1. Introduction: Why Convergence Tests?

Many improper integrals cannot be evaluated explicitly using elementary functions. For example:

No Elementary Antiderivative

\int_1^{+\infty} \frac{\sin x}{x}\,dx

Cannot find F(x) in closed form

Complex Expression

\int_0^{+\infty} \frac{e^{-x^2}}{1+x^4}\,dx

Explicit evaluation is very difficult

Convergence tests allow us to determine whether an integral converges or diverges without computing its exact value. This is analogous to convergence tests for infinite series.

Remark 6.7: Strategy Overview

The main approaches for testing convergence:

Comparison Test: Compare with a known convergent/divergent integral
Limit Comparison: Compare asymptotic behavior at the problem point
Cauchy Criterion: Check if tail integrals become arbitrarily small
Special Tests: p-test, logarithmic tests, exponential decay

2. Direct Comparison Test

Theorem 6.6: Direct Comparison Test (Type I)

Let $0 \leq f(x) \leq g(x)$ for all $x \geq a$ . Then:

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

Proof:

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

To Show: Convergence of $\int g$ implies convergence of $\int f$

Proof:

Define $F(b) = \int_a^b f(x)\,dx$ and $G(b) = \int_a^b g(x)\,dx$ .

Since $f \geq 0$ , $F(b)$ is increasing in $b$ .

Since $f \leq g$ , we have $F(b) \leq G(b)$ for all $b$ .

If $\int_a^{+\infty} g$ converges to $L$ , then $F(b) \leq G(b) \leq L$ .

An increasing function bounded above has a limit, so $\lim_{b \to \infty} F(b)$ exists.

Therefore $\int_a^{+\infty} f(x)\,dx$ converges. ∎

∎

Remark 6.8: Key Insight

The comparison test requires $f(x) \geq 0$ (non-negative integrands). For oscillating functions, use the absolute convergence test or Dirichlet/Abel tests (covered in the next section).

Example 6.8: Comparison with p-Integral

Problem: Show $\int_1^{+\infty} \frac{1}{x^2 + 1}\,dx$ converges.

Solution:

For $x \geq 1$ :

0 \leq \frac{1}{x^2 + 1} < \frac{1}{x^2}

Since $\int_1^{+\infty} \frac{1}{x^2}\,dx$ converges (p = 2 > 1), by comparison test:

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

Example 6.9: Showing Divergence

Problem: Show $\int_1^{+\infty} \frac{1}{\sqrt{x^2 + 1}}\,dx$ diverges.

Solution:

For $x \geq 1$ :

\frac{1}{\sqrt{x^2 + 1}} > \frac{1}{\sqrt{x^2 + x^2}} = \frac{1}{\sqrt{2}x}

Since $\int_1^{+\infty} \frac{1}{x}\,dx$ diverges (p = 1), by comparison test:

$\int_1^{+\infty} \frac{1}{\sqrt{x^2 + 1}}\,dx$ diverges.

Theorem 6.7: Direct Comparison Test (Type II)

Let $0 \leq f(x) \leq g(x)$ for all $x \in (a, b]$ where both have a singularity at $x = a$ . Then:

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

Example 6.10: Type II Comparison

Problem: Show $\int_0^1 \frac{1}{\sqrt{x(1+x)}}\,dx$ converges.

Solution:

Near $x = 0$ , there's a singularity. For $x \in (0, 1]$ :

0 < \frac{1}{\sqrt{x(1+x)}} < \frac{1}{\sqrt{x \cdot 1}} = \frac{1}{\sqrt{x}}

Since $\int_0^1 \frac{1}{\sqrt{x}}\,dx = 2$ converges (p = 1/2 < 1), by comparison:

$\int_0^1 \frac{1}{\sqrt{x(1+x)}}\,dx$ converges.

3. Limit Comparison Test

Theorem 6.8: Limit Comparison Test (Type I)

Let $f(x) > 0$ and $g(x) > 0$ for $x \geq a$ . If

\lim_{x \to +\infty} \frac{f(x)}{g(x)} = L

where $0 < L < \infty$ (finite and positive), then:

$\int_a^{+\infty} f(x)\,dx$ and $\int_a^{+\infty} g(x)\,dx$ either both converge or both diverge.

