Evaluation of Improper Integrals using the Comparison Test

∞

Example 1: I = ∫( [ x / (x³ + 1)]dx

0

Note that the denominator dominates for large values of x. So it appears that the improper

integral probably converges.

Here g(x) = x / (x³ + 1). So pick an f(x) that is larger than g(x). Then if

∫ f(x) dx converges so will ∫ g(x) dx .

∞ 1 ∞

I = ∫ [ x / (x³ + 1)]dx = ∫( [ x / (x³ + 1)]dx + ∫( [ x / (x³ + 1)]dx

0 0 1

Note the first integral is definite and the second one is improper.

Pick f(x) = 1 / x² and g(x) = x / (x³ + 1) .

Note: 1 / x² > x (x³ + 1) for large values of x. So f(x) ≥ g(x) .

∞

I_c = ∫ 1/ x² dx

1

t

I_c = lim ( ˗ 1/x) | = 1 result: integral converges

t →∞ 1

and since f(x) ≥ g(x) for large x, the original integral converges.

∞

Example 2: I = ∫( [ 1 / (x – e^-x)]dx here f(x) = 1 / (x – e^-x)]

0

Will this original integral converge or diverge? Note that the term e^-x dominates in

the denominator. So it appears that the integral probably diverges. i.e.

Pick g(x) = 1/x Note that f(x) = 1 / (x – e^-x) > 1/x so f(x) > g(x)

∞ t

I = ∫( [ 1 /x]dx = lim ln x | = ∞ - ln 1 = ∞

1 t →∞ 1

Since ∫ g(x) diverges (smaller area), the larger area ∫ f(x) dx will also diverge

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