Explain why _1^∞1x\,dx diverges. [2 marks]

State the condition on p for _1^∞1x^p\,dx to converge. [1 mark]

Why is _0^11sqrt(x)\,dx improper? [1 mark]

SEAB Original-style practice: Evaluate _0^1 (1)/(sqrt(x))\,dx, explaining why it is improper, or show that it diverges.

The integrand 1sqrt(x) = x^-1/2 is unbounded as x → 0^+ (a vertical asymptote at x = 0), so the integral is improper at the lower limit. Replace the lower limit by a > 0 and take the limit as a → 0^+: $_0^1 x^-1/2\,dx = _a→ 0^+_a^1 x^-1/2\,dx = _a→ 0^+[2x^1/2]_a^1 = _a→ 0^+(2 - 2sqrt(a)).$ As a → 0^+, sqrt(a) → 0, so the limit is 2. The integral converges to 2. Markers reward identifying the singularity at x = 0, the limit with lower limit a, integrating to 2sqrt(x), and the value 2.

SingaporeFurther MathsSyllabus dot point

What is an improper integral, and how do we decide whether one converges and find its value?

Evaluate improper integrals with infinite limits or integrands with a singularity, determining convergence by means of limits

A focused answer to the H2 Further Mathematics outcome on improper integrals. Integrals over an infinite interval, integrands with a vertical asymptote, evaluating them as limits, deciding convergence, and the standard p-integral results.

Generated by Claude Opus 4.810 min answerUpdated 2026-06-06

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

SEAB wants you to evaluate improper integrals: those with an infinite limit of integration, or whose integrand becomes infinite (a vertical asymptote) somewhere on the interval. You must rewrite the integral as a limit of a proper integral, evaluate the limit, and decide whether the integral converges (a finite value) or diverges (no finite value).

The answer

Two kinds of improper integral

An integral is improper if either:

a limit is infinite, for example $\displaystyle\int_a^{\infty}\mathrm{f}(x)\,dx$ ; or
the integrand has a singularity (becomes unbounded) at an endpoint or interior point of the interval, for example $\displaystyle\int_0^1 \dfrac{1}{\sqrt{x}}\,dx$ .

In both cases the ordinary definition of a definite integral does not directly apply, so we use a limiting process.

Evaluating by a limit

Replace the troublesome limit with a variable and take a limit:

\int_a^{\infty}\mathrm{f}(x)\,dx = \lim_{b\to\infty}\int_a^{b}\mathrm{f}(x)\,dx, \qquad \int_a^{b}\mathrm{f}(x)\,dx = \lim_{t\to c^+}\int_t^{b}\mathrm{f}(x)\,dx \;\text{(singularity at }c = a\text{)}.

If the limit exists and is finite, the integral converges to that value; if the limit is infinite or does not exist, the integral diverges.

Singularities inside the interval

If the integrand blows up at an interior point, split the integral at that point into two improper integrals and require both to converge; if either diverges, the whole integral diverges. You may not integrate straight through a singularity.

The standard p-integrals

Two benchmark results worth knowing:

\int_1^{\infty}\frac{1}{x^p}\,dx \;\text{converges} \iff p > 1, \qquad \int_0^{1}\frac{1}{x^p}\,dx \;\text{converges} \iff p < 1.

The borderline $p = 1$ diverges in both: $\int_1^\infty \tfrac{1}{x}\,dx$ and $\int_0^1 \tfrac{1}{x}\,dx$ both give a logarithm that runs off to infinity.

Worked example

Evaluate $\displaystyle\int_0^{\infty} x\,\mathrm{e}^{-x}\,dx$ .

Step 1: Recognise the improper feature

The upper limit is infinite, so write the integral as a limit with finite upper limit $b$ .

Step 2: Set up the limit

\int_0^{\infty} x\mathrm{e}^{-x}\,dx = \lim_{b\to\infty}\int_0^{b} x\mathrm{e}^{-x}\,dx.

