Worked examples · Sheet 1

Explicit integrals over unbounded intervals

Theory used: Integral over an unbounded interval Benchmark integrals of powers

Exercise 1

Compute

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

\int_{1}^{t} \frac{d x}{x ^{3}} = [- \frac{1}{2 x ^{2}}]_{1}^{t} = \frac{1}{2} - \frac{1}{2 t ^{2}}

\int_{1}^{+ \infty} \frac{d x}{x ^{3}} = t \to + \infty lim (\frac{1}{2} - \frac{1}{2 t ^{2}}) = \frac{1}{2}

The only critical point is $+ \infty$ , therefore by the definition one integrates over $[1, t]$ and then computes the limit as $t \to + \infty$ . An antiderivative of $x^{- 3}$ is $\frac{x ^{- 2}}{- 2} = - \frac{1}{2 x ^{2}}$ , hence the integral over $[1, t]$ equals $\frac{1}{2} - \frac{1}{2 t ^{2}}$ . As $t \to + \infty$ the term $\frac{1}{2 t ^{2}}$ tends to $0$ , therefore the limit exists and is finite, and the integral converges to the value $\frac{1}{2}$ . The result is consistent with the benchmark of powers, since the exponent $α = 3$ is greater than $1$ and the predicted value is $\frac{1}{α - 1} = \frac{1}{2}$ .

Exercise 2

Compute

\int_{0}^{+ \infty} e^{- 2 x} d x .

\int_{0}^{t} e^{- 2 x} d x = [- \frac{e ^{- 2 x}}{2}]_{0}^{t} = \frac{1}{2} - \frac{e ^{- 2 t}}{2}

\int_{0}^{+ \infty} e^{- 2 x} d x = t \to + \infty lim (\frac{1}{2} - \frac{e ^{- 2 t}}{2}) = \frac{1}{2}

The integrand is continuous on $[0, + \infty)$ , therefore the only critical point is $+ \infty$ . An antiderivative of $e^{- 2 x}$ is $- \frac{e ^{- 2 x}}{2}$ , as one checks by differentiating, since the derivative of $- \frac{e ^{- 2 x}}{2}$ is $e^{- 2 x}$ . The integral over $[0, t]$ therefore equals $\frac{1}{2} - \frac{e ^{- 2 t}}{2}$ . As $t \to + \infty$ the exponential $e^{- 2 t}$ tends to $0$ , hence the integral converges and equals $\frac{1}{2}$ .

Exercise 3

Compute

\int_{0}^{+ \infty} \frac{d x}{1 + x ^{2}} .

\int_{0}^{t} \frac{d x}{1 + x ^{2}} = [arctan x]_{0}^{t} = arctan t

\int_{0}^{+ \infty} \frac{d x}{1 + x ^{2}} = t \to + \infty lim arctan t = \frac{π}{2}

y = \dfrac{1}{1 + x^{2}}

The unbounded region under

y = \frac{1}{1 + x ^{2}}

for

x \geq 0

has finite area, equal to

\frac{π}{2}

An antiderivative of $\frac{1}{1 + x ^{2}}$ is $arctan x$ , and $arctan 0 = 0$ , therefore the integral over $[0, t]$ equals $arctan t$ . As $t \to + \infty$ the arctangent tends to its supremum $\frac{π}{2}$ , hence the integral converges and equals $\frac{π}{2}$ . The integrand is asymptotic to $\frac{1}{x ^{2}}$ as $x \to + \infty$ , therefore convergence was also predicted by comparison with the benchmark of exponent $2$ ; the explicit computation additionally provides the value.