Comparison Test for Improper Integrals for MATH 141
Exam Relevance for MATH 141
Comparison for improper integrals shows up 1-2 times per MATH 141 exam. Know comparisons with 1/x^p.
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Understanding the Comparison Test for Improper Integrals
Sometimes you need to know if an improper integral converges or diverges, but actually computing it is difficult or impossible. For example, $\int_1^{\infty} e^{-x^2} \, dx$ has no elementary antiderivative!
The comparison test lets you determine convergence without computing the integral. The idea: compare your integral to one you already know converges or diverges.
The Direct Comparison Test
Suppose $f(x) \geq 0$ and $g(x) \geq 0$ on the interval.
For convergence: If $0 \leq f(x) \leq g(x)$ and $\int g(x) \, dx$ converges, then $\int f(x) \, dx$ also converges.
For divergence: If $f(x) \geq g(x) \geq 0$ and $\int g(x) \, dx$ diverges, then $\int f(x) \, dx$ also diverges.
In plain English:
- Smaller than a convergent integral → converges
- Bigger than a divergent integral → diverges
Common Comparison Functions
⚠️ Important: The p-test works opposite ways for $[1, \infty)$ vs $[0, 1]$!
For integrals from 1 to $\infty$ (infinite limit)
| Integral | Behavior |
|---|---|
| $\int_1^{\infty} \frac{1}{x^p} \, dx$ | Converges if $p > 1$, diverges if $p \leq 1$ |
| $\int_1^{\infty} \frac{1}{x^2} \, dx$ | Converges |
| $\int_1^{\infty} \frac{1}{x} \, dx$ | Diverges |
| $\int_1^{\infty} \frac{1}{\sqrt{x}} \, dx$ | Diverges |
| $\int_1^{\infty} e^{-x} \, dx$ | Converges |
For integrals from 0 to 1 (discontinuity at 0)
| Integral | Behavior |
|---|---|
| $\int_0^{1} \frac{1}{x^p} \, dx$ | Converges if $p < 1$, diverges if $p \geq 1$ |
| $\int_0^{1} \frac{1}{\sqrt{x}} \, dx$ | Converges ($p = \frac{1}{2} < 1$) |
| $\int_0^{1} \frac{1}{x} \, dx$ | Diverges ($p = 1$) |
| $\int_0^{1} \frac{1}{x^2} \, dx$ | Diverges ($p = 2 > 1$) |
Why the difference?
- At $\infty$: larger powers shrink faster → converge
- At $0$: larger powers blow up faster → diverge
Problem: Does $\int_1^{\infty} \frac{1}{x^2 + 1} \, dx$ converge or diverge?
Step 1: Find a comparison function
For large $x$: $x^2 + 1 > x^2$
So: $\frac{1}{x^2 + 1} < \frac{1}{x^2}$
Step 2: Check the comparison integral
We know $\int_1^{\infty} \frac{1}{x^2} \, dx$ converges (p-test, $p = 2 > 1$).
Step 3: Apply comparison test
Since $0 < \frac{1}{x^2+1} < \frac{1}{x^2}$ and the bigger one converges...
$$\boxed{\text{Converges by comparison with } \frac{1}{x^2}}$$
Problem: Does $\int_1^{\infty} \frac{1}{\sqrt{x} + 1} \, dx$ converge or diverge?
Step 1: Find a comparison function
For $x \geq 1$: $\sqrt{x} + 1 \leq \sqrt{x} + \sqrt{x} = 2\sqrt{x}$
So: $\frac{1}{\sqrt{x} + 1} \geq \frac{1}{2\sqrt{x}}$
Step 2: Check the comparison integral
We know $\int_1^{\infty} \frac{1}{2\sqrt{x}} \, dx$ diverges.
(It's $\frac{1}{2} \int_1^{\infty} \frac{1}{x^{1/2}} \, dx$, and $p = \frac{1}{2} < 1$.)
Step 3: Apply comparison test
Since $\frac{1}{\sqrt{x}+1} \geq \frac{1}{2\sqrt{x}} > 0$ and the smaller one diverges...
$$\boxed{\text{Diverges by comparison with } \frac{1}{\sqrt{x}}}$$
The Limit Comparison Test
Sometimes direct comparison is tricky. The limit comparison test is often easier:
If $f(x) > 0$ and $g(x) > 0$ for $x \geq a$, and
$$\lim_{x \to \infty} \frac{f(x)}{g(x)} = L$$
where $0 < L < \infty$ (a finite positive number), then:
$$\int_a^{\infty} f(x) \, dx \text{ and } \int_a^{\infty} g(x) \, dx \text{ both converge or both diverge}$$
In plain English: If two functions have the same "growth rate" at infinity, their integrals behave the same way.
