Improper Integrals
Improper integrals extend the idea of a definite integral to cases where the integrand or the interval of integration behaves badly — the function becomes infinite, or the interval is unbounded.
Despite these issues, many improper integrals still have well-defined finite values using limits.
What Are Improper Integrals?
A definite integral is called improper when either:
The interval of integration is infinite
Examples:
-
$$\int_{1}^{\infty}\frac{1}{x^{2}}dx$$
-
$$\int_{-\infty}^{5}e^{x}dx$$
The integrand becomes infinite within the interval
Examples:
-
$$\int_{0}^{1}\frac{1}{\sqrt{x}}$$
-
$$\int_{-1}^{1}\frac{1}{x^{2}}dx$$(blows up at x=0 inside the interval)
Improper Integrals with Infinite Limits
These are evaluated using limits:
Case 1: Infinite Upper Limit
$$\int_{a}^{\infty}f(x)dx=\lim_{t\to\infty}\int_{a}^{t}f(x)dx$$
Case 2: Infinite Lower Limit
$$\int_{-\infty}^{a}f(x)dx=\lim_{t\to-\infty}\int_{t}^{a}f(x)dx$$
Case 3: Infinite on Both Sides
$$\int_{-\infty}^{\infty}f(x)dx=\lim_{t\to-\infty}\int_{t}^{c}f(x)dx+\lim_{k\to\infty}\int_{c}^{k}f(x)dx$$
Convergence vs Divergence
An improper integral:
-
Converges if the limiting value exists and is finite.
-
Diverges if the limit does not exist or is infinite.
Example: Infinite Interval
$$\int_{1}^{\infty}\frac{1}{x^{2}}dx$$
Solve:
$$\lim_{t\to\infty}\int_{1}^{t}x^{-2}dx=\lim_{t\to\infty}\left[-\frac{1}{x}\right]_{1}^{t}=\lim_{t\to\infty}\left(-\frac{1}{t}+1\right)$$
Then:
$$\lim_{t\to\infty}\left(-\frac{1}{t}+1\right)=1$$
Converges
Value = 1
Improper Integrals with Vertical Asymptotes (Infinite Discontinuities)
If the integrand blows up at a point, use limit from left/right.
Case 1: Discontinuity at Lower Limit
Example:
$$\int_{0}^{1}\frac{1}{\sqrt{x}}dx$$
Rewrite using limits:
$$\lim_{t\to0^{+}}\int_{t}^{1}x^{-1/2}dx$$
Compute:
$$=$$
📌 Convergent because the blow-up is mild.
Case 2: Discontinuity at Upper Limit
$$\int_{2}^{3}\frac{1}{(x-2)^{3/2}}dx$$
then let t approach to $$2^{+}$$
$$\int_{2}^{3}\frac{1}{(x-2)^{3/2}}dx=\lim_{t\to2^{+}}\int_{t}^{3}(x-2)^{-3/2}dx$$
Evaluate:
At the lower limit, $$(t-2)^{-1/2} \longrightarrow \infty$$. so → ∞.
❌ Divergent
Interior Discontinuities
If the integrand is undefined at a point inside [a,b][a,b], you must break the integral.
Example:
$$\int_{0}^{2}\frac{1}{x-1}dx$$
Split at the discontinuity:
$$\int_{0}^{1}\frac{1}{x-1}dx+\int_{1}^{2}\frac{1}{x-1}dx$$
Both pieces diverge.
So the whole integral ➜ diverges.