Comparison Test for Improper Integrals

TL;DR

The Comparison Test helps you figure out if an improper integral converges or diverges by comparing it to another integral you already understand. If your integral is "smaller" than a known convergent integral, it converges; if it's "bigger" than a known divergent integral, it diverges. This is especially useful when direct integration is too hard.

1. The Mental Model

Imagine you have two functions, one clearly above the other. If the "bigger" function has a finite area under its curve (converges), then the "smaller" one must also have a finite area. Conversely, if the "smaller" function has an infinite area (diverges), the "bigger" one must also have an infinite area.

2. The Core Material

The Comparison Test is a powerful tool for improper integrals, particularly those from $a$ to $\infty$ or when there's an infinite discontinuity within the integration interval. It works best when your integral looks similar to a p-integral or an exponential integral, whose convergence/divergence properties are already known.

Conditions for the Comparison Test

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For the test to work, you need two functions, $f(x)$ and $g(x)$, that are continuous and positive over the interval $[a, \infty)$.

• Case 1: Convergence
  If $0 \le f(x) \le g(x)$ for all $x \ge a$, AND $\int_a^\infty g(x) \, dx$ converges, then $\int_a^\infty f(x) \, dx$ also converges. Think of $f(x)$ as "trapped" beneath a finite area.

• Case 2: Divergence
  If $0 \le g(x) \le f(x)$ for all $x \ge a$, AND $\int_a^\infty g(x) \, dx$ diverges, then $\int_a^\infty f(x) \, dx$ also diverges. Here, $f(x)$ is "bigger" than an infinite area, so its own area must also be infinite.

Reference Integrals (The ones you compare against)

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1. p-integrals: $\int_1^\infty \frac{1}{x^p} \, dx$

   • Converges if $p > 1$.
   • Diverges if $p \le 1$.
     (Note: The lower limit can be any $a \ge 1$, or even $a > 0$ if the function is well-behaved at $x=1$).

2. Exponential integrals: $\int_a^\infty e^{-kx} \, dx$

   • Converges if $k > 0$.
   • Diverges if $k \le 0$.

Strategy for Applying the Comparison Test

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graph TD
    A["Identify the improper integral"] --> B{"Is direct integration easy?"};
    B -- No --> C["Find a 'similar' function g(x)"];
    B -- Yes --> Z["Integrate directly"];
    C --> D{"Choose g(x) whose convergence is known (p-integral, exponential)"};
    D --> E{"Establish inequality: 0 <= f(x) <= g(x) OR 0 <= g(x) <= f(x)"};
    E --> F{"Does known integral g(x) converge or diverge?"};
    F -- g(x) converges AND f(x) <= g(x) --> G["Integral f(x) converges"];
    F -- g(x) diverges AND g(x) <= f(x) --> H["Integral f(x) diverges"];
    F -- Other case (e.g., f(x) > g(x) and g(x) converges) --> I["Comparison Test inconclusive. Try another g(x) or another test."];

The key is often to simplify the integrand $f(x)$ by dropping "less significant" terms in the denominator or numerator as $x \to \infty$.

For example, if you have $\int_1^\infty \frac{x}{x^3 + 1} \, dx$, for large $x$, $\frac{x}{x^3 + 1}$ behaves like $\frac{x}{x^3} = \frac{1}{x^2}$. Since $\int_1^\infty \frac{1}{x^2} \, dx$ converges (p=2 > 1), you'd try to show that $\frac{x}{x^3 + 1} \le \frac{1}{x^2}$.

3. Worked Example

Let's determine if $\int_1^\infty \frac{\sin^2(x)}{x^2} \, dx$ converges or diverges.

1. Identify $f(x)$: Here, $f(x) = \frac{\sin^2(x)}{x^2}$.
2. Check conditions: For $x \ge 1$, $\sin^2(x) \ge 0$, so $f(x) \ge 0$. It's continuous for $x \ge 1$.
3. Find a comparable $g(x)$: We know that $0 \le \sin^2(x) \le 1$ for all $x$.
   So, if we divide by $x^2$ (which is positive for $x \ge 1$), we get:
   $0 \le \frac{\sin^2(x)}{x^2} \le \frac{1}{x^2}$.
   Let $g(x) = \frac{1}{x^2}$.
4. Evaluate $\int_1^\infty g(x) \, dx$: This is a p-integral with $p=2$.
   Since $p=2 > 1$, the integral $\int_1^\infty \frac{1}{x^2} \, dx$ converges.
5. Apply the Comparison Test: We have $0 \le f(x) \le g(x)$, and $\int_1^\infty g(x) \, dx$ converges.
   Therefore, by the Comparison Test, $\int_1^\infty \frac{\sin^2(x)}{x^2} \, dx$ converges.

4. Key Takeaways

• The Comparison Test is for improper integrals where direct integration is difficult.
• You need to compare your function $f(x)$ to a known function $g(x)$ (like p-integrals or exponential integrals).
• Ensure both $f(x)$ and $g(x)$ are positive and continuous on the interval of integration.
• For convergence, $f(x)$ must be smaller than a known convergent $g(x)$.
• For divergence, $f(x)$ must be larger than a known divergent $g(x)$.
• If your inequality goes the "wrong way" (e.g., $f(x) \ge g(x)$ and $g(x)$ converges), the test is inconclusive.
• For large $x$, focus on the dominant terms in $f(x)$ to choose your $g(x)$.

Common Mistakes to Avoid

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• Not checking positivity: Both functions must be non-negative on the integration interval.
• Incorrect inequalities: Make sure you establish the correct $f(x) \le g(x)$ or $g(x) \le f(x)$ relationship.
• Using an inconclusive comparison: If $f(x)$ is larger than a convergent integral, or smaller than a divergent integral, the test tells you nothing. You need to try a different comparison or a different test (like the Limit Comparison Test).
• Misidentifying p-values: Double-check whether $p > 1$ (converges) or $p \le 1$ (diverges) for p-integrals.

5. Now Try It

Determine if the integral $\int_2^\infty \frac{1}{\sqrt{x^3 - x}} \, dx$ converges or diverges using the Comparison Test.

What to do:
1. Identify $f(x)$.
2. Find a simpler $g(x)$ that behaves similarly for large $x$.
3. Establish an inequality between $f(x)$ and $g(x)$ that's useful for the Comparison Test.
4. Evaluate the convergence of $\int_2^\infty g(x) \, dx$.
5. State your conclusion for $\int_2^\infty f(x) \, dx$.

What success looks like: You correctly choose $g(x) = \frac{1}{x^{3/2}}$ (or something similar), correctly establish $f(x) \le g(x)$ for $x \ge 2$, determine that $\int_2^\infty g(x) \, dx$ converges (because $p = 3/2 > 1$), and conclude that the original integral converges.