Unit 5, lesson 61

Improper Integrals over Unbounded Intervals

How an integral over an infinite interval is defined as a limit, when it converges, which power functions have finite area under their tails, and why a symmetric limit is not enough.

  • Theory part 121
  • Practice part 122
  • Improper Integrals

Theory, part 121

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Practice, part 122

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Can a region that stretches forever have finite area? The region under y=1/x2y = 1/x^2 to the right of x=1x = 1 never ends, yet its area is exactly 11. The region under y=1/xy = 1/x looks almost the same and has infinite area. Integrals over unbounded intervals make this precise, and they appear whenever a quantity is spread over all time or all space, such as total radiation from a decaying sample or a probability on the whole real line.

Key terms

Limit definition at infinity

Definition Let ff be integrable on [a,t][a, t] for every t>at > a (for example, continuous on [a,∞)[a, \infty)). The improper integral of ff over [a,∞)[a, \infty) is

∫a∞f(x) dx=lim⁡t→∞∫atf(x) dx,\int_a^\infty f(x) \, dx = \lim_{t \to \infty} \int_a^t f(x) \, dx,

provided this limit exists as a finite number. In the same way, if ff is integrable on [t,b][t, b] for every t<bt < b,

∫−∞bf(x) dx=lim⁡t→−∞∫tbf(x) dx.\int_{-\infty}^b f(x) \, dx = \lim_{t \to -\infty} \int_t^b f(x) \, dx.

In plain words You cannot integrate all the way to infinity in one step. Integrate up to a finite point tt, then ask what happens as tt runs off to infinity.

Example To find the area under y=1/x2y = 1/x^2 to the right of x=1x = 1, first stop at a finite point t>1t > 1.

∫1tdxx2=1−1t\int_1^t \frac{dx}{x^2} = 1 - \frac{1}{t}

The table shows these partial integrals for growing tt.

tt∫1tx−2 dx\int_1^t x^{-2} \, dx
220.50.5
10100.90.9
1001000.990.99
100010000.9990.999

As t→∞t \to \infty these approach 11, so

∫1∞dxx2=1.\int_1^\infty \frac{dx}{x^2} = 1.

Convergence and divergence

Definition An improper integral converges if the limit in its definition exists and is a finite real number, and that number is its value. Otherwise it diverges. Divergence happens in two ways. The partial integrals can grow without bound (toward ∞\infty or −∞-\infty), or they can fail to settle on any value at all.

In plain words Convergent means the running total settles on one finite number. Anything else, whether it blows up or keeps wandering, is divergent.

Example Two integrals diverge for different reasons. For 1/x1/x on [1,∞)[1, \infty) the partial integrals are

∫1tdxx=ln⁡t→∞,\int_1^t \frac{dx}{x} = \ln t \to \infty,

so the integral diverges to ∞\infty. For cos⁡x\cos x on [0,∞)[0, \infty) they are

∫0tcos⁡x dx=sin⁡t,\int_0^t \cos x \, dx = \sin t,

which keeps swinging between −1-1 and 11. That integral diverges even though it never blows up.

p-integrals

Definition For a real number pp,

∫1∞dxxp  converges if p>1, with value 1p−1, and diverges if p≤1.\int_1^\infty \frac{dx}{x^p} \ \text{ converges if } p > 1, \text{ with value } \frac{1}{p - 1}, \text{ and diverges if } p \le 1.

For p≠1p \neq 1 this follows from

∫1tx−p dx=t1−p−11−p,\int_1^t x^{-p} \, dx = \frac{t^{1-p} - 1}{1 - p},

where t1−p→0t^{1-p} \to 0 when p>1p > 1 and t1−p→∞t^{1-p} \to \infty when p<1p < 1. For p=1p = 1 the partial integral is ln⁡t→∞\ln t \to \infty.

In plain words Power functions decaying faster than 1/x1/x have a finite tail area, and 1/x1/x itself is exactly the borderline that fails. These integrals are the yardsticks you compare other integrals against.

Example With p=32p = \frac{3}{2},

∫1∞x−3/2 dx=lim⁡t→∞(2−2t)=2=132−1.\int_1^\infty x^{-3/2} \, dx = \lim_{t \to \infty} \left(2 - \frac{2}{\sqrt{t}}\right) = 2 = \frac{1}{\frac{3}{2} - 1}.

