Improper Integrals

In A-Level Further Mathematics you meet integrals that push beyond the familiar territory of the standard definite integral. An improper integral is one where either the interval of integration is infinite, or the integrand has a discontinuity (a vertical asymptote) somewhere on the interval. Rather than evaluating such integrals blindly, we redefine them carefully as limits of ordinary definite integrals. This lesson develops both types rigorously, the test for convergence versus divergence, the comparison test, and the exam technique examiners reward.

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1 Spec mapping (AQA 7367)

Improper integrals sit in the compulsory Pure content assessed on Paper 1 and Paper 2 (each 2 h, 100 marks, 33⅓%, with per-paper weighting AO1 55% / AO2 25% / AO3 20%). They form part of the "Further calculus" strand alongside volumes of revolution, the mean value of a function and reduction formulae. The skill is predominantly AO1 (selecting and carrying out the limit-evaluation technique correctly), but a well-set question — "explain why this integral is improper" or "determine whether it converges" — also tests AO2 (reasoning and rigorous communication), and a problem-in-context version can reach into AO3. Improper integrals reappear synoptically: the normal-distribution constant in Statistics, Laplace transforms beyond A-Level, and the integral test for series. Mastering them now pays off repeatedly across the rest of Further calculus.

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2 Core theory — why "improper", and the limit definition

A standard definite integral $\int_a^b f(x)\,dx$∫ab​f(x)dx is guaranteed to make sense provided both of the following hold:

1. the limits $a$a and $b$b are finite;
2. the integrand $f$f is continuous (hence bounded) on the whole closed interval $[a,b]$[a,b].

When either condition fails, the symbol $\int_a^b f(x)\,dx$∫ab​f(x)dx is not yet defined — we cannot simply apply the Fundamental Theorem of Calculus. We give it meaning through a limit.

Type 1 — an infinite limit of integration

$\int_a^{\infty} f(x)\,dx = \lim_{t \to \infty} \int_a^{t} f(x)\,dx$∫a∞​f(x)dx=limt→∞​∫at​f(x)dx

If this limit exists and is finite, the integral converges to that value; otherwise it diverges. The mirror-image definitions are

$\int_{-\infty}^{b} f(x)\,dx = \lim_{t \to -\infty} \int_t^{b} f(x)\,dx, \qquad \int_{-\infty}^{\infty} f(x)\,dx = \int_{-\infty}^{c} f(x)\,dx + \int_c^{\infty} f(x)\,dx$∫−∞b​f(x)dx=limt→−∞​∫tb​f(x)dx,∫−∞∞​f(x)dx=∫−∞c​f(x)dx+∫c∞​f(x)dx

for any convenient real $c$c; the doubly-infinite integral converges only if both pieces converge independently.

Type 2 — a discontinuity (asymptote) in the interval

If $f$f has a vertical asymptote at an endpoint, we again retreat to a limit. With the asymptote at the lower limit $x=a$x=a:

$\int_a^{b} f(x)\,dx = \lim_{t \to a^{+}} \int_t^{b} f(x)\,dx$∫ab​f(x)dx=limt→a+​∫tb​f(x)dx

and symmetrically $\displaystyle \int_a^b f(x)\,dx = \lim_{t \to b^{-}} \int_a^t f(x)\,dx$∫ab​f(x)dx=t→b−lim​∫at​f(x)dx when the asymptote is at $x=b$x=b. If the asymptote is at an interior point $c$c with $a<c<b$a<c<b, you must split the integral at $c$c and treat each piece as its own limit:

$\int_a^{b} f(x)\,dx = \lim_{t \to c^{-}} \int_a^{t} f(x)\,dx + \lim_{s \to c^{+}} \int_s^{b} f(x)\,dx$∫ab​f(x)dx=limt→c−​∫at​f(x)dx+lims→c+​∫sb​f(x)dx

Why the limit definition is not optional

It is tempting to treat an improper integral as "just another definite integral" and slot the bad limit straight into the antiderivative. The definition above exists precisely because that shortcut can give nonsense. The cleanest cautionary tale is the function $f(x)=\dfrac{1}{x^2}$f(x)=x21​ on $[-1,1]$[−1,1]. It is positive everywhere it is defined, so any sensible "area" must be positive. Yet ignoring the asymptote at $x=0$x=0 and applying the Fundamental Theorem blindly gives

$\int_{-1}^{1} \frac{1}{x^2}\,dx \;\overset{?}{=}\; \Big[ -\frac{1}{x} \Big]_{-1}^{1} = (-1) - (1) = -2,$∫−11​x21​dx=?[−x1​]−11​=(−1)−(1)=−2,

a negative answer for a positive integrand — an immediate contradiction that flags the error. Handled properly, the integral must be split at $x=0$x=0, and each half is then examined as a one-sided limit:

$\int_0^{1} \frac{1}{x^2}\,dx = \lim_{t\to 0^{+}}\Big[ -\frac1x \Big]_t^{1} = \lim_{t\to 0^{+}}\Big( -1 + \frac1t \Big) = +\infty,$∫01​x21​dx=limt→0+​[−x1​]t1​=limt→0+​(−1+t1​)=+∞,

so the right-hand half diverges, and (by symmetry) so does the left; the whole integral therefore diverges. The moral: always scan the integrand for points where it blows up inside the interval before you integrate. The limit machinery is what keeps the answer honest.

