Section 8.3, Improper Integrals: Infinite Limits of Integration

Section 8.3, Improper Integrals: Inﬁnite Limits of Integration

Homework: 8.3 #1–19 odds, 25, 27, 31

In this section and the next, we will look at improper integrals, which are definite integrals where either one or both of the limits of integration are infinite, or the integrand is infinite. In this section, we will focus on definite integrals with infinite limits. Definition of Improper Integrals with Infinite Limits

Z b Z b f(x) dx = lim f(x) dx −∞ a→−∞ a Z ∞ Z b f(x) dx = lim f(x) dx a b→∞ a

If the limit exists and have ﬁnite values, we say that the improper integral converges. Otherwise, we say that the integral diverges. Examples Evaluate each of the following integrals or show that it diverges: Z 2 1. ex dx −∞

Z 2 Z 2 ex dx = lim ex dx −∞ a→−∞ a 2 x = lim e a→−∞ a = lim e2 − ea) = e2 a→−∞

Z ∞ dx 2. √ 1 πx

Z ∞ dx Z b √ = lim (πx)−1/2 dx 1 πx b→∞ 1 b 2 1/2 = lim (πx) b→∞ π 1 2 = lim (bπ)1/2 − π1/2 b→∞ π = ∞,

so this integral diverges.

1 Z ∞ x 3. 2 2 dx 1 (1 + x ) Z ∞ Z b x 2 −2 2 2 dx = lim x(1 + x ) dx 1 (1 + x ) b→∞ 1 b 1 2 −1 = lim − (1 + x ) b→∞ 2 1 1 1 1 = lim − (1 + b2)−1 + · b→∞ 2 2 2 1 = 4

Z ∞ 1 4. p dx, where p is a real number 1 x We will consider every possible value of p. First, for p = 1:

Z ∞ 1 Z b 1 dx = lim dx 1 x b→∞ 1 x b

= lim ln x b→∞ 1 = lim ln b = ∞, b→∞ so the integral diverges when p = 1. Now, for p 6= 1, the power rule applies:

Z ∞ Z b 1 −p p dx = lim x dx 1 x b→∞ 1 b 1 1−p = lim x b→∞ 1 − p 1 1 1 = lim b1−p − b→∞ 1 − p 1 − p ( ∞ if p < 1 = 1 p−1 if p > 1

1 This means that for p ≤ 1, the integral diverges, and for p > 1, it equals p−1 Deﬁnition of Improper Integrals with Two Inﬁnite Limits R ∞ R 0 R ∞ If 0 f(x) dx and −∞ f(x) dx converge, then −∞ f(x) dx converges and Z ∞ Z 0 Z ∞ f(x) dx = f(x) dx + f(x) dx −∞ −∞ 0 R ∞ Otherwise, −∞ f(x) dx diverges. Example Z ∞ dx 2 −∞ x + 16

Z ∞ dx Z 0 dx Z b dx 2 = lim 2 + lim 2 −∞ x + 16 a→−∞ a x + 16 b→∞ 0 x + 16 0 b 1 −1 x 1 −1 x = lim tan + lim tan a→−∞ 4 4 a b→∞ 4 4 0 1 a 1 b = lim 0 − tan−1 + lim tan−1 − 0 a→−∞ 4 4 b→∞ 4 4 1 π 1 π π = · + · = 4 2 4 2 4

2 1 Probability Density Functions

A probability density function (PDF) is a non-negative function f(x), deﬁned on the real num- R ∞ bers such that −∞ f(x) dx = 1. R b The probability that a random variable X is between a and b is a f(x) dx. The mean of the random variable with the PDF of f(x) is Z ∞ µ = xf(x) dx. −∞ This is a weighted average of all the value that the random variable can take. The variance is Z ∞ σ2 = (x − µ2)f(x) dx. −∞ This describes how spread out the distribution is. Example A continuous random variable X has an exponential distribution with parameter θ for θ > 0 if its PDF is: ( 1 e−x/θ if x > 0 f(x) = θ 0 if x < 0

1. Show that this is a valid PDF. Z ∞ Z 0 Z ∞ 1 f(x) dx = 0 dx + e−x/θ dx −∞ −∞ 0 θ b −x/θ = 0 + lim (−e ) b→∞ 0 = lim − e−b/θ + e0 = 1 b→∞

2. Find the mean µ and variance σ2 of the distribution.

Z ∞ x µ = e−x/θ dx 0 θ ∞ Z ∞ −x/θ −x/θ = −xe + e dx 0 0 ∞ −x/θ = 0 + −θe 0 = θ,

where we used integration by parts.

Z ∞ 1 σ2 = (x − θ)2 e−x/θ dx 0 θ ∞ Z ∞ 2 −x/θ −x/θ = −(x − θ) e + 2(x − θ)e dx 0 0 ∞ Z ∞ 2 −x/θ −x/θ = θ + (−2)θ(x − θ)e + 2θe dx 0 0 ∞ 2 2 2 −x/θ = θ − 2θ − 2θ e 0 = θ2 − 2θ2 + 2θ2 = θ2

3 3. If θ = 2, ﬁnd the probability that X is less than ln 4.

Z ln 4 1 P (X < ln 4) = e−x/2 dx 0 2 ln 4 −x/2 −(ln 4)/2 = −e = −e + 1 0 1 = 2