Unit 5, lesson 62

Improper Integrals with Singularities and Comparison Tests

How to integrate a function that blows up inside or at the edge of an interval, and how to decide convergence by comparing with integrals you already know, without computing anything exactly.

  • Theory part 123
  • Practice part 124
  • Improper Integrals

Theory, part 123

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Practice, part 124

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Apply the fundamental theorem blindly to 1/x21/x^2 on [−1,1][-1, 1] and you get

∫−11dxx2=[−1x]−11=−2.\int_{-1}^1 \frac{dx}{x^2} = \left[-\frac{1}{x}\right]_{-1}^1 = -2.

A positive function with a negative area is a clear sign something went wrong. The integrand is unbounded at x=0x = 0, so the ordinary definite integral does not apply. This page handles integrands that blow up, and gives tests that decide convergence when no antiderivative is available.

Key terms

Unbounded integrands

Definition Let ff be continuous on [a,b)[a, b) and unbounded near bb. Then

∫abf(x) dx=lim⁡t→b−∫atf(x) dx,\int_a^b f(x) \, dx = \lim_{t \to b^-} \int_a^t f(x) \, dx,

provided the limit exists as a finite number, in which case the integral converges. Otherwise it diverges. Likewise, if ff is continuous on (a,b](a, b] and unbounded near aa, then

∫abf(x) dx=lim⁡t→a+∫tbf(x) dx,\int_a^b f(x) \, dx = \lim_{t \to a^+} \int_t^b f(x) \, dx,

with the same meaning of convergence. For powers,

∫01dxxp  converges if p<1, with value 11−p, and diverges if p≥1.\int_0^1 \frac{dx}{x^p} \ \text{ converges if } p < 1, \text{ with value } \frac{1}{1 - p}, \text{ and diverges if } p \ge 1.

In plain words Stop just short of the bad point, integrate, then let the stopping point slide toward it. Near zero, powers are the reverse of the tail case. Mild blow up such as x−1/2x^{-1/2} is fine, and 1/x1/x or anything worse is not.

Example For 1/x1/\sqrt{x} on (0,1](0, 1] the partial integrals are

∫t1x−1/2 dx=2−2t→2as t→0+,\int_t^1 x^{-1/2} \, dx = 2 - 2\sqrt{t} \to 2 \quad \text{as } t \to 0^+,

so the integral converges to 22, matching the formula 1/(1−p)1/(1 - p) with p=1/2p = 1/2.

Endpoint and interior singularities

Definition Let ff be continuous on [a,b][a, b] except at one point cc with a<c<ba < c < b, where it is unbounded. Then

∫abf(x) dx=∫acf(x) dx+∫cbf(x) dx,\int_a^b f(x) \, dx = \int_a^c f(x) \, dx + \int_c^b f(x) \, dx,

and the left side converges only if both improper integrals on the right converge. A singularity at an endpoint needs one limit. A singularity inside the interval needs a split and two limits.

In plain words Always look for points inside the interval where the integrand is undefined or blows up. Applying the fundamental theorem across such a point gives nonsense.

Example For 1/x21/x^2 on [−1,1][-1, 1] split at 00. The right half has partial integrals

∫t1x−2 dx=1t−1→∞as t→0+.\int_t^1 x^{-2} \, dx = \frac{1}{t} - 1 \to \infty \quad \text{as } t \to 0^+.

One half diverges, so the whole integral diverges. The value −2-2 from the opening paragraph is wrong.

Separate limits

Definition When an integral is improper for more than one reason, such as singularities at both endpoints, or a singularity together with an infinite interval, choose points that split it into pieces with exactly one improper feature each. The integral converges if and only if every piece converges, each through its own independent limit, and the value is the sum of the pieces. The result does not depend on where you split.

In plain words One limit per problem spot. Never let two limits share a single variable, because that can hide a divergence by cancellation.