Convergence intuition. Picture the region under $y=f(x)$y=f(x) stretching off to infinity. Convergence means that, although the region is unbounded in extent, its area settles to a finite total — successive slivers added on the far right shrink fast enough that the running total approaches a ceiling. Divergence means the running total grows without bound. Everything in this lesson is a precise way of deciding which of these two pictures applies.

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3 Worked examples with M1/A1 mark scheme

Worked Example 1 — a convergent Type-1 integral

Evaluate $\displaystyle \int_1^{\infty} \frac{1}{x^2}\,dx.$∫1∞​x21​dx.

$\int_1^{t} x^{-2}\,dx = \left[ -x^{-1} \right]_1^{t} = -\frac{1}{t} - \left( -1 \right) = 1 - \frac{1}{t}$∫1t​x−2dx=[−x−1]1t​=−t1​−(−1)=1−t1​

(M1 replacing the infinite limit with $t$t and integrating; A1 for $1 - \tfrac{1}{t}$1−t1​.)

$\int_1^{\infty} \frac{1}{x^2}\,dx = \lim_{t \to \infty} \left( 1 - \frac{1}{t} \right) = 1 - 0 = 1$∫1∞​x21​dx=limt→∞​(1−t1​)=1−0=1

(M1 taking the limit; A1 for the value $1$1 with the conclusion "converges".) The integral converges to $1$1.

Worked Example 2 — a divergent Type-1 integral

Evaluate $\displaystyle \int_1^{\infty} \frac{1}{x}\,dx.$∫1∞​x1​dx.

$\int_1^{t} \frac{1}{x}\,dx = \Big[ \ln |x| \Big]_1^{t} = \ln t - \ln 1 = \ln t$∫1t​x1​dx=[ln∣x∣]1t​=lnt−ln1=lnt

As $t \to \infty$t→∞, $\ln t \to \infty$lnt→∞, so the integral diverges.

Key insight: $\dfrac{1}{x^2}$x21​ converges but $\dfrac{1}{x}$x1​ diverges. The rate at which the integrand decays as $x \to \infty$x→∞ is what decides the outcome — and $\tfrac1x$x1​ simply does not decay fast enough.

Worked Example 3 — exponential decay

Evaluate $\displaystyle \int_0^{\infty} e^{-2x}\,dx.$∫0∞​e−2xdx.

$\int_0^{t} e^{-2x}\,dx = \left[ -\tfrac{1}{2} e^{-2x} \right]_0^{t} = -\tfrac{1}{2} e^{-2t} + \tfrac{1}{2}$∫0t​e−2xdx=[−21​e−2x]0t​=−21​e−2t+21​

(M1 for the integral with the $-\tfrac12$−21​ factor; A1 for the bracket.) As $t \to \infty$t→∞, $e^{-2t} \to 0$e−2t→0, so

$\int_0^{\infty} e^{-2x}\,dx = \frac{1}{2}.$∫0∞​e−2xdx=21​.

More generally $\displaystyle \int_0^{\infty} e^{-kx}\,dx = \frac{1}{k}$∫0∞​e−kxdx=k1​ for any $k>0$k>0 — a result worth memorising.

Worked Example 4 — a Type-2 (asymptote) integral

Evaluate $\displaystyle \int_0^{1} \frac{1}{\sqrt{x}}\,dx.$∫01​x​1​dx. Note $\dfrac{1}{\sqrt{x}} \to \infty$x​1​→∞ as $x \to 0^{+}$x→0+, so the lower limit is a singularity.

$\int_t^{1} x^{-1/2}\,dx = \left[ 2x^{1/2} \right]_t^{1} = 2 - 2\sqrt{t}$∫t1​x−1/2dx=[2x1/2]t1​=2−2t​

(M1 introducing $t \to 0^{+}$t→0+ and integrating to $2\sqrt{x}$2x​; A1 for $2 - 2\sqrt{t}$2−2t​.) As $t \to 0^{+}$t→0+, $2\sqrt{t} \to 0$2t​→0, so

$\int_0^{1} \frac{1}{\sqrt{x}}\,dx = 2 \quad \text{(converges).}$∫01​x​1​dx=2(converges).

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4 Specimen-style exam question

(Specimen-style — not a real past paper.)

