What makes an integral improper
A definite integral is improper if either of these is true:
- Type 1 — the interval is infinite. One or both limits of integration are ∞ or −∞.
- Type 2 — the function is unbounded somewhere on the interval. There's a vertical asymptote at an endpoint, or somewhere in the middle.
Both types break the ordinary definition of a definite integral, which assumes a finite interval and a bounded function. So we don't evaluate them directly — we define them as limits of ordinary integrals.
Every improper integral has exactly one of two outcomes. If the limit exists and is a finite number, the integral converges to that number. If the limit is infinite or doesn't exist, the integral diverges, and "diverges" is the answer — there's no number to find.
The method: swap infinity for a limit
- Name the problem spot. Replace the infinite limit (or the point where the function blows up) with a variable — b for an upper limit, a for a lower one.
- Integrate as normal. Find the antiderivative and evaluate it between the finite limits. No infinities involved yet.
- Take the limit. Let your variable approach the value you replaced. Finite answer → converges. Infinite or undefined → diverges.
That's the whole technique. The difficulty in these problems is almost never the limit step — it's the antiderivative in step 2.
Type 1: infinite interval
Start with a classic pair that look nearly identical but behave completely differently.
Example 1: a convergent one
Rewrite with a limit, then integrate:
As b grows, 1/b shrinks to nothing, so the area settles on 1. The region is infinitely long but has finite area — the function drops off fast enough to make up for it.
Example 2: a divergent one
Same shape of region, same infinite interval — but 1/x doesn't decay fast enough, so the area is infinite. The integral diverges.
Those two examples are the whole intuition behind improper integrals: convergence is a race between how long the interval is and how fast the function shrinks. Nothing about "infinite length" alone decides it.
The p-test shortcut
That pattern generalises, and it's worth memorising because it answers a whole family of problems instantly:
Check it against what we just did: p = 2 converged, p = 1 diverged. And when it converges, the value is 1/(p−1) — for example, with p = 3/2 you get exactly 2.
The mirror-image rule applies near zero: ∫01 1/xp dx converges if p < 1, which is the reverse of the rule above. The exponent that kills you at infinity is the one that saves you at zero.
Type 2: the function blows up
Here the interval is perfectly ordinary — but the function isn't defined at one end.
Example 3
At x = 0 the integrand is undefined and shoots to infinity, so approach it from the right:
It converges to 2. Here p = 1/2 < 1, exactly as the near-zero rule predicts.
If the blow-up is inside the interval rather than at an endpoint, you must split the integral at that point and test each piece separately. Both pieces have to converge for the whole thing to converge.
Integrating straight through a vertical asymptote as if it weren't there is the single most common way to get a confidently wrong answer — you can even end up with a tidy finite number for an integral that actually diverges.
A harder example, worked in full
Here's one where the limit step is trivial and the antiderivative is the entire challenge:
- Complete the square under the root. x² + 2x − 2 = (x+1)² − 3. That's the move the problem is built around — the same (x+1) appears outside the root, which is a strong hint.
- Substitute. Let u = x + 1, so du = dx. The limits shift from x: 1→∞ to u: 2→∞, and the integral becomes ∫2∞ du / (u√(u²−3)).
- Recognise the standard form. ∫ du / (u√(u²−a²)) = (1/a) arcsec(u/a) + C, with a = √3 here.
- Evaluate the limit. (1/√3)[arcsec(u/√3)] from 2 to ∞. As u→∞, arcsec → π/2. At u = 2, arcsec(2/√3) = arccos(√3/2) = π/6.
√3 π / 9 ≈ 0.6046
Notice how little of that work was about the improper part. Once you've handled the substitution, the ∞ contributed one easy limit at the very end.
Four mistakes that cost marks
- Writing ∞ inside the antiderivative. You can't substitute ∞ into an expression — it isn't a number. Always carry the limit notation until the final step.
- Missing an interior discontinuity. Check whether the integrand is defined everywhere on the interval before you start, not after.
- Dropping the limit notation. Many mark schemes award the method separately from the answer. Write "limb→∞" at every stage and you bank those marks even if the algebra slips.
- Answering "∞" when the limit doesn't exist at all. ∫0∞ cos x dx doesn't grow without bound — it oscillates forever. It still diverges, but for a different reason, and saying "= ∞" is wrong.
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