Proof:

Given: $\lim_{x \to +\infty} f(x)/g(x) = L$ with $0 < L < \infty$

To Show: Both integrals have the same convergence behavior

Proof:

Choose $\epsilon = L/2$ . There exists $M$ such that for $x > M$ :

\frac{L}{2} < \frac{f(x)}{g(x)} < \frac{3L}{2}

This gives: $\frac{L}{2}g(x) < f(x) < \frac{3L}{2}g(x)$

By direct comparison test applied in both directions, $\int f$ and $\int g$ have the same behavior. ∎

∎

Remark 6.9: Edge Cases

If $L = 0$ : $f$ decreases faster than $g$ . If $\int g$ converges, so does $\int f$ .
If $L = +\infty$ : $f$ decreases slower than $g$ . If $\int g$ diverges, so does $\int f$ .

Example 6.11: Limit Comparison at Infinity

Problem: Determine convergence of $\int_1^{+\infty} \frac{x^2 + 3x + 1}{x^4 + x^2 + 7}\,dx$

Solution:

As $x \to +\infty$ , the integrand behaves like:

\frac{x^2 + 3x + 1}{x^4 + x^2 + 7} \sim \frac{x^2}{x^4} = \frac{1}{x^2}

Check the limit:

\lim_{x \to +\infty} \frac{(x^2 + 3x + 1)/(x^4 + x^2 + 7)}{1/x^2} = \lim_{x \to +\infty} \frac{x^4 + 3x^3 + x^2}{x^4 + x^2 + 7} = 1

Since $\int_1^{+\infty} 1/x^2\,dx$ converges (p = 2 > 1), the original integral converges.

Theorem 6.9: Limit Comparison Test (Type II)

Let $f(x) > 0$ and $g(x) > 0$ near $x = a$ (singularity). If

\lim_{x \to a^+} \frac{f(x)}{g(x)} = L

where $0 < L < \infty$ , then $\int_a^b f$ and $\int_a^b g$ have the same convergence behavior.

Example 6.12: Limit Comparison at Singularity

Problem: Determine convergence of $\int_0^1 \frac{\sqrt{x}}{\sin x}\,dx$

Solution:

Near $x = 0$ : $\sin x \approx x$ , so:

\frac{\sqrt{x}}{\sin x} \approx \frac{\sqrt{x}}{x} = \frac{1}{\sqrt{x}} = x^{-1/2}

Verify the limit:

\lim_{x \to 0^+} \frac{\sqrt{x}/\sin x}{1/\sqrt{x}} = \lim_{x \to 0^+} \frac{x}{\sin x} = 1

Since $\int_0^1 x^{-1/2}\,dx$ converges (p = 1/2 < 1), the original integral converges.

Example 6.13: Rational Function Near Zero

Problem: Analyze $\int_0^1 \frac{1}{x^2 - x}\,dx$

Solution:

Factor: $\frac{1}{x^2 - x} = \frac{1}{x(x-1)}$

Singularities at $x = 0$ and $x = 1$ . Split at $x = 1/2$ .

Near x = 0: $\frac{1}{x(x-1)} \approx \frac{-1}{x}$ → p = 1, diverges

The integral diverges (no need to check x = 1).

4. Cauchy Convergence Criterion

Theorem 6.10: Cauchy Criterion for Improper Integrals

The improper integral $\int_a^{+\infty} f(x)\,dx$ converges if and only if:

\forall \epsilon > 0, \exists M > a : \forall c, d > M, \left|\int_c^d f(x)\,dx\right| < \epsilon

In words: the integral over any sufficiently far interval can be made arbitrarily small.

Proof:

(⇒) If $\int_a^{+\infty} f$ converges to $L$ , then $F(b) = \int_a^b f \to L$ .

For $c, d > M$ : $\left|\int_c^d f\right| = |F(d) - F(c)| < \epsilon$ by Cauchy criterion for limits.

(⇐) If the Cauchy condition holds, then $F(b)$ satisfies Cauchy criterion, so the limit exists.