The integrand $f(x) = (4-x)^{-2/3}$f(x)=(4−x)−2/3 is undefined at $x=4$x=4. (a) Explain why $\displaystyle \int_0^{4} \frac{1}{(4-x)^{2/3}}\,dx$∫04​(4−x)2/31​dx is an improper integral. (b) Determine, using a limit, whether it converges, and if so find its exact value.

(a) As $x \to 4^{-}$x→4−, $(4-x)^{2/3} \to 0^{+}$(4−x)2/3→0+, so $f(x) \to \infty$f(x)→∞: the integrand has a vertical asymptote at the upper limit $x=4$x=4, making the integral improper of Type 2.

(b) Substitute $u = 4-x$u=4−x, $du = -dx$du=−dx:

$\int (4-x)^{-2/3}\,dx = -\int u^{-2/3}\,du = -3u^{1/3} + C = -3(4-x)^{1/3} + C.$∫(4−x)−2/3dx=−∫u−2/3du=−3u1/3+C=−3(4−x)1/3+C.

$\int_0^{t} (4-x)^{-2/3}\,dx = \Big[ -3(4-x)^{1/3} \Big]_0^{t} = -3(4-t)^{1/3} + 3 \cdot 4^{1/3}$∫0t​(4−x)−2/3dx=[−3(4−x)1/3]0t​=−3(4−t)1/3+3⋅41/3

As $t \to 4^{-}$t→4−, $(4-t)^{1/3} \to 0$(4−t)1/3→0, so

$\int_0^{4} \frac{1}{(4-x)^{2/3}}\,dx = 3 \cdot 4^{1/3} = 3\sqrt[3]{4} \quad \text{(converges).}$∫04​(4−x)2/31​dx=3⋅41/3=334​(converges).

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5 The p-test (the key convergence criterion)

The two single most useful results compare powers of $x$x:

$\int_1^{\infty} \frac{1}{x^{p}}\,dx \ \text{converges} \iff p > 1, \qquad \int_0^{1} \frac{1}{x^{p}}\,dx \ \text{converges} \iff p < 1.$∫1∞​xp1​dx converges⟺p>1,∫01​xp1​dx converges⟺p<1.

When $p>1$p>1, $\displaystyle \int_1^{\infty} x^{-p}\,dx = \frac{1}{p-1}$∫1∞​x−pdx=p−11​. These are worth proving once, because the proof shows exactly where the borderline $p=1$p=1 comes from. For $p\neq 1$p=1,

$\int_1^{t} x^{-p}\,dx = \left[ \frac{x^{1-p}}{1-p} \right]_1^{t} = \frac{t^{1-p} - 1}{1-p}.$∫1t​x−pdx=[1−px1−p​]1t​=1−pt1−p−1​.

As $t\to\infty$t→∞, the term $t^{1-p}$t1−p tends to $0$0 iff the exponent $1-p<0$1−p<0, i.e. $p>1$p>1; then the limit is $\dfrac{0-1}{1-p}=\dfrac{1}{p-1}$1−p0−1​=p−11​. If $p<1$p<1, then $t^{1-p}\to\infty$t1−p→∞ and the integral diverges. The case $p=1$p=1 is excluded from this algebra (the antiderivative is $\ln x$lnx, not a power) and we saw in Worked Example 2 that it diverges. The mirror result near $0$0 follows the same way: $\int_t^{1} x^{-p}\,dx = \dfrac{1-t^{1-p}}{1-p}$∫t1​x−pdx=1−p1−t1−p​, and now it is $t^{1-p}\to 0$t1−p→0 as $t\to 0^{+}$t→0+ that needs $1-p>0$1−p>0, i.e. $p<1$p<1. The conditions are opposite at the two ends, and they dovetail neatly:

$p$p	$\displaystyle\int_0^1 x^{-p}\,dx$∫01​x−pdx (near $0$0)	$\displaystyle\int_1^{\infty} x^{-p}\,dx$∫1∞​x−pdx (at $\infty$∞)
$p<1$p<1	converges	diverges
$p=1$p=1	diverges	diverges
$p>1$p>1	diverges	converges

The borderline $p=1$p=1 (the function $\tfrac1x$x1​) diverges at both ends — a fact worth internalising, because it is the single most common case examiners use to test whether you really understand the boundary.

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6 The comparison test, and mark-scheme literacy

When no antiderivative is available you can still settle convergence by comparison.

Comparison test. Suppose $0 \le f(x) \le g(x)$0≤f(x)≤g(x) for all $x \ge a$x≥a. If $\int_a^{\infty} g\,dx$∫a∞​gdx converges then so does $\int_a^{\infty} f\,dx$∫a∞​fdx; contrapositively, if $\int_a^{\infty} f\,dx$∫a∞​fdx diverges then so does $\int_a^{\infty} g\,dx$∫a∞​gdx.

