FAD1014 L20 — Improper Integrals

Lecture covering improper integrals: types, convergence/divergence criteria, the comparison test, and worked examples.

Summary

This lecture extends the definite integral to cases where either (1) the interval of integration is infinite, or (2) the integrand has a vertical asymptote (infinite discontinuity) within or at the boundary of the integration interval. Both situations produce an improper integral, which is evaluated as a limit of proper integrals. The lecture introduces the p-test for both types, the comparison test for determining convergence/divergence, and multiple worked examples.

Key Concepts

1. Type 1: Infinite Limits of Integration

An integral is improper of Type 1 when at least one limit of integration is infinite.

Definition (Infinite Upper Limit)

If $\int_a^t f(x),dx$ exists for every $t \ge a$, then:

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

provided this limit exists (as a finite number).

Definition (Infinite Lower Limit)

If $\int_t^b f(x),dx$ exists for every $t \le b$, then:

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

Definition (Both Limits Infinite)

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

where $c$ is any real number. The integral converges only if both integrals on the right converge independently.

[!important] Convergence Terminology If the limit exists (is finite), the improper integral converges. If the limit does not exist or is infinite, the improper integral diverges.

2. Type 2: Discontinuous Integrand

An integral is improper of Type 2 when the integrand $f(x)$ has an infinite discontinuity (vertical asymptote) at a point within or at the boundary of $[a, b]$.

Definition (Discontinuity at Lower Limit)

If $f$ is continuous on $(a, b]$ and discontinuous at $a$:

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

Definition (Discontinuity at Upper Limit)

If $f$ is continuous on $[a, b)$ and discontinuous at $b$:

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

Definition (Discontinuity in the Interior)

If $f$ has a discontinuity at $c \in (a, b)$:

$$\int_a^b f(x),dx = \int_a^c f(x),dx + \int_c^b f(x),dx$$

The integral converges only if both improper integrals on the right converge independently.

3. The p-Test

Type 1 (p-Test for Infinite Limits)

$$\int_1^\infty \frac{1}{x^p},dx \quad \text{is} \quad \begin{cases} \displaystyle\text{convergent,} & p > 1 \[8pt] \displaystyle\text{divergent,} & p \le 1 \end{cases}$$

Type 2 (p-Test for Discontinuity at Zero)

$$\int_0^1 \frac{1}{x^p},dx \quad \text{is} \quad \begin{cases} \displaystyle\text{convergent,} & p < 1 \[8pt] \displaystyle\text{divergent,} & p \ge 1 \end{cases}$$

[!note] Notice the Reversal The conditions for convergence are reversed between the two types. For Type 1, the integral converges only when $p$ is large enough ($p > 1$). For Type 2, it converges only when $p$ is small enough ($p < 1$).

Worked Examples

Example 1 (Type 1 — Convergent)

Evaluate $\displaystyle \int_1^\infty \frac{1}{x^2},dx$.

Solution:

The integral is improper because the upper limit is infinite.

$$\int_1^\infty \frac{1}{x^2},dx = \lim_{t \to \infty} \int_1^t \frac{1}{x^2},dx = \lim_{t \to \infty} \left[ -\frac{1}{x} \right]1^t = \lim{t \to \infty} \left( -\frac{1}{t} + 1 \right) = 0 + 1 = 1$$

Since the limit exists and equals $1$, the integral converges to 1.

Geometrically: the region under $y = 1/x^2$ from $x = 1$ to $\infty$ has finite area, exactly $1$.

Example 2 (Type 1 — Divergent)

Evaluate $\displaystyle \int_1^\infty \frac{1}{x},dx$.

Solution:

$$\int_1^\infty \frac{1}{x},dx = \lim_{t \to \infty} \int_1^t \frac{1}{x},dx = \lim_{t \to \infty} \bigl[ \ln|x| \bigr]1^t = \lim{t \to \infty} (\ln t - 0) = \infty$$

The limit does not exist (tends to infinity), so the integral diverges.

This is the $p = 1$ case of the p-test. Contrast with Example 1 ($p = 2$), which converged. The borderline $p = 1$ diverges for Type 1.

Example 3 (Type 2 — Convergent)

Evaluate $\displaystyle \int_0^1 \frac{1}{\sqrt{x}},dx$.

