that appears to cancel but diverges
Consider ∫−11 1/x dx. The integrand is odd, so one might expect the positive and negative areas to cancel, giving 0. They do not.
The left half already diverges. The integral does not exist — the apparent cancellation is an illusion. Symmetry does not rescue a divergent integral.
Try it yourself
Does ∫−11 x−1/3 dx converge? If so, evaluate it. Split at x = 0 and treat each half separately.
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Singularity at x = 0. For each half, the exponent 1/3 < 1, so both halves converge.
The second half is negative because x−1/3 is negative for x < 0. Unlike 1/x, both halves converge and symmetry does give the total integral = 0.
Example 8 — an interior singularity
∫03 1/(x−1)2/3 dx has a singularity at x = 1, inside the interval. Split there:
The antiderivative is 3(x−1)1/3. Evaluating each half:
The singularity at x = 1 looks alarming but the exponent 2/3 < 1 means it is mild enough to integrate — directly analogous to the 1/√x case in Example 3.
Try it yourself
Evaluate ∫−11 x−2/3 dx. The integrand has a singularity at x = 0; split there.
Show answer
p = 2/3 < 1, so both halves converge. The antiderivative of x−2/3 is 3x1/3.
Summary
Type
Condition for convergence
Watch out for
∫1∞ 1/xp dx
p > 1
p = 1 is the critical borderline — diverges
∫01 1/xp dx
p < 1
Condition flips relative to the infinite-limit case
∫ polynomial × e−x
Always converges on [0, ∞)
Boundary term tne−t → 0: use L'Hôpital
∫ ln(x)/xp, p > 0
Converges on [1, ∞)
Boundary term ln(t)/tp → 0: use L'Hôpital
Two problem points
Both halves converge independently
Do not let symmetry replace a proper limit calculation
Interior singularity
Both halves converge independently
Split at the singularity — do not integrate across it blindly