Worked Example 5 — comparison test

Does $\displaystyle \int_1^{\infty} \frac{1}{x^2 + 1}\,dx$∫1∞​x2+11​dx converge? For $x \ge 1$x≥1, $x^2 + 1 > x^2$x2+1>x2, so $0 < \dfrac{1}{x^2+1} < \dfrac{1}{x^2}$0<x2+11​<x21​. Since $\int_1^{\infty} x^{-2}\,dx$∫1∞​x−2dx converges ($p=2>1$p=2>1), the smaller integral converges too. Here we can even evaluate it directly:

$\int_1^{\infty} \frac{1}{x^2+1}\,dx = \Big[ \arctan x \Big]_1^{\infty} = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}.$∫1∞​x2+11​dx=[arctanx]1∞​=2π​−4π​=4π​.

Worked Example 6 — comparison test with a divergent benchmark

Show that $\displaystyle \int_2^{\infty} \frac{1}{\ln x}\,dx$∫2∞​lnx1​dx diverges. For $x \ge 2$x≥2, $\ln x < x$lnx<x, so $\dfrac{1}{\ln x} > \dfrac{1}{x} > 0$lnx1​>x1​>0. Since the smaller integral $\int_2^{\infty}\dfrac{1}{x}\,dx = \lim_{t\to\infty}[\ln x]_2^{t} = \lim_{t\to\infty}(\ln t - \ln 2)$∫2∞​x1​dx=limt→∞​[lnx]2t​=limt→∞​(lnt−ln2) already diverges, the larger one must diverge too. This is the other direction of the comparison test: a function lying above a divergent benchmark inherits the divergence.

Choosing the benchmark. The whole skill is picking a comparison function $g$g whose behaviour you already know — almost always a power $\dfrac{1}{x^p}$xp1​ or an exponential $e^{-kx}$e−kx. To prove convergence, bound your integrand above by a convergent $g$g; to prove divergence, bound it below by a divergent $g$g. Getting the inequality the wrong way round proves nothing, so state it carefully.

Mark-scheme literacy. The command word matters. "Show that the integral converges" rewards the limit being taken correctly and an explicit "converges" conclusion — a numerical answer with no limit written down loses the method marks. "Hence" signals you must reuse a result just derived (e.g. a p-test value). Follow-through (ft) marks are available: if you mis-integrate but then take the limit correctly on your expression, the limit mark can still be earned. Always write the limit symbol $\lim_{t\to\infty}$limt→∞​ explicitly — examiners do not award "plugging in $\infty$∞". For a comparison-test answer, the marks are for (i) the stated inequality, (ii) naming the known convergence/divergence of the benchmark, and (iii) the correct conclusion in the correct direction.

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7 Grade-band model answers

Question: Determine whether $\displaystyle \int_1^{\infty} \frac{1}{x}\,dx$∫1∞​x1​dx converges, justifying your answer fully.

Mid-band response. "$\int \frac1x\,dx = \ln x$∫x1​dx=lnx. So $\int_1^{\infty}\frac1x\,dx = \ln\infty - \ln 1 = \infty$∫1∞​x1​dx=ln∞−ln1=∞, so it diverges."

Examiner-style commentary. The conclusion is correct and the antiderivative is right, but "$\ln\infty$ln∞" is not legitimate notation — no limit process is shown, so the reasoning (AO2) marks are at risk even though the final word "diverges" is right.

Stronger response. "$\int_1^{t}\frac1x\,dx = [\ln x]_1^{t} = \ln t$∫1t​x1​dx=[lnx]1t​=lnt. As $t\to\infty$t→∞, $\ln t \to \infty$lnt→∞, so the limit does not exist and the integral diverges."

Examiner-style commentary. This is essentially complete: the integral is replaced by a proper definite integral up to $t$t, evaluated, and the limit examined. It would gain full marks. The only refinement is naming why the limit fails.

Top-band response. "By definition $\int_1^{\infty}\frac1x\,dx = \lim_{t\to\infty}\int_1^{t}\frac1x\,dx = \lim_{t\to\infty}[\ln x]_1^{t} = \lim_{t\to\infty}\ln t$∫1∞​x1​dx=limt→∞​∫1t​x1​dx=limt→∞​[lnx]1t​=limt→∞​lnt. Since $\ln t$lnt increases without bound as $t\to\infty$t→∞, the limit does not exist (is not finite); hence the integral diverges. This is the boundary case $p=1$p=1 of the p-test, consistent with convergence requiring $p>1$p>1."

Examiner-style commentary. Exemplary. It quotes the definition, executes the limit cleanly, states precisely why the limit fails, and connects the result to the general p-test — communicating mathematical reasoning, not merely computing. This is what "Top-band" means: correctness plus articulate justification plus synoptic awareness.