∎

Remark 6.10: When to Use Cauchy Criterion

The Cauchy criterion is useful when:

No explicit antiderivative exists
You need to prove convergence without finding the value
Dealing with oscillating integrands

Corollary 6.1: Necessary Condition for Convergence

If $\int_a^{+\infty} f(x)\,dx$ converges, then:

\lim_{b \to +\infty} \int_b^{+\infty} f(x)\,dx = 0

Warning: This is necessary but NOT sufficient. The integrand need not go to 0.

Example 6.14: Cauchy Criterion Application

Problem: Show $\int_1^{+\infty} \frac{\sin x}{x^2}\,dx$ converges using Cauchy criterion.

Solution:

For $d > c > 1$ :

\left|\int_c^d \frac{\sin x}{x^2}\,dx\right| \leq \int_c^d \frac{|\sin x|}{x^2}\,dx \leq \int_c^d \frac{1}{x^2}\,dx

= \frac{1}{c} - \frac{1}{d} < \frac{1}{c}

Given $\epsilon > 0$ , choose $M = 1/\epsilon$ . Then for $c, d > M$ :

\left|\int_c^d \frac{\sin x}{x^2}\,dx\right| < \frac{1}{c} < \epsilon

By Cauchy criterion, the integral converges.

Theorem 6.11: Comparison with Convergent Integral

If $|f(x)| \leq g(x)$ for $x \geq a$ and $\int_a^{+\infty} g(x)\,dx$ converges, then $\int_a^{+\infty} f(x)\,dx$ converges absolutely.

5. Logarithmic Tests

Theorem 6.12: Logarithmic p-Test (Type I)

The integral with logarithmic factors:

\int_e^{+\infty} \frac{1}{x(\ln x)^p}\,dx \begin{cases} \text{converges} & \text{if } p > 1 \\ \text{diverges} & \text{if } p \leq 1 \end{cases}

Proof:

Substitute $u = \ln x$ , so $du = dx/x$ :

\int_e^{+\infty} \frac{1}{x(\ln x)^p}\,dx = \int_1^{+\infty} \frac{1}{u^p}\,du

This is the standard p-integral, which converges iff $p > 1$ . ∎

∎

Theorem 6.13: Logarithmic p-Test (Type II)

For the integral near 1 (where $\ln x \to 0$ ):

\int_1^e \frac{1}{x|\ln x|^p}\,dx \begin{cases} \text{converges} & \text{if } p < 1 \\ \text{diverges} & \text{if } p \geq 1 \end{cases}

Remark 6.11: Logarithms Grow Slowly

Key fact: $(\ln x)^k / x^\epsilon \to 0$ as $x \to +\infty$ for any $k > 0$ and $\epsilon > 0$ .

This means logarithmic factors don't change the convergence threshold of the p-test; they only affect borderline cases.

Example 6.15: Logarithmic Comparison

Problem: Determine convergence of $\int_2^{+\infty} \frac{1}{x(\ln x)^2}\,dx$

Solution:

This matches the logarithmic p-test with p = 2 > 1.

Alternatively, substitute $u = \ln x$ :

\int_2^{+\infty} \frac{1}{x(\ln x)^2}\,dx = \int_{\ln 2}^{+\infty} \frac{1}{u^2}\,du = \frac{1}{\ln 2}

The integral converges with value $1/\ln 2$ .

Example 6.16: Divergent Logarithmic Integral

Problem: Show $\int_2^{+\infty} \frac{1}{x\ln x}\,dx$ diverges.

Solution:

Logarithmic p-test with p = 1 (boundary case).

Substitute $u = \ln x$ :

\int_2^{+\infty} \frac{1}{x\ln x}\,dx = \int_{\ln 2}^{+\infty} \frac{1}{u}\,du = +\infty

The integral diverges.

Example 6.17: Double Logarithm

Problem: Determine convergence of $\int_e^{+\infty} \frac{1}{x\ln x(\ln\ln x)^2}\,dx$

Solution:

Substitute $u = \ln x$ , then $v = \ln u$ :

\int_e^{+\infty} \frac{1}{x\ln x(\ln\ln x)^2}\,dx = \int_1^{+\infty} \frac{1}{u(\ln u)^2}\,du

This is logarithmic p-test with p = 2 > 1.

The integral converges.