Example The integral

∫0∞dxx(1+x)\int_0^\infty \frac{dx}{\sqrt{x}(1 + x)}

is improper at 00 and at ∞\infty. Split at 11. An antiderivative is 2arctan⁡x2\arctan\sqrt{x}, as you can check with u=xu = \sqrt{x}. Then

∫01dxx(1+x)=lim⁡s→0+(2⋅π4−2arctan⁡s)=π2,\int_0^1 \frac{dx}{\sqrt{x}(1 + x)} = \lim_{s \to 0^+} \left(2 \cdot \frac{\pi}{4} - 2\arctan\sqrt{s}\right) = \frac{\pi}{2},∫1∞dxx(1+x)=lim⁡t→∞(2arctan⁡t−π2)=π−π2=π2.\int_1^\infty \frac{dx}{\sqrt{x}(1 + x)} = \lim_{t \to \infty} \left(2\arctan\sqrt{t} - \frac{\pi}{2}\right) = \pi - \frac{\pi}{2} = \frac{\pi}{2}.

Both converge, so the integral equals π\pi.

Direct comparison test

Definition Let ff and gg be continuous on [a,∞)[a, \infty) with 0≤f(x)≤g(x)0 \le f(x) \le g(x) for all x≥ax \ge a.

  1. If ∫a∞g(x) dx\int_a^\infty g(x) \, dx converges, then ∫a∞f(x) dx\int_a^\infty f(x) \, dx converges.
  2. If ∫a∞f(x) dx\int_a^\infty f(x) \, dx diverges, then ∫a∞g(x) dx\int_a^\infty g(x) \, dx diverges.

The same holds for integrals over [a,b)[a, b) or (a,b](a, b] with a singularity at an endpoint, when 0≤f≤g0 \le f \le g on that interval.

In plain words A nonnegative function sitting under one with finite area also has finite area. A function sitting above one with infinite area also has infinite area. The test says nothing if ff sits below a function with infinite area, or gg sits above one with finite area.

Example For x≥1x \ge 1 you have x2≥xx^2 \ge x, so 0<e−x2≤e−x0 < e^{-x^2} \le e^{-x}. The larger function has a convergent integral, so

∫1∞e−x2 dx≤∫1∞e−x dx=1e≈0.368,\int_1^\infty e^{-x^2} \, dx \le \int_1^\infty e^{-x} \, dx = \frac{1}{e} \approx 0.368,

and the integral of e−x2e^{-x^2} converges. No antiderivative was needed.

Limit comparison test

Definition Let ff and gg be continuous and positive on [a,∞)[a, \infty), and suppose

lim⁡x→∞f(x)g(x)=Lwith0<L<∞.\lim_{x \to \infty} \frac{f(x)}{g(x)} = L \quad \text{with} \quad 0 < L < \infty.

Then the integrals of ff and gg over [a,∞)[a, \infty) either both converge or both diverge. For functions continuous and positive on [a,b)[a, b) with a singularity at bb, the same conclusion holds with the limit taken as x→b−x \to b^-, and similarly at a left endpoint.

In plain words If two positive functions behave alike far out, up to a constant factor, their tails share the same fate. Pick gg as the simplest function with the same leading behavior.

Example For f(x)=1/x3+1f(x) = 1/\sqrt{x^3 + 1} on [1,∞)[1, \infty) compare with g(x)=x−3/2g(x) = x^{-3/2}. The ratio is

f(x)g(x)=x3/2x3+1=11+x−3→1.\frac{f(x)}{g(x)} = \frac{x^{3/2}}{\sqrt{x^3 + 1}} = \frac{1}{\sqrt{1 + x^{-3}}} \to 1.

The p-integral of x−3/2x^{-3/2} over [1,∞)[1, \infty) converges, so the integral of ff over [1,∞)[1, \infty) converges too.