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8 Common misconceptions

1. "Just substitute $\infty$∞." $\infty$∞ is not a number, so $F(\infty)$F(∞) is meaningless and writing "$\ln\infty$ln∞" or "$=\infty - 1$=∞−1" is not legitimate working. You must introduce a variable upper limit $t$t, evaluate the definite integral as a function of $t$t, and then write $\lim_{t\to\infty}F(t)$limt→∞​F(t). This is where the method marks live.
2. Missing an interior asymptote. Evaluating $\int_{-1}^{1} x^{-2}\,dx$∫−11​x−2dx as $[-x^{-1}]_{-1}^{1} = -2$[−x−1]−11​=−2 gives a negative answer for a positive integrand — absurd, and a red flag that a singularity was missed. The asymptote at $x=0$x=0 must be detected, the integral split at $0$0, and each half tested separately (here each half diverges, so the whole thing diverges). Always inspect the integrand's denominator for interior zeros first.
3. Confusing the two p-tests. Convergence needs $p>1$p>1 at infinity but $p<1$p<1 near zero — opposite conditions. A useful mnemonic: near $0$0 a mild blow-up (small $p$p) is survivable; at $\infty$∞ you need fast decay (large $p$p).
4. Thinking convergence means the area is "small". A convergent improper integral can equal any finite number — $7$7, $\pi$π, $10^6$106. "Converges" means the limit is finite, not that the value is small or close to zero.
5. Assuming $f(x)\to 0$f(x)→0 guarantees convergence. $\tfrac1x \to 0$x1​→0 as $x\to\infty$x→∞, yet $\int_1^{\infty}\tfrac1x\,dx$∫1∞​x1​dx diverges. The integrand must tend to $0$0 for an infinite-interval integral to have a chance, but that condition is necessary, not sufficient — the decay must be fast enough (faster than $\tfrac1x$x1​).
6. Mishandling a doubly-improper integral. $\int_{-\infty}^{\infty} f\,dx$∫−∞∞​fdx converges only if each half (split at any finite $c$c) converges on its own. You may not pair off the $+\infty$+∞ and $-\infty$−∞ ends by symmetry and hope cancellation rescues a divergent integral — for instance $\int_{-\infty}^{\infty} x\,dx$∫−∞∞​xdx diverges even though it "looks" antisymmetric.
7. Dropping the modulus in $\int \frac1x\,dx = \ln|x|$∫x1​dx=ln∣x∣. It rarely changes a value when the limits are positive, but examiners expect $\ln|x|$ln∣x∣, and the modulus is essential whenever the interval includes negative $x$x.

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9 Common errors

• Sign slips integrating negative powers: $\int x^{-2}\,dx = -x^{-1}$∫x−2dx=−x−1, not $+x^{-1}$+x−1; a single sign error here flips a convergent answer to a nonsensical negative.
• Lost constant factors with exponentials: $\int e^{-2x}\,dx = -\tfrac12 e^{-2x}$∫e−2xdx=−21​e−2x, not $-e^{-2x}$−e−2x — the chain-rule factor $\tfrac12$21​ is easy to drop and changes the final value.
• Forgetting to state the conclusion ("converges"/"diverges") — the question explicitly asks for it, and the conclusion is usually worth a mark in its own right.
• Evaluating the limit too early, before integrating, instead of integrating up to $t$t first and then letting $t$t tend to its limit.
• Wrong one-sided limit: for an asymptote at the lower limit you need $t\to a^{+}$t→a+ (approaching from inside the interval), not $t\to a^{-}$t→a−; the side matters because the function is only defined on one side there.
• Substitution limits left in the old variable: after a substitution like $u=4-x$u=4−x, either change the limits to $u$u-values or convert back to $x$x before substituting the $x$x-limits — mixing the two is a frequent slip.

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9a A doubly-improper integral worked in full

Some integrals are improper for two reasons at once and must be split. Consider $\displaystyle \int_0^{\infty} \frac{1}{\sqrt{x}\,(1+x)}\,dx$∫0∞​x​(1+x)1​dx. At the lower limit the integrand behaves like $\dfrac{1}{\sqrt x}\to\infty$x​1​→∞ (a Type-2 singularity at $x=0$x=0), and the upper limit is infinite (Type 1). Split at a convenient point, say $x=1$x=1, and treat each piece independently:

$\int_0^{\infty} \frac{dx}{\sqrt{x}\,(1+x)} = \int_0^{1} \frac{dx}{\sqrt{x}\,(1+x)} + \int_1^{\infty} \frac{dx}{\sqrt{x}\,(1+x)}.$∫0∞​x​(1+x)dx​=∫01​x​(1+x)dx​+∫1∞​x​(1+x)dx​.

The substitution $u=\sqrt{x}$u=x​, so $x=u^2$x=u2 and $dx = 2u\,du$dx=2udu, simplifies the integrand beautifully:

$\int \frac{dx}{\sqrt{x}\,(1+x)} = \int \frac{2u\,du}{u\,(1+u^2)} = \int \frac{2}{1+u^2}\,du = 2\arctan u = 2\arctan\sqrt{x}.$∫x​(1+x)dx​=∫u(1+u2)2udu​=∫1+u22​du=2arctanu=2arctanx​.

