Improper Integrals
Improper Integrals
An improper integral is a definite integral where either the interval of integration is infinite, or the integrand is unbounded (has a discontinuity) within the interval. Despite their name, many improper integrals have perfectly well-defined finite values.
Types of Improper Integral
Type 1: Infinite limits of integration
- ∫₁^∞ 1/x² dx (upper limit is infinite)
- ∫₋∞^0 eˣ dx (lower limit is infinite)
- ∫₋∞^∞ e^(−x²) dx (both limits infinite)
Type 2: Discontinuous integrand
- ∫₀^1 1/√x dx (integrand unbounded at x = 0)
- ∫₀^1 ln x dx (integrand unbounded at x = 0)
- ∫₀^π tan x dx (integrand unbounded at x = π/2, inside the interval)
Evaluating Type 1: Infinite Limits
Replace the infinite limit with a finite variable, evaluate the integral, then take the limit.
∫ₐ^∞ f(x) dx = lim(t→∞) ∫ₐ^t f(x) dx
If the limit exists and is finite, the integral converges. Otherwise, it diverges.
Example 1: ∫₁^∞ 1/x² dx
- = lim(t→∞) [−1/x]₁^t = lim(t→∞) (−1/t + 1) = 1 ✓ (converges)
Example 2: ∫₁^∞ 1/x dx
- = lim(t→∞) [ln x]₁^t = lim(t→∞) ln t = ∞ ✗ (diverges)
Example 3: ∫₁^∞ 1/x^p dx converges if and only if p > 1.
- For p ≠ 1: = lim(t→∞) [x^(1−p)/(1−p)]₁^t
- If p > 1: exponent 1−p < 0, so t^(1−p) → 0, giving 1/(p−1)
- If p < 1: exponent 1−p > 0, so t^(1−p) → ∞, diverges
This is the p-test, one of the most important convergence results.
Evaluating Type 2: Discontinuous Integrand
Replace the problematic endpoint with a variable and take the limit.
If f is unbounded at x = a:
∫ₐ^b f(x) dx = lim(ε→0⁺) ∫_(a+ε)^b f(x) dx
Example: ∫₀^1 1/√x dx
- = lim(ε→0⁺) [2√x]_ε^1 = lim(ε→0⁺) (2 − 2√ε) = 2 ✓ (converges)
If f is unbounded at an interior point c (a < c < b):
Split into two integrals: ∫ₐ^c f(x) dx + ∫_c^b f(x) dx, and evaluate each as a separate improper integral. Both must converge for the whole integral to converge.
Key Convergence Tests
| Test | Statement |
|---|---|
| p-test (at ∞) | ∫₁^∞ 1/x^p dx converges ⟺ p > 1 |
| p-test (at 0) | ∫₀^1 1/x^p dx converges ⟺ p < 1 |
| Comparison | If 0 ≤ f(x) ≤ g(x) and ∫g converges, then ∫f converges |
| Comparison | If f(x) ≥ g(x) ≥ 0 and ∫g diverges, then ∫f diverges |
| Limit comparison | If f, g > 0 and lim f(x)/g(x) = L (0 < L < ∞), then ∫f and ∫g both converge or both diverge |
Important Convergent Improper Integrals
| Integral | Value |
|---|---|
| ∫₀^∞ e^(−x) dx | 1 |
| ∫₀^∞ e^(−x²) dx | √π/2 (the Gaussian integral) |
| ∫₀^∞ x^(n−1) e^(−x) dx | (n−1)! = Γ(n) |
| ∫₀^1 ln x dx | −1 |
| ∫₀^1 x^p ln x dx (p > −1) | −1/(p+1)² |
Comparison Test: Worked Examples
Show ∫₁^∞ e^(−x²) dx converges.
For x ≥ 1: x² ≥ x, so e^(−x²) ≤ e^(−x).
Since ∫₁^∞ e^(−x) dx = 1/e (converges), by comparison, ∫₁^∞ e^(−x²) dx also converges.
Show ∫₁^∞ 1/(x² + sin²x) dx converges.
x² + sin²x ≥ x², so 1/(x² + sin²x) ≤ 1/x².
Since ∫₁^∞ 1/x² dx converges (p = 2 > 1), by comparison the integral converges.
Gamma Function (Extension)
The Gamma function Γ(n) = ∫₀^∞ x^(n−1) e^(−x) dx extends the factorial to non-integers:
- Γ(n) = (n−1)! for positive integers n
- Γ(1/2) = √π
- Γ(n+1) = nΓ(n) (the recurrence relation)
Common Mistakes
- Forgetting to check for discontinuities inside the interval — always look for where the denominator is zero or the function is undefined
- Writing ∫₋₁^1 1/x² dx = [−1/x]₋₁^1 = −1 − 1 = −2. This is wrong! The integrand has a discontinuity at x = 0, and both halves diverge, so the integral diverges.
- Applying the fundamental theorem of calculus without checking continuity on the entire interval
Exam Tips
- Always explicitly write the limit definition — don't skip to the antiderivative
- State whether the integral converges or diverges as a clear conclusion
- For the p-test, remember the boundary is p = 1 (and p = 1 diverges in both cases)
- When using comparison, state which function bounds which and confirm the comparator converges/diverges
- If both limits are infinite, split at any convenient point (often x = 0) and handle each half separately