6. Using p-Integrals as Comparison Standards

Remark 6.12: The Power of p-Integrals

The p-integrals serve as the primary comparison standards because:

Type I: $\int_1^{+\infty} 1/x^p\,dx$ converges ⟺ $p > 1$
Type II: $\int_0^1 1/x^p\,dx$ converges ⟺ $p < 1$

Most rational and algebraic functions can be compared to appropriate p-integrals.

Theorem 6.14: Asymptotic p-Comparison

If $f(x) \sim C/x^p$ as $x \to +\infty$ for some constant $C > 0$ , then:

\int_a^{+\infty} f(x)\,dx \text{ converges} \iff p > 1

Example 6.18: Finding the Equivalent p

Problem: Determine convergence of $\int_1^{+\infty} \frac{3x^2 + 5}{x^4 - 2x + 1}\,dx$

Solution:

Dominant terms as $x \to +\infty$ :

\frac{3x^2 + 5}{x^4 - 2x + 1} \sim \frac{3x^2}{x^4} = \frac{3}{x^2}

This is equivalent to p = 2 > 1.

The integral converges.

Example 6.19: Square Root Comparison

Problem: Determine convergence of $\int_1^{+\infty} \frac{1}{\sqrt{x^3 + x}}\,dx$

Solution:

As $x \to +\infty$ :

\frac{1}{\sqrt{x^3 + x}} \sim \frac{1}{\sqrt{x^3}} = \frac{1}{x^{3/2}}

This is p = 3/2 > 1.

The integral converges.

Example 6.20: Type II p-Comparison

Problem: Determine convergence of $\int_0^1 \frac{1}{\sqrt[3]{x^2(1-x)}}\,dx$

Solution:

Two singularities: x = 0 and x = 1. Split at x = 1/2.

Near x = 0: $\frac{1}{\sqrt[3]{x^2(1-x)}} \approx \frac{1}{x^{2/3}}$

This is p = 2/3 < 1 ✓

Near x = 1: $\frac{1}{\sqrt[3]{x^2(1-x)}} \approx \frac{1}{(1-x)^{1/3}}$

This is p = 1/3 < 1 ✓

Both parts converge, so the integral converges.

Remark 6.13: Decision Procedure

To analyze $\int f(x)\,dx$ at a problem point:

Identify the dominant behavior: $f(x) \sim C \cdot |x - a|^{-p}$ or $f(x) \sim C \cdot x^{-p}$
Extract the exponent p
Apply the p-test: Type I needs p > 1; Type II needs p < 1

7. More Worked Examples

Example 6.9: Exponential Decay

Problem: Evaluate $\int_0^{+\infty} e^{-x}\,dx$

Solution:

\int_0^{+\infty} e^{-x}\,dx = \lim_{b \to +\infty} [-e^{-x}]_0^b = \lim_{b \to +\infty} (-e^{-b} + 1) = 1

Example 6.10: Rational Function

Problem: Evaluate $\int_0^{+\infty} \frac{x}{(1+x^2)^2}\,dx$

Solution:

Substitute $u = 1 + x^2$ , $du = 2x\,dx$ :

\int_0^{+\infty} \frac{x}{(1+x^2)^2}\,dx = \frac{1}{2}\int_1^{+\infty} \frac{du}{u^2} = \frac{1}{2} \cdot 1 = \frac{1}{2}

Example 6.11: Logarithmic Singularity

Problem: Evaluate $\int_0^1 \ln x\,dx$

Solution:

Note $\ln x \to -\infty$ as $x \to 0^+$ , so this is Type II.

\int_0^1 \ln x\,dx = [x\ln x - x]_0^1 = (0 - 1) - \lim_{x \to 0^+}(x\ln x - x)

Since $\lim_{x \to 0^+} x\ln x = 0$ (by L'Hôpital):

= -1 - (0 - 0) = -1

Example 6.12: Double Singularity

Problem: Evaluate $\int_0^1 \frac{1}{\sqrt{x(1-x)}}\,dx$

Solution:

Singularities at both $x = 0$ and $x = 1$ .

Near $x = 0$ : $\frac{1}{\sqrt{x(1-x)}} \approx \frac{1}{\sqrt{x}}$

Near $x = 1$ : $\frac{1}{\sqrt{x(1-x)}} \approx \frac{1}{\sqrt{1-x}}$

Both are p = 1/2 < 1, so both parts converge.