Now evaluate each piece as a limit. For the singular lower piece,

$\int_0^{1}\frac{dx}{\sqrt x(1+x)} = \lim_{t\to 0^{+}}\Big[ 2\arctan\sqrt x \Big]_t^{1} = 2\arctan 1 - 0 = 2\cdot\frac{\pi}{4} = \frac{\pi}{2},$∫01​x​(1+x)dx​=limt→0+​[2arctanx​]t1​=2arctan1−0=2⋅4π​=2π​,

and for the infinite upper piece,

$\int_1^{\infty}\frac{dx}{\sqrt x(1+x)} = \lim_{t\to\infty}\Big[ 2\arctan\sqrt x \Big]_1^{t} = 2\cdot\frac{\pi}{2} - 2\cdot\frac{\pi}{4} = \pi - \frac{\pi}{2} = \frac{\pi}{2}.$∫1∞​x​(1+x)dx​=limt→∞​[2arctanx​]1t​=2⋅2π​−2⋅4π​=π−2π​=2π​.

Both halves converge, so the original integral converges to $\dfrac{\pi}{2} + \dfrac{\pi}{2} = \pi$2π​+2π​=π. The lesson here is procedural: identify every reason the integral is improper, split so that each piece has a single difficulty, and resolve each with its own limit.

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9b Where improper integrals appear

Improper integrals are not an academic curiosity — they underpin several later topics and university mathematics:

• The normal distribution. The constant $\dfrac{1}{\sqrt{2\pi}}$2π​1​ in the standard normal density exists so that $\displaystyle \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}}\,e^{-x^2/2}\,dx = 1$∫−∞∞​2π​1​e−x2/2dx=1 — a (doubly-infinite) improper integral that makes the total probability $1$1.
• Laplace transforms. The transform $\mathcal{L}\{f\}(s) = \displaystyle\int_0^{\infty} e^{-st} f(t)\,dt$L{f}(s)=∫0∞​e−stf(t)dt is an improper integral in $t$t; its convergence for large enough $s$s is exactly a p-test/exponential-decay argument.
• The integral test for series. A positive decreasing sequence $a_n = f(n)$an​=f(n) gives a series $\sum a_n$∑an​ that converges iff $\int_1^{\infty} f(x)\,dx$∫1∞​f(x)dx converges — directly linking improper integrals to the convergence of infinite series.
• Physics. Total work done against an inverse-square force from a finite radius out to infinity, and electrostatic/gravitational potentials, are computed as improper integrals.

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10 Going further (STEP / MAT flavour)

A beautiful divergent case is Torricelli's trumpet (Gabriel's horn): rotate $y = \tfrac1x$y=x1​ for $x \ge 1$x≥1 about the $x$x-axis. Its volume is finite,

$V = \pi \int_1^{\infty} \frac{1}{x^2}\,dx = \pi,$V=π∫1∞​x21​dx=π,

yet its surface area is infinite, because

$S = 2\pi \int_1^{\infty} \frac{1}{x}\sqrt{1 + \frac{1}{x^4}}\,dx > 2\pi \int_1^{\infty} \frac{1}{x}\,dx = \infty.$S=2π∫1∞​x1​1+x41​​dx>2π∫1∞​x1​dx=∞.

A solid you could fill with a finite amount of paint, but never paint! The paradox dissolves once you realise "painting" assumes a layer of constant thickness (an infinite-volume coat), whereas the filling paint thins indefinitely.

A STEP-style follow-up generalises the factorial. Define the Gamma function

$\Gamma(n) = \int_0^{\infty} x^{n-1} e^{-x}\,dx \qquad (n>0).$Γ(n)=∫0∞​xn−1e−xdx(n>0).

Integrate by parts with $u = x^{n-1}$u=xn−1 and $dv = e^{-x}\,dx$dv=e−xdx (so $du = (n-1)x^{n-2}\,dx$du=(n−1)xn−2dx and $v = -e^{-x}$v=−e−x):

$\Gamma(n) = \Big[ -x^{n-1}e^{-x} \Big]_0^{\infty} + (n-1)\int_0^{\infty} x^{n-2}e^{-x}\,dx.$Γ(n)=[−xn−1e−x]0∞​+(n−1)∫0∞​xn−2e−xdx.

The boundary term vanishes at both ends (at $\infty$∞ because $e^{-x}$e−x crushes the polynomial; at $0$0 because $x^{n-1}\to 0$xn−1→0 for $n>1$n>1), leaving the recurrence $\Gamma(n) = (n-1)\,\Gamma(n-1)$Γ(n)=(n−1)Γ(n−1). Since $\Gamma(1) = \int_0^{\infty} e^{-x}\,dx = 1$Γ(1)=∫0∞​e−xdx=1, iterating gives $\Gamma(n) = (n-1)!$Γ(n)=(n−1)! for positive integers — a convergent improper integral that is the factorial, and which extends it smoothly to non-integers (famously $\Gamma(\tfrac12) = \sqrt{\pi}$Γ(21​)=π​). This single identity ties together improper integrals, integration by parts and the factorial, and is exactly the kind of synthesis STEP rewards.

