Worked examples · Sheet 2

Explicit integrals of unbounded functions

Theory used: Integral of a function unbounded near an endpoint Benchmark integrals of powers

Exercise 1

Compute

\int_{0}^{1} \frac{d x}{x} .

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

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

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

Exercise 2

Compute

\int_{0}^{4} \frac{d x}{4 - x} .

\int_{0}^{t} \frac{d x}{4 - x} = [- 2 4 - x]_{0}^{t} = 4 - 2 4 - t

\int_{0}^{4} \frac{d x}{4 - x} = t \to 4^{-} lim (4 - 2 4 - t) = 4

y = \dfrac{1}{\sqrt{4 - x}}

Near

x = 4

the function

\frac{1}{4 - x}

is unbounded, but the shaded area on

[0, 4)

is finite and equals

4

The critical point is the upper endpoint $4$ , where the denominator vanishes and the integrand tends to $+ \infty$ , therefore one integrates over $[0, t]$ with $t < 4$ and passes to the limit as $t \to 4^{-}$ . An antiderivative of $(4 - x)^{- \frac{1}{2}}$ is $- 2 4 - x$ , because differentiating with the chain rule gives $- 2 \cdot \frac{- 1}{2 4 - x} = \frac{1}{4 - x}$ . At $x = 0$ the antiderivative equals $- 4$ , therefore the integral over $[0, t]$ equals $4 - 2 4 - t$ . As $t \to 4^{-}$ the root $4 - t$ tends to $0$ , hence the integral converges and equals $4$ .

Exercise 3

Compute

\int_{0}^{1} ln x d x .

\int_{t}^{1} ln x d x = [x ln x - x]_{t}^{1} = - 1 - t ln t + t

t \to 0^{+} lim t ln t = s \to + \infty lim (- \frac{ln s}{s}) = 0 (s = \frac{1}{t})

\int_{0}^{1} ln x d x = t \to 0^{+} lim (- 1 - t ln t + t) = - 1

y = \ln x

(0, 1]

the graph of

y = ln x

lies below the

x

-axis and is unbounded near

0

: the shaded area is finite and equals

1

, therefore the integral equals

- 1

The logarithm tends to $- \infty$ as $x \to 0^{+}$ , therefore the integrand is unbounded near $0$ and the integral is improper with critical point $0$ . An antiderivative of $ln x$ is $x ln x - x$ , as one checks by differentiating, since the derivative is $ln x + 1 - 1 = ln x$ . At the endpoint $1$ the antiderivative equals $- 1$ , therefore it remains to compute the limit of $t ln t$ as $t \to 0^{+}$ . With the substitution $s = \frac{1}{t}$ the product becomes $- \frac{ln s}{s}$ , which tends to $0$ by the hierarchy of infinities (chapter on limits). The integral therefore converges to the value $- 1$ ; the negative sign corresponds to the sign of the logarithm, which is negative on $(0, 1)$ .