Absolute and conditional convergence

Definition Let ff be continuous on [a,∞)[a, \infty). The improper integral of ff over [a,∞)[a, \infty) converges absolutely if

∫a∞∣f(x)∣ dx\int_a^\infty \lvert f(x) \rvert \, dx

converges. Absolute convergence implies convergence, and then

∣∫a∞f(x) dx∣≤∫a∞∣f(x)∣ dx.\left\lvert \int_a^\infty f(x) \, dx \right\rvert \le \int_a^\infty \lvert f(x) \rvert \, dx.

The integral converges conditionally if it converges but the integral of ∣f∣\lvert f \rvert diverges. The same definitions apply to integrals with singularities.

In plain words The comparison tests need nonnegative functions. For a function that changes sign, apply them to ∣f∣\lvert f \rvert. If that converges, you are done. If not, the integral might still converge thanks to cancellation, and that is the conditional case.

Example For x≥1x \ge 1 the bound

∣sin⁡xx2∣≤1x2\left\lvert \frac{\sin x}{x^2} \right\rvert \le \frac{1}{x^2}

holds, so the integral of sin⁡x/x2\sin x / x^2 over [1,∞)[1, \infty) converges absolutely. In contrast, the integral of sin⁡x/x\sin x / x over [1,∞)[1, \infty) converges only conditionally. Integration by parts gives

∫1tsin⁡xx dx=cos⁡1−cos⁡tt−∫1tcos⁡xx2 dx,\int_1^t \frac{\sin x}{x} \, dx = \cos 1 - \frac{\cos t}{t} - \int_1^t \frac{\cos x}{x^2} \, dx,

and the right side has a limit as t→∞t \to \infty, since cos⁡t/t→0\cos t / t \to 0 and the last integral converges absolutely. Yet ∣sin⁡x∣≥sin⁡2x\lvert \sin x \rvert \ge \sin^2 x, so for x≥1x \ge 1

∣sin⁡x∣x≥sin⁡2xx=1−cos⁡2x2x≥0.\frac{\lvert \sin x \rvert}{x} \ge \frac{\sin^2 x}{x} = \frac{1 - \cos 2x}{2x} \ge 0.

The integral of (1−cos⁡2x)/(2x)(1 - \cos 2x)/(2x) over [1,∞)[1, \infty) diverges, because the integral of 1/(2x)1/(2x) diverges while the integral of cos⁡2x/(2x)\cos 2x / (2x) converges by the same parts argument. Direct comparison then shows that the integral of ∣sin⁡x∣/x\lvert \sin x \rvert / x over [1,∞)[1, \infty) diverges.

Check yourself

  1. Evaluate ∫03dx(x−1)2/3\int_0^3 \frac{dx}{(x - 1)^{2/3}}.
Show the answer

The singularity is at x=1x = 1, inside the interval, so split there. An antiderivative is 3x−133\sqrt[3]{x - 1}. The left piece is 0−3−13=30 - 3\sqrt[3]{-1} = 3 and the right piece is 3⋅21/3−03 \cdot 2^{1/3} - 0, both found as limits at 11. The integral converges to 3+323≈6.7803 + 3\sqrt[3]{2} \approx 6.780.

  1. Evaluate ∫01ln⁡x dx\int_0^1 \ln x \, dx.
Show the answer

For 0<t<10 < t < 1 the partial integral is

∫t1ln⁡x dx=[xln⁡x−x]t1=−1−tln⁡t+t.\int_t^1 \ln x \, dx = \left[x \ln x - x\right]_t^1 = -1 - t\ln t + t.

As t→0+t \to 0^+, tln⁡t→0t \ln t \to 0, so the integral equals −1-1.

  1. Does ∫1∞2+sin⁡xx dx\int_1^\infty \frac{2 + \sin x}{x} \, dx converge?
Show the answer

No. Since 2+sin⁡x≥12 + \sin x \ge 1, the integrand is at least 1/x≥01/x \ge 0, and the integral of 1/x1/x over [1,∞)[1, \infty) diverges. By direct comparison the integral diverges.

How this lesson connects

It also uses comparisons.