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11 Additional A* practice (with worked answers)

P1. Evaluate $\displaystyle \int_0^{\infty} x e^{-x}\,dx.$∫0∞​xe−xdx. Answer. By parts with $u=x,\ dv=e^{-x}dx$u=x, dv=e−xdx: $\int_0^{t} x e^{-x}\,dx = [-x e^{-x}]_0^{t} + \int_0^{t} e^{-x}\,dx = -t e^{-t} + [-e^{-x}]_0^{t} = -t e^{-t} - e^{-t} + 1$∫0t​xe−xdx=[−xe−x]0t​+∫0t​e−xdx=−te−t+[−e−x]0t​=−te−t−e−t+1. As $t\to\infty$t→∞, both $t e^{-t}\to 0$te−t→0 and $e^{-t}\to 0$e−t→0 (exponential beats polynomial), so the value is $\boxed{1}$1​.

P2. Show that $\displaystyle \int_0^{\infty} \frac{1}{1+x^2}\,dx = \frac{\pi}{2}.$∫0∞​1+x21​dx=2π​. Answer. $\int_0^{t} \frac{dx}{1+x^2} = [\arctan x]_0^{t} = \arctan t$∫0t​1+x2dx​=[arctanx]0t​=arctant. As $t\to\infty$t→∞, $\arctan t \to \tfrac{\pi}{2}$arctant→2π​. $\;\square$□

P3. For which values of $k>0$k>0 does $\displaystyle \int_0^{1} \frac{1}{x^{k}}\,dx$∫01​xk1​dx converge, and what is its value when it does? Answer. For $k\neq 1$k=1, $\int_t^{1} x^{-k}\,dx = \left[\frac{x^{1-k}}{1-k}\right]_t^{1} = \frac{1 - t^{1-k}}{1-k}$∫t1​x−kdx=[1−kx1−k​]t1​=1−k1−t1−k​. As $t\to 0^{+}$t→0+, $t^{1-k}\to 0$t1−k→0 iff $1-k>0$1−k>0, i.e. $k<1$k<1. So it converges $\iff k<1$⟺k<1, to $\dfrac{1}{1-k}$1−k1​. (At $k=1$k=1 it diverges.)

P4. Determine whether $\displaystyle \int_1^{\infty} \frac{1}{\sqrt{x^3 + 1}}\,dx$∫1∞​x3+1​1​dx converges. Answer. For $x\ge 1$x≥1, $x^3+1 > x^3$x3+1>x3, so $0 < \frac{1}{\sqrt{x^3+1}} < \frac{1}{x^{3/2}}$0<x3+1​1​<x3/21​. Since $\int_1^{\infty} x^{-3/2}\,dx$∫1∞​x−3/2dx converges ($p=\tfrac32>1$p=23​>1), by comparison the given integral converges.

P5. Evaluate $\displaystyle \int_0^{1} \ln x\,dx$∫01​lnxdx (improper at $x=0$x=0). Answer. $\int_t^{1} \ln x\,dx = [x\ln x - x]_t^{1} = (0-1) - (t\ln t - t) = -1 - t\ln t + t$∫t1​lnxdx=[xlnx−x]t1​=(0−1)−(tlnt−t)=−1−tlnt+t. As $t\to 0^{+}$t→0+, $t\ln t \to 0$tlnt→0, so the value is $\boxed{-1}$−1​.

P6. Evaluate $\displaystyle \int_0^{\infty} x^2 e^{-x}\,dx.$∫0∞​x2e−xdx. Answer. Integrating by parts twice (or quoting $\Gamma(3) = 2! = 2$Γ(3)=2!=2): with $I_n = \int_0^{\infty} x^n e^{-x}\,dx$In​=∫0∞​xne−xdx one has $I_n = n\,I_{n-1}$In​=nIn−1​ and $I_0 = 1$I0​=1, so $I_2 = 2\cdot 1\cdot I_0 = 2$I2​=2⋅1⋅I0​=2. Value $= \boxed{2}$=2​.

P7. Determine whether $\displaystyle \int_0^{1} \frac{1}{\sqrt{1-x}}\,dx$∫01​1−x​1​dx converges, and find its value if so. Answer. Singularity at the upper limit $x=1$x=1. With $u=1-x$u=1−x: $\int_0^{t}(1-x)^{-1/2}\,dx = \big[-2\sqrt{1-x}\big]_0^{t} = -2\sqrt{1-t} + 2$∫0t​(1−x)−1/2dx=[−21−x​]0t​=−21−t​+2. As $t\to 1^{-}$t→1−, $\sqrt{1-t}\to 0$1−t​→0, so the integral converges to $\boxed{2}$2​.

