Improper Integrals

Types of Improper Integrals

Type 1: Infinite Limits

$\int_a^\infty f(x) \, dx = \lim_{t \to \infty} \int_a^t f(x) \, dx$

$\int_{-\infty}^b f(x) \, dx = \lim_{t \to -\infty} \int_t^b f(x) \, dx$

$\int_{-\infty}^\infty f(x) \, dx = \int_{-\infty}^c f(x) \, dx + \int_c^\infty f(x) \, dx$

Type 2: Discontinuous Integrands

If $f$ has a discontinuity at $c$ in $[a, b]$:

$\int_a^b f(x) \, dx = \lim_{t \to c^-} \int_a^t f(x) \, dx + \lim_{s \to c^+} \int_s^b f(x) \, dx$

If $f$ is discontinuous at $a$:

$\int_a^b f(x) \, dx = \lim_{t \to a^+} \int_t^b f(x) \, dx$

If $f$ is discontinuous at $b$:

$\int_a^b f(x) \, dx = \lim_{t \to b^-} \int_a^t f(x) \, dx$

Convergence and Divergence

• Converges: if the limit exists and is finite
• Diverges: if the limit is $\pm\infty$ or doesn't exist

Important Result: p-Integral

$\int_1^\infty \frac{1}{x^p} \, dx = \begin{cases} \frac{1}{p-1} & \text{if } p > 1 \text{ (converges)} \\ \text{diverges} & \text{if } p \leq 1 \end{cases}$

$\int_0^1 \frac{1}{x^p} \, dx = \begin{cases} \frac{1}{1-p} & \text{if } p < 1 \text{ (converges)} \\ \text{diverges} & \text{if } p \geq 1 \end{cases}$

Comparison Test

If $0 \leq f(x) \leq g(x)$ for $x \geq a$:

• If $\int_a^\infty g(x) \, dx$ converges, then $\int_a^\infty f(x) \, dx$ converges
• If $\int_a^\infty f(x) \, dx$ diverges, then $\int_a^\infty g(x) \, dx$ diverges

Examples

Example 1: Infinite upper limit

$\int_1^\infty \frac{1}{x^2} \, dx = \lim_{t \to \infty} \int_1^t x^{-2} \, dx$

$= \lim_{t \to \infty} \left[-\frac{1}{x}\right]_1^t$

$= \lim_{t \to \infty} \left(-\frac{1}{t} + 1\right)$

$= 0 + 1 = 1$

Converges to 1.

Example 2: Infinite lower limit

$\int_{-\infty}^0 e^x \, dx = \lim_{t \to -\infty} \int_t^0 e^x \, dx$

$= \lim_{t \to -\infty} \left[e^x\right]_t^0$

$= \lim_{t \to -\infty} (1 - e^t)$

$= 1 - 0 = 1$

Converges to 1.

Example 3: Both limits infinite

$\int_{-\infty}^\infty \frac{1}{1+x^2} \, dx$

$= \int_{-\infty}^0 \frac{1}{1+x^2} \, dx + \int_0^\infty \frac{1}{1+x^2} \, dx$

$= \lim_{t \to -\infty} \left[\arctan(x)\right]_t^0 + \lim_{s \to \infty} \left[\arctan(x)\right]_0^s$

$= \left(0 - \left(-\frac{\pi}{2}\right)\right) + \left(\frac{\pi}{2} - 0\right)$

$= \pi$

Example 4: Discontinuity at endpoint

$\int_0^1 \frac{1}{\sqrt{x}} \, dx = \lim_{t \to 0^+} \int_t^1 x^{-1/2} \, dx$

$= \lim_{t \to 0^+} \left[2\sqrt{x}\right]_t^1$

$= \lim_{t \to 0^+} (2 - 2\sqrt{t})$

$= 2 - 0 = 2$

Example 5: Discontinuity in interior

$\int_0^3 \frac{1}{(x-1)^{2/3}} \, dx$

Split at $x = 1$:

$= \int_0^1 \frac{1}{(x-1)^{2/3}} \, dx + \int_1^3 \frac{1}{(x-1)^{2/3}} \, dx$

First: $\lim_{t \to 1^-} \left[3(x-1)^{1/3}\right]_0^t = \lim_{t \to 1^-} (3(t-1)^{1/3} + 3) = 3$

Second: $\lim_{s \to 1^+} \left[3(x-1)^{1/3}\right]_s^3 = 3(2)^{1/3} - 0 = 3\sqrt[3]{2}$

Total: $3 + 3\sqrt[3]{2}$

Example 6: Divergent integral

$\int_1^\infty \frac{1}{x} \, dx = \lim_{t \to \infty} \left[\ln|x|\right]_1^t$

$= \lim_{t \to \infty} (\ln(t) - 0)$

$= \infty$

Diverges.

Example 7: Using comparison

$\int_1^\infty e^{-x^2} \, dx$

Since $e^{-x^2} < e^{-x}$ for $x > 1$:

$\int_1^\infty e^{-x} \, dx = \lim_{t \to \infty} \left[-e^{-x}\right]_1^t = 0 + e^{-1} = \frac{1}{e} \text{ (converges)}$

Therefore $\int_1^\infty e^{-x^2} \, dx$ also converges.