Step 3: Integrate by parts

With $u = x$ , $dv = \mathrm{e}^{-x}dx$ , so $du = dx$ , $v = -\mathrm{e}^{-x}$ :

\int_0^{b} x\mathrm{e}^{-x}\,dx = \left[-x\mathrm{e}^{-x}\right]_0^{b} + \int_0^{b}\mathrm{e}^{-x}\,dx = -b\mathrm{e}^{-b} + \left[-\mathrm{e}^{-x}\right]_0^{b} = -b\mathrm{e}^{-b} - \mathrm{e}^{-b} + 1.

Step 4: Take the limit

As $b \to \infty$ , both $b\mathrm{e}^{-b} \to 0$ (exponential beats the linear factor) and $\mathrm{e}^{-b} \to 0$ , leaving

\lim_{b\to\infty}\left(-b\mathrm{e}^{-b} - \mathrm{e}^{-b} + 1\right) = 1.

Step 5: Conclude

The integral converges to $1$ . (This is the mean of the standard exponential distribution.)

Examples in context

Example 1. Total probability of a continuous distribution. Verifying that a probability density integrates to $1$ over an infinite range, as for the exponential density $\lambda\mathrm{e}^{-\lambda x}$ on $[0, \infty)$ , is an improper integral, linking convergence directly to the continuous random variables work.

Example 2. Escape energy. The work to move a mass infinitely far against an inverse-square force is $\int_R^{\infty}\dfrac{k}{x^2}\,dx$ , a convergent improper integral whose finite value is the escape energy, showing why a $\tfrac{1}{x^2}$ force allows escape with finite energy.

Try this

Q1. Explain why $\displaystyle\int_1^{\infty}\dfrac{1}{x}\,dx$ diverges. [2 marks]

Cue. $\int_1^b \tfrac{1}{x}\,dx = \ln b$ , which tends to infinity as $b \to \infty$ , so there is no finite value.

Q2. State the condition on $p$ for $\displaystyle\int_1^{\infty}\dfrac{1}{x^p}\,dx$ to converge. [1 mark]

Cue. $p > 1$ .

Q3. Why is $\displaystyle\int_0^{1}\dfrac{1}{\sqrt{x}}\,dx$ improper? [1 mark]

Cue. The integrand $x^{-1/2}$ is unbounded as $x \to 0^+$ , a singularity at the lower limit.

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original5 marksDetermine whether

\displaystyle\int_1^{\infty} \frac{1}{x^2}\,dx

converges, and find its value if it does.

Show worked answer →

Replace the infinite limit by a finite $b$ and take a limit:

\int_1^{\infty}\frac{1}{x^2}\,dx = \lim_{b\to\infty}\int_1^{b} x^{-2}\,dx = \lim_{b\to\infty}\left[-\frac{1}{x}\right]_1^{b} = \lim_{b\to\infty}\left(-\frac{1}{b} + 1\right).

b \to \infty

\dfrac{1}{b} \to 0

, so the limit is

1

The integral converges and its value is $1$ .

Markers reward writing the integral as a limit with finite upper limit $b$ , integrating to $-\tfrac{1}{x}$ , taking the limit, and concluding convergence to $1$ .

Original6 marksEvaluate

\displaystyle\int_0^{1} \frac{1}{\sqrt{x}}\,dx

, explaining why it is improper, or show that it diverges.

Show worked answer →

The integrand $\dfrac{1}{\sqrt{x}} = x^{-1/2}$ is unbounded as $x \to 0^+$ (a vertical asymptote at $x = 0$ ), so the integral is improper at the lower limit.

Replace the lower limit by $a > 0$ and take the limit as $a \to 0^+$ :

\int_0^{1} x^{-1/2}\,dx = \lim_{a\to 0^+}\int_a^{1} x^{-1/2}\,dx = \lim_{a\to 0^+}\left[2x^{1/2}\right]_a^{1} = \lim_{a\to 0^+}\left(2 - 2\sqrt{a}\right).

a \to 0^+

\sqrt{a} \to 0