An integral is improper when the interval of integration is unbounded, or when the integrand has a vertical asymptote within (or at the boundary of) the interval.
$$\int_a^{\infty} f(x)\,dx = \lim_{b \to \infty} \int_a^b f(x)\,dx$$
$$\int_{-\infty}^b f(x)\,dx = \lim_{a \to -\infty} \int_a^b f(x)\,dx$$
$$\int_{-\infty}^{\infty} f(x)\,dx = \int_{-\infty}^c f(x)\,dx + \int_c^{\infty} f(x)\,dx$$
Convergence and divergence:
| $\int_1^\infty x^{-p}\,dx$ | Converges if $p > 1$, diverges if $p \leq 1$ |
|---|---|
Example 1: Evaluate $\displaystyle\int_0^\infty e^{-2x}\,dx$.
$$= \lim_{b\to\infty}\left[-\frac{e^{-2x}}{2}\right]0^b = \lim{b\to\infty}\left(-\frac{e^{-2b}}{2}+\frac{1}{2}\right) = \frac{1}{2}$$
Example 2: Evaluate $\displaystyle\int_1^\infty \frac{1}{x}\,dx$.
$$= \lim_{b\to\infty}[\ln x]1^b = \lim{b\to\infty}\ln b = \infty. \quad \text{Diverges.}$$
Example 3: Evaluate $\displaystyle\int_{-\infty}^{\infty} \frac{1}{1+x^2}\,dx$.
$$= \lim_{a\to-\infty}\int_a^0 \frac{dx}{1+x^2} + \lim_{b\to\infty}\int_0^b \frac{dx}{1+x^2}$$
$$= [\arctan x]_{-\infty}^0 + [\arctan x]_0^\infty = (0-(-\pi/2)) + (\pi/2 - 0) = \pi$$
If $f$ is unbounded at $x = a$ (left endpoint):
$$\int_a^b f(x)\,dx = \lim_{c\to a^+}\int_c^b f(x)\,dx$$
Example 4: $\displaystyle\int_0^1 \frac{1}{\sqrt{1-x}}\,dx$.
Integrand $\to \infty$ as $x \to 1^-$.
$$= \lim_{c\to1^-}\int_0^c (1-x)^{-1/2}\,dx = \lim_{c\to1^-}\left[-2\sqrt{1-x}\right]0^c = \lim{c\to1^-}(-2\sqrt{1-c}+2) = 2$$
Example 5: $\displaystyle\int_0^1 \frac{1}{x}\,dx$.
$$= \lim_{c\to0^+}[\ln x]c^1 = \lim{c\to0^+}(0-\ln c) = \infty. \quad \text{Diverges.}$$
Before evaluating, identify:
1. Are limits finite? (If not, Type 1)
2. Does the integrand have a discontinuity in $[a,b]$? (Type 2)
3. If both, split appropriately.
Example 6: $\displaystyle\int_0^\infty \frac{e^{-x}}{\sqrt{x}}\,dx$ is improper at both ends ($x=0$ and $\infty$).
Split: $\displaystyle\int_0^1 \frac{e^{-x}}{\sqrt{x}}\,dx + \int_1^\infty \frac{e^{-x}}{\sqrt{x}}\,dx$.
Both converge (first by comparison with $x^{-1/2}$; second because $e^{-x}$ decays fast). Total $= \sqrt{\pi}$.
KEY TAKEAWAY: Always convert an improper integral into a limit before evaluating. Convergence and divergence must be established, not assumed.
EXAM TIP: After evaluating the antiderivative, write the limit explicitly before substituting. The examiner awards marks for the limit step.
COMMON MISTAKE: Writing $\int_1^\infty x^{-1}\,dx = [\ln x]1^\infty = \infty - 0$ without the limit notation. Always use $\lim{b\to\infty}$ explicitly.