Solution:

The integrand $1/\sqrt{x}$ is discontinuous at $x = 0$ (vertical asymptote). This is a Type 2 improper integral.

$$\int_0^1 \frac{1}{\sqrt{x}},dx = \lim_{t \to 0^+} \int_t^1 x^{-1/2},dx = \lim_{t \to 0^+} \bigl[ 2x^{1/2} \bigr]t^1 = \lim{t \to 0^+} (2 - 2\sqrt{t}) = 2$$

The limit exists, so the integral converges to 2.

Example 4 (Type 2 — Divergent)

Evaluate $\displaystyle \int_0^1 \frac{1}{x},dx$.

Solution:

The integrand $1/x$ is discontinuous at $x = 0$.

$$\int_0^1 \frac{1}{x},dx = \lim_{t \to 0^+} \int_t^1 \frac{1}{x},dx = \lim_{t \to 0^+} \bigl[ \ln|x| \bigr]t^1 = \lim{t \to 0^+} (0 - \ln t) = -(-\infty) = \infty$$

The integral diverges.

Example 5 (Comparison Test for Convergence)

Determine the convergence of $\displaystyle \int_1^\infty \frac{1}{x^2 + 1},dx$.

Solution (using comparison):

For $x \ge 1$, we have $x^2 + 1 \ge x^2$, so:

$$\frac{1}{x^2 + 1} \le \frac{1}{x^2}$$

From Example 1, $\displaystyle \int_1^\infty \frac{1}{x^2},dx$ converges. By the comparison test, since $0 \le \frac{1}{x^2 + 1} \le \frac{1}{x^2}$ for all $x \ge 1$ and the larger integral converges:

$$\int_1^\infty \frac{1}{x^2 + 1},dx \quad \text{converges}$$

Solution (direct evaluation):

$$\int_1^\infty \frac{1}{x^2 + 1},dx = \lim_{t \to \infty} \int_1^t \frac{1}{x^2 + 1},dx = \lim_{t \to \infty} \bigl[ \tan^{-1} x \bigr]1^t = \lim{t \to \infty} (\tan^{-1} t - \tan^{-1} 1) = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$$

The direct evaluation confirms convergence; the integral converges to $\pi/4$.

Example 6 (Comparison Test for Divergence)

Determine the convergence of $\displaystyle \int_1^\infty \frac{1}{\sqrt{x} + 1},dx$.

Solution:

For $x \ge 1$, $\sqrt{x} \ge 1$, so $\sqrt{x} + 1 \le \sqrt{x} + \sqrt{x} = 2\sqrt{x}$. Therefore:

$$\frac{1}{\sqrt{x} + 1} \ge \frac{1}{2\sqrt{x}}$$

Now, $\displaystyle \int_1^\infty \frac{1}{2\sqrt{x}},dx = \frac{1}{2}\int_1^\infty \frac{1}{x^{1/2}},dx$ has $p = \frac{1}{2} \le 1$, so it diverges (by the p-test).

Since $\frac{1}{\sqrt{x} + 1} \ge \frac{1}{2\sqrt{x}} > 0$ and the smaller integral diverges, by the comparison test:

$$\int_1^\infty \frac{1}{\sqrt{x} + 1},dx \quad \text{diverges}$$

The Comparison Test (Formal Statement)

[!theorem] Comparison Test for Improper Integrals Suppose $f$ and $g$ are continuous functions with $0 \le g(x) \le f(x)$ for all $x \ge a$.

  1. If $\displaystyle\int_a^\infty f(x),dx$ converges, then $\displaystyle\int_a^\infty g(x),dx$ also converges.
  2. If $\displaystyle\int_a^\infty g(x),dx$ diverges, then $\displaystyle\int_a^\infty f(x),dx$ also diverges.

[!tip] Memory Aid

  • A smaller positive function under a convergent integral must also converge.
  • A larger positive function over a divergent integral must also diverge.

Key Equations

Type Form Convergence Condition
Type 1 $\displaystyle\int_1^\infty \frac{1}{x^p},dx$ Converges iff $p > 1$
Type 2 $\displaystyle\int_0^1 \frac{1}{x^p},dx$ Converges iff $p < 1$
Type 1 $\displaystyle\int_a^\infty e^{-kx},dx$ ($k > 0$) Always converges to $\dfrac{e^{-ka}}{k}$
Type 1 $\displaystyle\int_1^\infty \frac{\ln x}{x^p},dx$ Converges iff $p > 1$

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