Worked examples · Sheet 3

Integrals by limit comparison at infinity

Theory used: Limit comparison test Benchmark integrals of powers Comparison test Order of infinitesimal and of infinity

Exercise 1

Determine the behavior of

\int_{1}^{+ \infty} \frac{x + 3}{x ^{3} + x + 1} d x .

x \to + \infty lim \frac{\frac{x + 3}{x ^{3} + x + 1}}{\frac{1}{x ^{2}}} = x \to + \infty lim \frac{x ^{3} + 3 x ^{2}}{x ^{3} + x + 1} = 1

\frac{x + 3}{x ^{3} + x + 1} \sim \frac{1}{x ^{2}} (x \to + \infty), the integral converges

The integrand is positive on $[1, + \infty)$ , because numerator and denominator are positive for $x \geq 1$ , therefore the comparison tests apply. Numerator and denominator are polynomials, hence at infinity the terms of highest degree are what matter and the candidate benchmark is $\frac{x}{x ^{3}} = \frac{1}{x ^{2}}$ . The limit of the ratio between the integrand and $\frac{1}{x ^{2}}$ equals $1$ , therefore the two functions are asymptotic. The order of infinitesimal is $2 > 1$ , the benchmark converges, hence by limit comparison the integral converges.

Exercise 2

Determine the behavior of

\int_{1}^{+ \infty} \frac{d x}{x ^{2} + 1} .

x \to + \infty lim \frac{\frac{1}{x ^{2} + 1}}{\frac{1}{x}} = x \to + \infty lim \frac{x}{x ^{2} + 1} = x \to + \infty lim \frac{1}{1 + \frac{1}{x ^{2}}} = 1

\frac{1}{x ^{2} + 1} \sim \frac{1}{x} (x \to + \infty), the integral diverges

The integrand is positive. Under the root the dominant term is $x^{2}$ , therefore the candidate benchmark is $\frac{1}{x}$ . To compute the limit of the ratio one brings $x = x^{2}$ inside the root (allowed because $x > 0$ ), and the ratio becomes the reciprocal of $1 + \frac{1}{x ^{2}}$ , which tends to $1$ . The integrand is thus asymptotic to $\frac{1}{x}$ , the benchmark with $α = 1$ diverges, hence the integral diverges by limit comparison.

Exercise 3

Determine the behavior of

\int_{1}^{+ \infty} x^{2} e^{- x} d x .

x \to + \infty lim \frac{x ^{2} e ^{- x}}{\frac{1}{x ^{2}}} = x \to + \infty lim \frac{x ^{4}}{e ^{x}} = 0

x^{2} e^{- x} < \frac{1}{x ^{2}} eventually, the integral converges

The integrand is positive. The exponential $e^{- x}$ tends to $0$ faster than every power, therefore no power is asymptotic to the integrand and one uses the form of the comparison with limit $L = 0$ . The ratio between the integrand and the benchmark $\frac{1}{x ^{2}}$ equals $\frac{x ^{4}}{e ^{x}}$ and tends to $0$ by the hierarchy of infinities (chapter on limits). Since the limit is $0$ and the integral of $\frac{1}{x ^{2}}$ converges, the integrand is eventually less than the benchmark and, by the comparison test, the integral converges.