Using substitution $x = \sin^2\theta$ : the integral equals $\pi$ .

8. Common Mistakes to Avoid

Mistake 1: Ignoring Singularities

Wrong: Treating $\int_0^1 \frac{1}{x}\,dx$ as a regular integral and computing $[\ln x]_0^1$ .

Right: Recognize the singularity at $x = 0$ and evaluate as $\lim_{\epsilon \to 0^+} [\ln x]_\epsilon^1$ .

Mistake 2: Confusing p-Test Conditions

Wrong: Thinking $\int_0^1 1/x^p$ converges for $p > 1$ .

Right: For Type II (singular), it's $p < 1$ . The conditions are opposite for Type I and Type II!

Mistake 3: Symmetric Cancellation Fallacy

Wrong: Claiming $\int_{-\infty}^{+\infty} x\,dx = 0$ by symmetry.

Right: Both halves diverge independently. The integral is undefined, not zero. (The symmetric limit gives the Cauchy Principal Value, which is different.)

Mistake 4: Forgetting to Split Mixed Integrals

Wrong: Evaluating $\int_0^{+\infty} 1/x^2\,dx$ as one limit.

Right: Split at $x = 1$ . Check $\int_0^1$ (singular) and $\int_1^{+\infty}$ (infinite) separately.

Mistake 5: Missing Interior Singularities

Wrong: Computing $\int_{-1}^1 1/x^2\,dx$ directly.

Right: There's a singularity at $x = 0$ inside the interval. Split into $\int_{-1}^0 + \int_0^1$ .

9. Additional Worked Examples

Example 6.13: Trigonometric Integral

Problem: Evaluate $\int_0^{+\infty} e^{-x}\cos x\,dx$

Solution:

Use integration by parts twice, or recognize this as the real part of:

\int_0^{+\infty} e^{-x} e^{ix}\,dx = \int_0^{+\infty} e^{(-1+i)x}\,dx

= \left[\frac{e^{(-1+i)x}}{-1+i}\right]_0^{+\infty} = 0 - \frac{1}{-1+i} = \frac{1}{1-i} = \frac{1+i}{2}

Taking the real part: $\int_0^{+\infty} e^{-x}\cos x\,dx = \frac{1}{2}$

Example 6.14: Power Times Exponential

Problem: Evaluate $\int_0^{+\infty} x^2 e^{-x}\,dx$

Solution:

This is related to the Gamma function. Using integration by parts twice:

\int_0^{+\infty} x^2 e^{-x}\,dx = [-x^2 e^{-x}]_0^{+\infty} + 2\int_0^{+\infty} x e^{-x}\,dx

The boundary term vanishes. For the remaining integral:

2\int_0^{+\infty} x e^{-x}\,dx = 2\left([-x e^{-x}]_0^{+\infty} + \int_0^{+\infty} e^{-x}\,dx\right) = 2(0 + 1) = 2

So $\int_0^{+\infty} x^2 e^{-x}\,dx = 2 = 2!$

Example 6.15: Rational with Multiple Singularities

Problem: Analyze $\int_0^2 \frac{1}{\sqrt{x}\sqrt{2-x}}\,dx$

Solution:

Singularities at both $x = 0$ and $x = 2$ .

Near $x = 0$ : $\frac{1}{\sqrt{x}\sqrt{2}} \sim \frac{1}{\sqrt{2x}}$ → p = 1/2 < 1 ✓

Near $x = 2$ : $\frac{1}{\sqrt{2}\sqrt{2-x}}$ → p = 1/2 < 1 ✓

Both singularities are integrable. Using substitution $x = 2\sin^2\theta$ : the integral equals $\pi$ .