Problem: Does $\int_1^{\infty} \frac{x}{x^3 + 5} \, dx$ converge or diverge?
Step 1: Identify dominant behavior
For large $x$: $\frac{x}{x^3 + 5} \approx \frac{x}{x^3} = \frac{1}{x^2}$
Let's compare to $g(x) = \frac{1}{x^2}$.
Step 2: Compute the limit
$$\lim_{x \to \infty} \frac{f(x)}{g(x)} = \lim_{x \to \infty} \frac{\frac{x}{x^3+5}}{\frac{1}{x^2}}$$
$$= \lim_{x \to \infty} \frac{x \cdot x^2}{x^3 + 5}$$
$$= \lim_{x \to \infty} \frac{x^3}{x^3 + 5}$$
$$= \lim_{x \to \infty} \frac{1}{1 + \frac{5}{x^3}} = 1$$
Step 3: Conclude
Since the limit is $1$ (finite and positive) and $\int_1^{\infty} \frac{1}{x^2} \, dx$ converges...
$$\boxed{\text{Converges by limit comparison with } \frac{1}{x^2}}$$
Problem: Does $\int_2^{\infty} \frac{1}{\sqrt{x^2 - 1}} \, dx$ converge or diverge?
Step 1: Identify dominant behavior
For large $x$: $\sqrt{x^2 - 1} \approx \sqrt{x^2} = x$
So $\frac{1}{\sqrt{x^2-1}} \approx \frac{1}{x}$.
Compare to $g(x) = \frac{1}{x}$.
Step 2: Compute the limit
$$\lim_{x \to \infty} \frac{\frac{1}{\sqrt{x^2-1}}}{\frac{1}{x}} = \lim_{x \to \infty} \frac{x}{\sqrt{x^2-1}}$$
$$= \lim_{x \to \infty} \frac{x}{x\sqrt{1 - \frac{1}{x^2}}}$$
$$= \lim_{x \to \infty} \frac{1}{\sqrt{1 - \frac{1}{x^2}}} = 1$$
Step 3: Conclude
Since the limit is $1$ and $\int_2^{\infty} \frac{1}{x} \, dx$ diverges...
$$\boxed{\text{Diverges by limit comparison with } \frac{1}{x}}$$
Problem: Does $\int_1^{\infty} e^{-x^2} \, dx$ converge or diverge?
Step 1: Find a comparison
For $x \geq 1$: $x^2 \geq x$, so $-x^2 \leq -x$
Therefore: $e^{-x^2} \leq e^{-x}$
Step 2: Check the comparison integral
$$\int_1^{\infty} e^{-x} \, dx = \lim_{t \to \infty} [-e^{-x}]_1^t$$
$$= \lim_{t \to \infty} (-e^{-t} + e^{-1}) = e^{-1}$$
This converges!
Step 3: Conclude
Since $0 < e^{-x^2} \leq e^{-x}$ and $\int_1^{\infty} e^{-x} \, dx$ converges...
$$\boxed{\text{Converges by comparison with } e^{-x}}$$
Problem: Does $\int_0^{1} \frac{1}{\sqrt{x + x^2}} \, dx$ converge or diverge?
Step 1: Identify the problem
As $x \to 0^+$: $\sqrt{x + x^2} \to 0$, so the integrand blows up.
Step 2: Find dominant behavior near 0
For small $x$: $x + x^2 \approx x$ (the $x^2$ term is tiny)
So: $\frac{1}{\sqrt{x + x^2}} \approx \frac{1}{\sqrt{x}}$
Step 3: Check the comparison integral
$\int_0^{1} \frac{1}{\sqrt{x}} \, dx$ converges (p-test at 0: $p = \frac{1}{2} < 1$).
Step 4: Make it rigorous
For $0 < x \leq 1$: $x + x^2 \geq x$
So: $\frac{1}{\sqrt{x + x^2}} \leq \frac{1}{\sqrt{x}}$
Since our function is smaller than a convergent integral...
$$\boxed{\text{Converges by comparison with } \frac{1}{\sqrt{x}}}$$
Problem: Does $\int_0^{1} \frac{1}{x\sqrt{x+1}} \, dx$ converge or diverge?
Step 1: Identify the problem
As $x \to 0^+$: we have $\frac{1}{x \cdot 1} = \frac{1}{x} \to \infty$.
