Evaluation of Improper Integrals (continued)

Let’s return to Type 3 for further discussion.

∞

Example: I = ∫ [1 /(x + x²)]dx

0

Note both the limits of integration and the integrand are unbounded.

Here upper limit is unbounded ( +∞) and also the integrand is unbounded at x = 0.

In this example one needs to break region of integration into two parts. i.e.

1 1

I₁ = ∫[1 /(x +x²)]dx ; I₁ = lim ∫[1 /(x +x²)]dx

0 t →0 t

Here the upper limit is arbitrary but select a convenient value ( in this case 1).

Pick the variable, t, for the lower limit since the integrand is unbounded at x = 0.

∞ t

I₂ = ∫[1 /(x +x²)]dx ; I₂ = lim ∫[1 /(x +x²)]dx

1 t →∞ 1

Retain the lower limit of 1 in this second integral (since 1 was selected for the first

integral) and pick the variable, t, for the upper limit in the second integral since

the upper limit of integration in this integral is unbounded.

The original integral now becomes the sum of two integrals.

I = I₁ + I₂

Note: If either integral diverges (no limit) then the original integral diverges.

So if you evaluate either integral and find that it diverges, then there is no need

to evaluate the other integral. The original integral diverges.

Click here for examples of Types 1, 2, and 3.

Return to Notes for Calculus 2

Copyright © 2017 Richard C. Coddington

All rights reserved.