Example 6.16: Polynomial Decay

Problem: Evaluate $\int_1^{+\infty} \frac{1}{x^2 + 3x + 2}\,dx$

Solution:

Factor: $x^2 + 3x + 2 = (x+1)(x+2)$

Partial fractions: $\frac{1}{(x+1)(x+2)} = \frac{1}{x+1} - \frac{1}{x+2}$

\int_1^{+\infty} \left(\frac{1}{x+1} - \frac{1}{x+2}\right)dx = \lim_{b \to +\infty} [\ln(x+1) - \ln(x+2)]_1^b

= \lim_{b \to +\infty} \ln\frac{b+1}{b+2} - \ln\frac{2}{3} = \ln 1 - \ln\frac{2}{3} = \ln\frac{3}{2}

Example 6.17: Infinite Interval Both Directions

Problem: Evaluate $\int_{-\infty}^{+\infty} \frac{1}{e^x + e^{-x}}\,dx$

Solution:

Note $\frac{1}{e^x + e^{-x}} = \frac{e^x}{e^{2x} + 1}$

Substitute $u = e^x$ , $du = e^x dx$ :

\int_{-\infty}^{+\infty} \frac{e^x}{e^{2x} + 1}\,dx = \int_0^{+\infty} \frac{1}{u^2 + 1}\,du = [\arctan u]_0^{+\infty} = \frac{\pi}{2}

10. Study Tips

1. Always Check for Singularities

Before evaluating any integral, scan the integrand for points where it might blow up. Check denominators, square roots, and logarithms.

2. Memorize p-Test Conditions

$\int_1^\infty 1/x^p$ : needs p > 1. $\int_0^1 1/x^p$ : needs p < 1. Opposite!

3. Practice Limit Evaluation

Many improper integrals require L'Hôpital's rule or known limits like $\lim_{x \to 0^+} x\ln x = 0$ .

4. Split Strategically

For mixed integrals, choose splitting points that separate singularities and infinite limits. Often x = 1 works well.

5. Use Substitutions Wisely

Substitutions can convert Type I to Type II or vice versa. For example, $u = 1/x$ swaps 0 and ∞.

6. Connect to Series

Many convergence tests for improper integrals parallel those for series. This helps with intuition.

11. Quick Reference Table

Integral	Type	Converges?	Value
$\int_1^{+\infty} \frac{1}{x^2}dx$	Type I	Yes (p=2>1)	1
$\int_1^{+\infty} \frac{1}{x}dx$	Type I	No (p=1)	—
$\int_0^1 \frac{1}{\sqrt{x}}dx$	Type II	Yes (p=½<1)	2
$\int_0^1 \frac{1}{x}dx$	Type II	No (p=1)	—
$\int_0^{+\infty} e^{-x}dx$	Type I	Yes	1
$\int_{-\infty}^{+\infty} \frac{1}{1+x^2}dx$	Type I (both)	Yes	π
$\int_0^1 \ln x\,dx$	Type II	Yes	−1
$\int_0^{+\infty} \frac{1}{x^2}dx$	Mixed	No	—

12. Challenge Problems

Challenge 1

Determine convergence and evaluate if convergent:

\int_0^{+\infty} \frac{x}{(1+x^2)^2}\,dx

Show hint

Substitute u = 1 + x².

Challenge 2

For what values of α does the following converge?

\int_0^1 \frac{x^\alpha - 1}{\ln x}\,dx

Show hint

Near x = 1, use L'Hôpital. Near x = 0, compare with x^α.

Challenge 3

Evaluate the following integral:

\int_0^{+\infty} \frac{\ln x}{1+x^2}\,dx

Show hint

Split at x = 1 and use the substitution u = 1/x on one part.

Challenge 4

Prove that the following integral converges:

\int_0^1 \frac{1}{\sqrt[3]{x(1-x)(2-x)}}\,dx

Show hint

Find the singularities and check the exponent at each.

13. Convergence Test Summary

Test	When to Use	Key Requirement
Direct Comparison	Can establish $f \leq g$ or $f \geq g$	Non-negative functions
Limit Comparison	Functions with similar asymptotic behavior	$\lim f/g = L \in (0, \infty)$
p-Test (Type I)	Power functions at infinity	Converges iff p > 1
p-Test (Type II)	Power functions at singularity	Converges iff p < 1
Logarithmic Test	Products with $1/(x(\ln x)^p)$	Same as p-test after substitution
Cauchy Criterion	No explicit antiderivative	Tail integrals → 0
Absolute Convergence	Oscillating integrands	$\int \|f\|$ converges

14. Decision Flowchart

1

Identify Problem Points

Check for infinite limits (Type I) and singularities (Type II)

