Jim Lambers MAT 460/560 Fall Semeseter 2009-10 Lecture 33 Notes

These notes correspond to Section 4.9 in the text.

Improper Integrals

An improper integral is an integral in which either of the limits of integration, or the integrand itself, is unbounded. Although such integrals can not be defined as limits as Riemann sums, it is sometimes possible to use other limiting processes to obtain well-defined values. We first consider the integral Z ∞ I(f) = f(x) dx. a The value of this integral is given by

Z b I(f) = lim f(x) dx, b→∞ a provided f is integrable on any finite interval [a, b], and the above limit exists and is finite. If it does, then, for any  > 0, there exists a value of b such that

Z b Z ∞

I(f) − f(x) dx = f(x) dx < . a b Therefore, I(f) can be approximated by evaluating the finite integral obtained by choosing a sufficiently large value for b. An alternative approach is to use a transformation of the integration variable to obtain an integral that has finite limits. Example We will try to approximate the improper integral

Z ∞ 1 2 3 dx. 1 (1 + x ) There are two ways in which this can be accomplished. The first involves using the definition of an integral over an infinite interval,

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

1 just as the definition of a derivative or an integral can be used to approximate these quantities. Specifically, we drop the limit, and assume Z ∞ Z b f(x) dx ≈ f(x) dx, a a where the value b is chosen in order to achieve sufficient accuracy. Certainly, the larger b is, the more accurate this approximation will be. We can determine an appropriate value of b using the fact that Z ∞ Z b Z ∞ f(x) dx = f(x) dx + f(x) dx, a a b and bounding f(x), for large x, by a function that can be integrated analytically from b to ∞ using the Fundamental Theorem of Calculus. This allows us to obtain a bound for the error we are introducing by truncating the interval of integration at b. In this case, we note that 1 2 < (1 + x2)3 x6 for all real x. Therefore, Z ∞ Z ∞ ∞ 1 2 2 −5 2 2 3 dx < 6 dx = x = 5 . b (1 + x ) b x 5 b 5b Therefore, we can ensure that the integral from b to ∞ is sufficiently small simply by choosing b to be sufficiently large. Then, we can approximate the integral from a to b using any quadrature rule. The second approach is to use a change of variable to obtain a new integral over a finite interval. In this case, if we define t = x−1, then t = 1 when x = 1, and t approaches 0 as x approaches ∞, and therefore Z ∞ 1 Z 0 1  1  Z 1 t−2 2 3 dx = − −2 3 − 2 dt = −2 3 dt. 1 (1 + x ) 1 (1 + t ) t 0 (1 + t ) If we multiply the numerator and denominator by t6, we obtain Z 1 t6t−2 Z 1 t4 Z 1 t4 6 −2 3 dt = 2 −2 3 dt = 2 3 dt. 0 t (1 + t ) 0 [t (1 + t )] 0 (t + 1) We can approximate this integral using any quadrature rule, such as the Composite Simpson’s Rule with n = 4. Our step size ℎ is given by ℎ = (1 − 0)/4 = 1/4, and therefore we have Z ∞ 1 Z 1 t4 2 3 dx = 2 3 dt 1 (1 + x ) 0 (t + 1) 1  04 (1/4)4 (1/2)4 (3/4)4 14  ≈ + 4 + 2 + 4 + 12 (02 + 1)3 ((1/4)2 + 1)3 ((1/2)2 + 1)3 ((3/4)2 + 1)3 (12 + 1)3 ≈ 0.04448355532940.

2 The exact value, to 14 significant digits, is 0.044524311274043, so the absolute error is −4.08×10−5. The relative error, however, is much larger, due to the small magnitude of the exact value; it is approximately 1 × 10−3. □ Next, we consider the integral Z b I(f) = f(x) dx a where the integrand f(x) has a singularity at some point c ∈ [a, b], and is therefore unbounded at c. We can define the value of I(f) to be

Z Z b I(f) = lim f(x) dx + lim f(x) dx, →c− a →c+  provided f is integrable on the indicated intervals, and the limits exist and are finite. Such an integral is most efficiently evaluated by using a transformation of the integration variable to obtain an integrand that does not have a singularity, or by subtracting the integral of an analytically integrable function that has the same singularity as f(x). Example Use the Composite Simpson’s Rule with n = 4 to evaluate the integral

Z 1 ex dx. 1/4 0 x This is an improper integral, due to the singularity at x = 0. To work around this problem, we can rewrite the integral as

Z 1 ex Z 1 ex − p (x) Z 1 p (x) dx = n dx + n dx, 1/4 1/4 1/4 0 x 0 x 0 x x where pn(x) is any Taylor polynomial for e centered at x = 0. Because the error term for the Composite Simpson’s Rule is only valid when the integrand is four times continuously differentiable, we choose n = 4, which yields

Z 1 ex Z 1 ex − 1 − x − x2/2 − x3/6 − x4/24 Z 1 1 + x + x2/2 + x3/6 + x4/24 dx = dx + dx. 1/4 1/4 1/4 0 x 0 x 0 x To evaluate the first integral, we define

( x e −p4(x) if 0 < x ≤ 1, G(x) = x1/4 . 0 if x = 0.

Then G(x) ∈ C4[0, 1], and Z 1 ex − p (x) Z 1 4 dx = G(x) dx. 1/4 0 x 0

3 We can approximate this integral using the Composite Simpson’s Rule

Z b ℎ f(x) dx ≈ [f(a) + 4f(x1) + 2f(x2) + 4f(x3) + ⋅ ⋅ ⋅ + 2f(xn−2) + 4f(xn−1) + f(b)] , a 3

1−0 with n = 4 and ℎ = 4 = 1/4, which yields

Z 1 1/4 G(x) dx ≈ [G(0) + 4G(0.25) + 2G(0.5) + 4G(0.75) + G(1)] 0 3 ≈ 0.00163047386619.

The second integral can be evaluated analytically. We have

Z 1 p (x) Z 1 1 + x + x2/2 + x3/6 + x4/24 4 dx = dx 1/4 1/4 0 x 0 x 1 4 3/4 4 7/4 2 11/4 2 15/4 1 19/4 = x + x + x + x + x 3 7 11 45 114 0 4 4 2 2 1 = + + + + 3 7 11 45 114 = 2.13979646084909, and therefore Z 1 ex Z 1 ex − p (x) Z 1 p (x) dx = 4 dx + 4 dx 1/4 1/4 1/4 0 x 0 x 0 x ≈ 0.00163047386619 + 2.13979646084909 ≈ 2.14142693471528. which has an error of −5.331 × 10−5. □

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