P8. Show that $\displaystyle \int_1^{\infty} \frac{2 + \sin x}{x^2}\,dx$∫1∞​x22+sinx​dx converges. Answer. Since $-1 \le \sin x \le 1$−1≤sinx≤1, we have $0 < \dfrac{2+\sin x}{x^2} \le \dfrac{3}{x^2}$0<x22+sinx​≤x23​ for $x\ge 1$x≥1. As $\int_1^{\infty}\dfrac{3}{x^2}\,dx = 3$∫1∞​x23​dx=3 converges (p-test, $p=2$p=2), the given integral converges by comparison — even though it has no elementary antiderivative. $\;\square$□

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12 Board-alignment footer

Aligned to AQA 7367 compulsory Pure (Papers 1 & 2), "Further calculus" strand. Edexcel (9FM0) and OCR (H245 / MEI H645) cover improper integrals identically — the limit definition, the convergence/divergence vocabulary and the p-test transfer verbatim, so this material serves all three boards.

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12a Standard results to know cold

A handful of improper integrals recur so often that you should recognise them instantly and be able to reproduce each in two or three lines. Memorise the values, but also rehearse the derivations — examiners frequently ask you to establish one of these rather than quote it.

Integral	Value	One-line reason
$\displaystyle\int_1^{\infty} \frac{1}{x^p}\,dx$∫1∞​xp1​dx	$\dfrac{1}{p-1}$p−11​ for $p>1$p>1 (else diverges)	$\big[\tfrac{x^{1-p}}{1-p}\big]_1^{\infty}$[1−px1−p​]1∞​
$\displaystyle\int_0^{\infty} e^{-kx}\,dx$∫0∞​e−kxdx	$\dfrac{1}{k}$k1​ for $k>0$k>0	$\big[-\tfrac1k e^{-kx}\big]_0^{\infty}$[−k1​e−kx]0∞​
$\displaystyle\int_0^{\infty} \frac{1}{1+x^2}\,dx$∫0∞​1+x21​dx	$\dfrac{\pi}{2}$2π​	$\big[\arctan x\big]_0^{\infty}$[arctanx]0∞​
$\displaystyle\int_0^{\infty} x^n e^{-x}\,dx$∫0∞​xne−xdx	$n!$n!	$\Gamma(n+1)$Γ(n+1), parts $n$n times
$\displaystyle\int_0^{1} \frac{1}{x^p}\,dx$∫01​xp1​dx	$\dfrac{1}{1-p}$1−p1​ for $p<1$p<1 (else diverges)	mirror p-test near $0$0
$\displaystyle\int_0^{1} \ln x\,dx$∫01​lnxdx	$-1$−1	$\big[x\ln x - x\big]_0^{1}$[xlnx−x]01​

Three habits turn these from memorised facts into reliable marks. First, always confirm why an integral is improper before quoting a value — a question may deliberately disguise a convergent-looking integrand that actually diverges. Second, when a value depends on a parameter ($p$p, $k$k), state the condition for convergence alongside it; an unqualified "$=\tfrac{1}{p-1}$=p−11​" is incomplete. Third, treat the exponential results $\int_0^{\infty} e^{-kx}\,dx=\tfrac1k$∫0∞​e−kxdx=k1​ and $\int_0^{\infty} x^n e^{-x}\,dx = n!$∫0∞​xne−xdx=n! as your default benchmarks for comparison tests, because exponential decay dominates every polynomial.

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13 Visual summary

$\boxed{\;\int_a^{\infty} f\,dx = \lim_{t\to\infty}\int_a^{t} f\,dx \quad\text{and}\quad \int_a^{b} f\,dx = \lim_{t\to a^{+}}\int_t^{b} f\,dx \ \text{(asymptote at } a)\;}$∫a∞​fdx=t→∞lim​∫at​fdxand∫ab​fdx=t→a+lim​∫tb​fdx (asymptote at a)​

Situation	Make improper because…	Evaluate as
$\int_a^{\infty}$∫a∞​	infinite upper limit	$\lim_{t\to\infty}\int_a^{t}$limt→∞​∫at​
$\int_{-\infty}^{b}$∫−∞b​	infinite lower limit	$\lim_{t\to -\infty}\int_t^{b}$limt→−∞​∫tb​
asymptote at $a$a	$f$f unbounded near $a$a	$\lim_{t\to a^{+}}\int_t^{b}$limt→a+​∫tb​
asymptote at interior $c$c	$f$f unbounded inside	split at $c$c, two limits

Recap. Improper $\Rightarrow$⇒ replace the bad limit by a variable, integrate, then take the limit. Convergence at infinity needs decay faster than $\tfrac1x$x1​ ($p>1$p>1); convergence near a singularity needs a milder blow-up than $\tfrac1x$x1​ ($p<1$p<1); and $\int_0^{\infty} e^{-kx}\,dx = \tfrac1k$∫0∞​e−kxdx=k1​.