Step 2: Find a comparison
For $0 < x \leq 1$: $\sqrt{x+1} \leq \sqrt{2}$
So: $\frac{1}{x\sqrt{x+1}} \geq \frac{1}{x\sqrt{2}} = \frac{1}{\sqrt{2}} \cdot \frac{1}{x}$
Step 3: Check the comparison integral
$\int_0^{1} \frac{1}{x} \, dx$ diverges (p-test at 0: $p = 1$).
Step 4: Conclude
Since our function is larger than a divergent integral...
$$\boxed{\text{Diverges by comparison with } \frac{1}{x}}$$
Problem: Does $\int_0^{1} \frac{1}{\sqrt{x}(1 + x)} \, dx$ converge or diverge?
Step 1: Identify dominant behavior near 0
As $x \to 0$: $\frac{1}{\sqrt{x}(1+x)} \approx \frac{1}{\sqrt{x}}$
Step 2: Compute the limit
$$\lim_{x \to 0^+} \frac{\frac{1}{\sqrt{x}(1+x)}}{\frac{1}{\sqrt{x}}} = \lim_{x \to 0^+} \frac{1}{1+x} = 1$$
Step 3: Conclude
The limit is 1 (finite and positive).
Since $\int_0^{1} \frac{1}{\sqrt{x}} \, dx$ converges ($p = \frac{1}{2} < 1$)...
$$\boxed{\text{Converges by limit comparison with } \frac{1}{\sqrt{x}}}$$
Quick Strategy Guide
For $\int_1^{\infty}$ (infinite limit):
- Look at dominant terms for large $x$
- Simplify: $\frac{x^2 + 3x}{x^4 + 1} \approx \frac{x^2}{x^4} = \frac{1}{x^2}$
- Remember: $p > 1$ converges, $p \leq 1$ diverges
For $\int_0^{1}$ (discontinuity at 0):
- Look at dominant terms for small $x$
- Simplify: $\frac{1}{\sqrt{x + x^2}} \approx \frac{1}{\sqrt{x}}$
- Remember: $p < 1$ converges, $p \geq 1$ diverges (opposite!)
General tips:
- Use limit comparison when direct comparison is awkward
- For exponentials: $e^{-x}$ and faster decay always converge
Common Mistakes and Misunderstandings
❌ Mistake: Comparing in the wrong direction
Wrong: "$\frac{1}{x^2+1} > \frac{1}{x^3}$ and $\int \frac{1}{x^3}$ converges, so our integral converges."
Why it's wrong: Being bigger than something convergent tells you nothing! You need to be smaller than a convergent integral.
Correct: Compare to something larger that still converges, like $\frac{1}{x^2}$.
❌ Mistake: Forgetting positivity requirement
Wrong: Using comparison test on functions that can be negative.
Why it's wrong: The comparison test requires $f(x) \geq 0$ and $g(x) \geq 0$.
Correct: Make sure both functions are non-negative on the interval.
❌ Mistake: Getting a limit of 0 or $\infty$ in limit comparison
What happens:
- If $\lim \frac{f}{g} = 0$: only know $f$ converges if $g$ converges
- If $\lim \frac{f}{g} = \infty$: only know $f$ diverges if $g$ diverges
Better approach: Choose a different comparison function so the limit is finite and positive.
Direct Comparison Test (Convergence)
If your function is smaller than a convergent integral, it also converges. Think: 'squeezed under something finite.'
Variables:
- $f(x)$:
- the function you're testing
- $g(x)$:
- a larger function with known convergent integral
Direct Comparison Test (Divergence)
If your function is bigger than a divergent integral, it also diverges. Think: 'at least as big as infinity.'
Variables:
- $f(x)$:
- the function you're testing
- $g(x)$:
- a smaller function with known divergent integral
Limit Comparison Test
If the limit is a finite positive number, then both integrals converge or both diverge. Works for limits at infinity OR at a discontinuity.
Variables:
- $f(x)$:
- the function you're testing
- $g(x)$:
- a function with known behavior
- $L$:
- the limit (must be finite and positive)
- $a$:
- infinity or the point of discontinuity
p-Test (1 to ∞)
Converges if p > 1, diverges if p ≤ 1. Use for integrals with infinite upper limit.
Variables:
- $p$:
- the exponent (p > 1 converges, p ≤ 1 diverges)
p-Test (0 to 1)
Converges if p < 1, diverges if p ≥ 1. OPPOSITE of the infinity case! Use for integrals with discontinuity at 0.
Variables:
- $p$:
- the exponent (p < 1 converges, p ≥ 1 diverges)
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