2

Split if Mixed

Separate each problem so each piece has only one issue

3

Find Dominant Behavior

Determine $f(x) \sim C/x^p$ or $f(x) \sim C/(x-a)^p$

4

Apply Appropriate Test

Use p-test, comparison, or limit comparison based on the form

5

Conclude

All pieces must converge for the whole integral to converge

15. More Practice Examples

Example 6.25: Exponential vs Polynomial

Problem: Determine convergence of $\int_1^{+\infty} \frac{x^{100}}{e^x}\,dx$

Solution:

Exponentials dominate any polynomial. For any $\epsilon > 0$ :

\frac{x^{100}}{e^x} < \frac{C}{e^{(1-\epsilon)x}} \text{ for large } x

Since $\int_1^{+\infty} e^{-cx}\,dx$ converges for any $c > 0$ :

The integral converges.

Example 6.26: Trigonometric Near Zero

Problem: Determine convergence of $\int_0^{\pi/2} \frac{1}{\sin x}\,dx$

Solution:

Singularity at $x = 0$ . Near 0: $\sin x \approx x$

\frac{1}{\sin x} \approx \frac{1}{x}

This is p = 1 (borderline). The integral diverges.

Example 6.27: Square Root of Polynomial

Problem: Determine convergence of $\int_0^1 \frac{1}{\sqrt{x^3 + x^4}}\,dx$

Solution:

Singularity at $x = 0$ . Near 0:

\frac{1}{\sqrt{x^3 + x^4}} = \frac{1}{x^{3/2}\sqrt{1 + x}} \approx \frac{1}{x^{3/2}}

This is p = 3/2 > 1. For Type II, we need p < 1.

The integral diverges.

Example 6.28: Mixed with Logarithm

Problem: Determine convergence of $\int_1^{+\infty} \frac{\ln x}{x^2}\,dx$

Solution:

Compare with a p-integral. For any $\epsilon > 0$ :

\frac{\ln x}{x^2} < \frac{x^\epsilon}{x^2} = \frac{1}{x^{2-\epsilon}}

Choose $\epsilon = 0.5$ . Then $2 - \epsilon = 1.5 > 1$ .

The integral converges.

16. Common Patterns to Recognize

Rational Functions

$\frac{P(x)}{Q(x)}$ where deg(Q) - deg(P) = k

Behaves like $1/x^k$ . Converges at ∞ iff k > 1.

Exponential Decay

$e^{-ax}$ for a > 0

Always converges at ∞. Dominates any polynomial.

Logarithmic Factors

$\frac{(\ln x)^k}{x^p}$

Log factors don't affect convergence threshold for p-test.

Trigonometric Near 0

$\sin x \approx x$ , $1 - \cos x \approx x^2/2$

Use Taylor expansions to find the equivalent power.

Convergence Tests Quiz

14

Questions

0

Correct

0%

Accuracy

1

To show

\int_1^{+\infty} \frac{1}{x^2+1}dx

converges, we compare with:

Medium

Not attempted

2

If

\lim_{x \to +\infty} \frac{f(x)}{g(x)} = L

where

0 < L < \infty

, then:

Medium

Not attempted

3

The integral

\int_1^{+\infty} \frac{1}{x\ln x}dx

:

Hard

Not attempted

4

The integral

\int_2^{+\infty} \frac{1}{x(\ln x)^2}dx

:

Hard

Not attempted

5

For

0 \leq f(x) \leq g(x)

and

\int_a^{+\infty} g(x)dx

converges, then:

Easy

Not attempted

6

The integral

\int_1^{+\infty} \frac{x+1}{x^3+2x}dx

:

Medium

Not attempted

7

To show

\int_0^1 \frac{1}{\sqrt{x(1-x)}}dx

converges, we check:

Medium

Not attempted

8

The Cauchy criterion states that

\int_a^{+\infty} f

converges iff:

Hard

Not attempted

9

The integral

\int_0^1 \frac{1}{x^{0.5}}dx

:

Easy

Not attempted

10

For

\int_1^{+\infty} \frac{\sin x}{x^2}dx

, which test is most appropriate?