With p=12p = \frac{1}{2} the partial integral is 2t−22\sqrt{t} - 2, which grows without bound, so the integral of x−1/2x^{-1/2} over [1,∞)[1, \infty) diverges.

Separate tails on the real line

Definition Let ff be integrable on every bounded closed interval. The integral of ff over the whole real line is defined by choosing any number cc and setting

∫−∞∞f(x) dx=∫−∞cf(x) dx+∫c∞f(x) dx.\int_{-\infty}^\infty f(x) \, dx = \int_{-\infty}^c f(x) \, dx + \int_c^\infty f(x) \, dx.

It converges only if both integrals on the right converge, each with its own independent limit. When they do, the sum does not depend on the choice of cc.

In plain words A line that is infinite in both directions has two tails. Each tail must have a finite integral by itself before you add them.

Example To integrate 1/(1+x2)1/(1 + x^2) over the whole line, split at c=0c = 0. Then

∫0∞dx1+x2=lim⁡t→∞arctan⁡t=π2,∫−∞0dx1+x2=lim⁡s→−∞(−arctan⁡s)=π2.\int_0^\infty \frac{dx}{1 + x^2} = \lim_{t \to \infty} \arctan t = \frac{\pi}{2}, \qquad \int_{-\infty}^0 \frac{dx}{1 + x^2} = \lim_{s \to -\infty} \left(-\arctan s\right) = \frac{\pi}{2}.

Both converge, so

∫−∞∞dx1+x2=π.\int_{-\infty}^\infty \frac{dx}{1 + x^2} = \pi.

Principal value versus an improper integral

Definition Let ff be integrable on every bounded closed interval. The principal value of its integral over the real line is the symmetric limit

PV⁡∫−∞∞f(x) dx=lim⁡t→∞∫−ttf(x) dx,\operatorname{PV} \int_{-\infty}^\infty f(x) \, dx = \lim_{t \to \infty} \int_{-t}^t f(x) \, dx,

when that limit exists as a finite number. If the improper integral of ff over the real line converges, the principal value exists and equals it. The converse is false. A principal value can exist while the improper integral diverges.

In plain words Letting both ends run out together at the same speed can make a positive tail and a negative tail cancel. That cancellation is not allowed in the real definition, which treats the two tails separately.

Example Take the odd function f(x)=x/(1+x2)f(x) = x/(1 + x^2). Oddness gives

∫−ttx1+x2 dx=0\int_{-t}^t \frac{x}{1 + x^2} \, dx = 0

for every tt, so the principal value is 00. But

∫0tx1+x2 dx=12ln⁡(1+t2)→∞,\int_0^t \frac{x}{1 + x^2} \, dx = \frac{1}{2} \ln(1 + t^2) \to \infty,

so the right tail diverges and the improper integral over the whole line diverges.

Check yourself

  1. Evaluate ∫2∞dxx(ln⁡x)2\int_2^\infty \frac{dx}{x (\ln x)^2}.
Show the answer

With u=ln⁡xu = \ln x the partial integral is 1ln⁡2−1ln⁡t\frac{1}{\ln 2} - \frac{1}{\ln t}, which tends to 1ln⁡2≈1.443\frac{1}{\ln 2} \approx 1.443 as t→∞t \to \infty. The integral converges to 1ln⁡2\frac{1}{\ln 2}.

  1. Does ∫1∞x−0.99 dx\int_1^\infty x^{-0.99} \, dx converge?
Show the answer

No. It is a p-integral with p=0.99≤1p = 0.99 \le 1, so it diverges. For x>1x > 1 its integrand is even slightly larger than 1x\frac{1}{x}, whose integral already diverges.

  1. Show that ∫−∞∞x dx\int_{-\infty}^\infty x \, dx diverges although its principal value is 00.
Show the answer

The symmetric integrals are ∫−ttx dx=0\int_{-t}^t x \, dx = 0 for every tt, so the principal value is 00. But ∫0tx dx=t2/2→∞\int_0^t x \, dx = t^2/2 \to \infty, so the right tail diverges and the improper integral diverges.

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