An improper integral is a definite integral where either the interval of integration is infinite (e.g., $\int_a^\infty f(x) dx$) or the integrand has an infinite discontinuity within the interval (e.g., $\int_0^1 \frac{1}{x} dx$). We evaluate improper integrals by taking limits of proper integrals.
🎯 In this section you will learn
- Improper integrals of Type I: infinite limits
- Improper integrals of Type II: unbounded integrands
- How to evaluate improper integrals using limits
- Convergence vs. divergence (when the limit exists as a finite number)
📌 Type I: Infinite Limits of Integration
$$ \int_a^\infty f(x) dx = \lim_{b \to \infty} \int_a^b f(x) dx $$
$$ \int_{-\infty}^b f(x) dx = \lim_{a \to -\infty} \int_a^b f(x) dx $$
$$ \int_{-\infty}^\infty f(x) dx = \int_{-\infty}^c f(x) dx + \int_c^\infty f(x) dx $$
💡 Convergence vs. Divergence
If the limit exists as a finite number, the improper integral converges. If the limit is infinite or does not exist, the improper integral diverges.
Evaluate $\displaystyle \int_1^\infty \frac{1}{x^2} dx$.
①
Write as a limit
$\int_1^\infty \frac{1}{x^2} dx = \lim_{b \to \infty} \int_1^b x^{-2} dx$
②
Evaluate the integral
$\int_1^b x^{-2} dx = \left[ -x^{-1} \right]_1^b = -\frac{1}{b} + 1$
③
Take the limit
$\lim_{b \to \infty} \left(1 - \frac{1}{b}\right) = 1$
④
Conclusion
$\int_1^\infty \frac{1}{x^2} dx = 1$ (converges)
Evaluate $\displaystyle \int_1^\infty \frac{1}{x} dx$.
- $\int_1^b \frac{1}{x} dx = [\ln x]_1^b = \ln b$
- $\lim_{b \to \infty} \ln b = \infty$ (diverges)
Evaluate $\displaystyle \int_{-\infty}^\infty \frac{1}{1+x^2} dx$.
①
Split at a convenient point (e.g., $0$)
$\int_{-\infty}^\infty \frac{1}{1+x^2} dx = \int_{-\infty}^0 \frac{1}{1+x^2} dx + \int_0^\infty \frac{1}{1+x^2} dx$
②
Evaluate $\int_0^\infty \frac{1}{1+x^2} dx$
$\lim_{b \to \infty} \int_0^b \frac{1}{1+x^2} dx = \lim_{b \to \infty} [\arctan x]_0^b = \frac{\pi}{2}$
③
By symmetry, $\int_{-\infty}^0 \frac{1}{1+x^2} dx = \frac{\pi}{2}$
Total = $\frac{\pi}{2} + \frac{\pi}{2} = \pi$
📌 Type II: Discontinuous Integrand
If $f(x)$ is unbounded at $x = a$, $x = b$, or at some $c$ in $(a, b)$:
$$ \int_a^b f(x) dx = \lim_{t \to a^+} \int_t^b f(x) dx \quad \text{(if unbounded at $a$)} $$
$$ \int_a^b f(x) dx = \lim_{t \to b^-} \int_a^t f(x) dx \quad \text{(if unbounded at $b$)} $$
Evaluate $\displaystyle \int_0^1 \frac{1}{\sqrt{x}} dx$.
①
Write as a limit (unbounded at $x=0$)
$\int_0^1 \frac{1}{\sqrt{x}} dx = \lim_{t \to 0^+} \int_t^1 x^{-1/2} dx$
②
Evaluate the integral
$\int_t^1 x^{-1/2} dx = \left[ 2x^{1/2} \right]_t^1 = 2 - 2\sqrt{t}$
③
Take the limit
$\lim_{t \to 0^+} (2 - 2\sqrt{t}) = 2$
④
Conclusion
$\int_0^1 \frac{1}{\sqrt{x}} dx = 2$ (converges)
Evaluate $\displaystyle \int_0^1 \frac{1}{x} dx$.
- $\int_t^1 \frac{1}{x} dx = [\ln x]_t^1 = -\ln t$
- $\lim_{t \to 0^+} (-\ln t) = \infty$ (diverges)
Evaluate $\displaystyle \int_0^2 \frac{1}{(x-1)^2} dx$.
①
Split at the discontinuity $x=1$
$\int_0^2 \frac{1}{(x-1)^2} dx = \int_0^1 \frac{1}{(x-1)^2} dx + \int_1^2 \frac{1}{(x-1)^2} dx$
②
Evaluate the first part
$\lim_{t \to 1^-} \int_0^t \frac{1}{(x-1)^2} dx = \lim_{t \to 1^-} \left[ -\frac{1}{x-1} \right]_0^t = \lim_{t \to 1^-} \left( -\frac{1}{t-1} + 1 \right) = \infty$
③
Conclusion
Since one part diverges, the whole integral diverges.
⚠️ Common Mistakes
- Forgetting the limit: Always write improper integrals as limits before evaluating.
- Ignoring discontinuities: If the integrand is unbounded inside the interval, split the integral at the discontinuity.
- Assuming convergence: Many improper integrals diverge — always check the limit.
- Mixing up limits: For Type II, the limit approaches the point of discontinuity from the direction where the function is defined.
🔍 Key Takeaways
- Type I: Infinite limits → $\int_a^\infty f(x) dx = \lim_{b \to \infty} \int_a^b f(x) dx$.
- Type II: Unbounded integrand → $\int_a^b f(x) dx = \lim_{t \to a^+} \int_t^b f(x) dx$ (if unbounded at $a$).
- If the limit is finite, the improper integral converges; otherwise, it diverges.
- For discontinuities inside the interval, split the integral at the discontinuity.
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