Medium

Not attempted

11

The integral

\int_0^{+\infty} \frac{1}{x^p}dx

:

Hard

Not attempted

12

If

\lim_{x \to +\infty} x^2 f(x) = 5

, then

\int_1^{+\infty} f(x)dx

:

Hard

Not attempted

13

The integral

\int_0^1 \frac{1}{x\sqrt{|\ln x|}}dx

:

Hard

Not attempted

14

For Type II integrals with singularity at x=a, the p-test says convergence occurs when:

Easy

Not attempted

Frequently Asked Questions

When should I use the comparison test vs limit comparison test?

Use direct comparison when you can easily show f(x) ≤ g(x) or f(x) ≥ g(x). Use limit comparison when the functions have similar asymptotic behavior but the inequality is hard to establish directly.

What are good comparison functions to know?

The p-integrals: ∫₁^∞ 1/xᵖ (converges for p>1), ∫₀¹ 1/xᵖ (converges for p<1), and exponentials like e^{-x} which decay faster than any polynomial.

How do I handle logarithmic factors?

Logarithms grow slower than any positive power: (ln x)^k / x^ε → 0 for any ε > 0. So ∫₁^∞ 1/(x(ln x)^p) converges iff p > 1, similar to the p-test.

What is the Cauchy criterion for improper integrals?

∫ₐ^∞ f converges iff for every ε > 0, there exists M such that |∫ᵦ^c f| < ε for all c > b > M. It's useful when you can't find the antiderivative explicitly.

Can I use the comparison test for oscillating functions?

Not directly. For oscillating functions like sin(x)/x, use the absolute value: if ∫|f| converges, then ∫f converges absolutely. For conditional convergence, use Dirichlet or Abel tests.

How do I determine the equivalent p for limit comparison?

Find the dominant terms as x→∞ (or x→a for singularities). For rational functions, compare the degrees of numerator and denominator. The difference gives you p.

What happens when the limit comparison gives L = 0 or L = ∞?

If L = 0 and ∫g converges, then ∫f converges (f is 'smaller'). If L = ∞ and ∫g diverges, then ∫f diverges (f is 'larger'). Other combinations are inconclusive.

How do I handle mixed integrals with both Type I and Type II issues?

Split the integral at a convenient point (like x=1) so each piece has only one problem. Then analyze each piece separately. The integral converges iff ALL pieces converge.

Why do the p-test conditions differ for Type I and Type II?

It's about how fast the function must decay. At ∞, you need f→0 fast enough (p>1). At a singularity, the blow-up can't be too severe (p<1). The threshold p=1 is borderline for both.

Can an integral converge even if the integrand doesn't approach 0?

Yes! For example, ∫₁^∞ sin(x²) dx converges even though sin(x²) oscillates. However, for non-negative integrands, f(x)→0 is necessary but not sufficient for convergence.

Convergence Tests

1. Introduction: Why Convergence Tests?

2. Direct Comparison Test

3. Limit Comparison Test

4. Cauchy Convergence Criterion

5. Logarithmic Tests

6. Using p-Integrals as Comparison Standards

7. More Worked Examples

8. Common Mistakes to Avoid

Mistake 1: Ignoring Singularities

Mistake 2: Confusing p-Test Conditions

Mistake 3: Symmetric Cancellation Fallacy

Mistake 4: Forgetting to Split Mixed Integrals

Mistake 5: Missing Interior Singularities

9. Additional Worked Examples

10. Study Tips

1. Always Check for Singularities

2. Memorize p-Test Conditions

3. Practice Limit Evaluation

4. Split Strategically

5. Use Substitutions Wisely

6. Connect to Series

11. Quick Reference Table

12. Challenge Problems

13. Convergence Test Summary

14. Decision Flowchart

15. More Practice Examples

16. Common Patterns to Recognize

Rational Functions

Exponential Decay

Logarithmic Factors

Trigonometric Near 0

Frequently Asked Questions

When should I use the comparison test vs limit comparison test?

What are good comparison functions to know?

How do I handle logarithmic factors?

What is the Cauchy criterion for improper integrals?

Can I use the comparison test for oscillating functions?

How do I determine the equivalent p for limit comparison?

What happens when the limit comparison gives L = 0 or L = ∞?

How do I handle mixed integrals with both Type I and Type II issues?

Why do the p-test conditions differ for Type I and Type II?

Can an integral converge even if the integrand doesn't approach 0?

Type I (Infinite Limits)

Type II (Singular)

p-Test (Type I)

p-Test (Type II)

Remember