Two kinds of improper integral
An ordinary definite integral \(\displaystyle\int_a^b f(x)\,dx\) has finite limits and a function that stays finite. An integral is improper when either of these fails:
First kind: an infinite limit
\[\int_a^{\infty} f(x)\,dx = \lim_{t \to \infty}\int_a^t f(x)\,dx\]
Second kind: the function blows up at a point
If \(f\) is unbounded at \(x = a\): \[\int_a^b f(x)\,dx = \lim_{\varepsilon \to 0^{+}}\int_{a + \varepsilon}^b f(x)\,dx\]
If the limit exists and is finite, the integral converges. Otherwise it diverges. If the trouble is in the middle of the interval, split the integral there; and if both limits are infinite, split at any convenient point. Each piece must converge separately.
The idea in one picture
Tests for convergence
| Benchmark | Converges when | Diverges when |
|---|---|---|
| \(\displaystyle\int_1^{\infty}\frac{dx}{x^p}\) | \(p > 1\) (value \(\tfrac{1}{p - 1}\)) | \(p \le 1\) |
| \(\displaystyle\int_0^1\frac{dx}{x^p}\) | \(p < 1\) (value \(\tfrac{1}{1 - p}\)) | \(p \ge 1\) |
| \(\displaystyle\int_0^{\infty}e^{-ax}\,dx\), with \(a > 0\) | always (value \(\tfrac{1}{a}\)) |
Comparison test: if \(0 \le f(x) \le g(x)\) and \(\int g\) converges, so does \(\int f\). If \(\int f\) diverges, so does \(\int g\). It works just like the comparison test for series.
Improper integrals matter here because they define the Gamma and Beta functions, the next three lessons.
Solved examples
Example 1: A convergent first-kind integral
EasyEvaluate \(\displaystyle\int_1^{\infty}\frac{dx}{x^2}\).
- \(\displaystyle\int_1^t x^{-2}\,dx = \left[-\frac{1}{x}\right]_1^t = 1 - \frac{1}{t}\).
- As \(t \to \infty\), this tends to 1.
AnswerConverges, value \(1\).
Example 2: A divergent one
EasyDoes \(\displaystyle\int_1^{\infty}\frac{dx}{x}\) converge?
- \(\displaystyle\int_1^t\frac{dx}{x} = \log t \to \infty\).
AnswerDiverges.
Example 3: Infinite range, inverse tangent
MediumEvaluate \(\displaystyle\int_0^{\infty}\frac{dx}{1 + x^2}\).
- \(\displaystyle\int_0^t\frac{dx}{1 + x^2} = \tan^{-1}t \to \frac{\pi}{2}\).
Answer\(\dfrac{\pi}{2}\)
Example 4: Second kind
MediumEvaluate \(\displaystyle\int_0^1\frac{dx}{\sqrt x}\).
- The integrand blows up at \(x = 0\). \(\displaystyle\int_\varepsilon^1 x^{-1/2}\,dx = \big[2\sqrt x\big]_\varepsilon^1 = 2 - 2\sqrt\varepsilon\).
- As \(\varepsilon \to 0^{+}\), this tends to 2.
AnswerConverges, value \(2\).
Example 5: By parts, towards Gamma
MediumEvaluate \(\displaystyle\int_0^{\infty}x\,e^{-x}\,dx\).
- By parts: \(\displaystyle\int_0^t x e^{-x}\,dx = \big[-xe^{-x}\big]_0^t + \int_0^t e^{-x}\,dx = -te^{-t} + 1 - e^{-t}\).
- As \(t \to \infty\), \(te^{-t} \to 0\) and \(e^{-t} \to 0\).
Answer\(1\). This is \(\Gamma(2)\), see the Gamma function.
Example 6: The hidden trap
Exam levelEvaluate \(\displaystyle\int_{-1}^{1}\frac{dx}{x^2}\).
- A careless calculation gives \(\left[-\dfrac{1}{x}\right]_{-1}^{1} = -1 - 1 = -2\). But the integrand is positive, so a negative answer must be wrong.
- The integrand blows up at \(x = 0\), inside the interval. Split: \(\displaystyle\int_0^1\frac{dx}{x^2}\) diverges (\(p = 2 \ge 1\)).
AnswerDiverges. Always check for points inside the interval where the function blows up.
Before integrating, scan the interval for any value that makes a denominator zero or a log or root undefined. If there is one, it's an improper integral of the second kind, and you must take a limit there.
Common mistakes
1. Missing a point where the function blows up inside the interval
As in Example 6, ignoring it can give a finite, even negative, answer to an integral that actually diverges.
2. Treating \(\infty\) like a number
Write the limit: \(\displaystyle\lim_{t \to \infty}\), not \([\dots]_0^{\infty}\) with \(\infty\) substituted directly. Examiners expect the limit.
3. Thinking that \(f(x) \to 0\) is enough
\(\dfrac{1}{x} \to 0\), yet \(\displaystyle\int_1^\infty\frac{dx}{x}\) diverges. Just like series, the function has to go to 0 fast enough.
Practice questions
Try each one on paper before you open the answer.
Q1\(\displaystyle\int_0^{\infty}e^{-x}\,dx\)
\(\big[-e^{-x}\big]_0^t = 1 - e^{-t} \to 1\).
Q2\(\displaystyle\int_2^{\infty}\frac{dx}{x^3}\)
\(\left[-\dfrac{1}{2x^2}\right]_2^t \to \dfrac{1}{8}\).
Q3\(\displaystyle\int_0^1\log x\,dx\)
By parts: \(\big[x\log x - x\big]_\varepsilon^1 = -1 - (\varepsilon\log\varepsilon - \varepsilon) \to -1\), since \(\varepsilon\log\varepsilon \to 0\).
Q4\(\displaystyle\int_0^2\frac{dx}{2 - x}\)
Blows up at \(x = 2\): \(\big[-\log(2 - x)\big]_0^{2 - \varepsilon} = -\log\varepsilon + \log 2 \to \infty\). Diverges.
Q5\(\displaystyle\int_0^{\infty}x^2 e^{-x}\,dx\)
By parts twice, or as \(\Gamma(3) = 2! = 2\).
Q6Does \(\displaystyle\int_1^{\infty}\frac{\sin^2 x}{x^2}\,dx\) converge?
Yes. \(0 \le \dfrac{\sin^2 x}{x^2} \le \dfrac{1}{x^2}\), and \(\displaystyle\int_1^\infty\frac{dx}{x^2}\) converges. So by comparison it converges.
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Frequently asked questions
What is an improper integral?
An integral with an infinite limit, or one whose integrand becomes infinite somewhere in the interval. It is defined as a limit of ordinary integrals.
What is the difference between the first and second kind?
First kind: the range is infinite. Second kind: the range is finite, but the function is unbounded at some point.
How do I know if an improper integral converges?
Evaluate the limit directly, or compare the integrand with a benchmark such as \(\dfrac{1}{x^p}\) or \(e^{-ax}\).
Where improper integrals lead next
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Vipul Sir has taught mathematics for over 15 years to HSC, CBSE, ICSE, IGCSE/IB, Diploma, Engineering and BSc students at VVS Classes in Goregaon and Vile Parle, Mumbai. He teaches every class himself, concepts first and exam technique second.