1. Introduction: Why Two Types of Convergence?

When dealing with improper integrals of oscillating functions, we encounter a fascinating phenomenon: some integrals converge due to cancellation effects, even though the absolute value of the integrand doesn't have a finite integral.

Classic Example

\int_1^{+\infty} \frac{\sin x}{x}\,dx = \frac{\pi}{2} - \text{Si}(1)

Converges (conditionally)

But its absolute value...

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

Diverges!

This distinction leads to two fundamentally different types of convergence with important implications for analysis.

Remark 6.14: Parallels with Series

The concepts of absolute and conditional convergence for integrals mirror those for infinite series:

The alternating harmonic series $\sum (-1)^n/n$ converges conditionally
The harmonic series $\sum 1/n$ diverges
Similarly, $\int \sin(x)/x\,dx$ converges while $\int |\sin(x)/x|\,dx$ diverges

2. Definitions of Absolute and Conditional Convergence

Definition 6.9: Absolute Convergence

An improper integral $\int_a^{+\infty} f(x)\,dx$ is said to converge absolutely if:

\int_a^{+\infty} |f(x)|\,dx \text{ converges}

That is, the integral of the absolute value of the function converges.

Definition 6.10: Conditional Convergence

An improper integral $\int_a^{+\infty} f(x)\,dx$ is said to converge conditionally if:

$\int_a^{+\infty} f(x)\,dx$ converges, AND
$\int_a^{+\infty} |f(x)|\,dx$ diverges

The convergence relies on cancellation between positive and negative parts.

Theorem 6.15: Absolute Convergence Implies Convergence

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

That is: Absolute convergence ⟹ Convergence

Proof:

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

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

Proof:

Note that $-|f(x)| \leq f(x) \leq |f(x)|$ , so:

0 \leq f(x) + |f(x)| \leq 2|f(x)|

By comparison test, $\int (f + |f|)$ converges.

Since $f = (f + |f|) - |f|$ and both integrals on the right converge:

$\int f$ converges by linearity. ∎

∎

Remark 6.15: The Converse is False!

Important:

Convergence does NOT imply absolute convergence.

Example: $\int_1^{+\infty} \frac{\sin x}{x}\,dx$ converges, but $\int_1^{+\infty} \frac{|\sin x|}{x}\,dx$ diverges.

Example 6.23: Absolutely Convergent Integral

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

Solution:

We have:

\left|\frac{\cos x}{x^2}\right| \leq \frac{1}{x^2}

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

By comparison test, $\int_1^{+\infty} \left|\frac{\cos x}{x^2}\right|\,dx$ converges.

Therefore the original integral converges absolutely.

Example 6.24: Testing for Absolute vs Conditional

Problem: Classify $\int_1^{+\infty} \frac{\sin x}{x^{3/2}}\,dx$

Solution:

Check absolute convergence:

\left|\frac{\sin x}{x^{3/2}}\right| \leq \frac{1}{x^{3/2}}

Since $\int_1^{+\infty} \frac{1}{x^{3/2}}\,dx$ converges (p = 3/2 > 1):

The integral converges absolutely.

3. The Dirichlet Test

Theorem 6.16: Dirichlet Test for Improper Integrals

Let $f$ and $g$ be defined on $[a, +\infty)$ . The integral $\int_a^{+\infty} f(x)g(x)\,dx$ converges if:

The partial integrals $F(b) = \int_a^b f(x)\,dx$ are uniformly bounded for all $b \geq a$
$g(x)$ is monotonically decreasing
$\lim_{x \to +\infty} g(x) = 0$

Proof:

Idea: Use integration by parts and show the boundary term vanishes while the remaining integral converges.

Let $F(x) = \int_a^x f(t)\,dt$ . Then $F'(x) = f(x)$ and $|F(x)| \leq M$ for all x.

\int_a^b f(x)g(x)\,dx = [F(x)g(x)]_a^b - \int_a^b F(x)g'(x)\,dx

As $b \to \infty$ : $F(b)g(b) \to 0$ since $g(b) \to 0$ and $F$ is bounded.

∎

Remark 6.16: Key Applications

The Dirichlet test is perfect for integrals involving:

$\sin x, \cos x$ with bounded partial integrals
Multiplied by a decreasing function like $1/x, 1/x^\alpha, e^{-x}$

Example 6.25: The Dirichlet Integral

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

Solution (Dirichlet Test):

Let $f(x) = \sin x$ and $g(x) = 1/x$ .

Check conditions:

$F(b) = \int_1^b \sin x\,dx = -\cos b + \cos 1$ is bounded: $|F(b)| \leq 2$
$g(x) = 1/x$ is monotonically decreasing
$\lim_{x \to +\infty} 1/x = 0$

By Dirichlet test, the integral converges.

Example 6.26: Cosine Variant

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

Solution:

Let $f(x) = \cos x$ and $g(x) = 1/\sqrt{x}$ .

$\int_1^b \cos x\,dx = \sin b - \sin 1$ is bounded: $|F(b)| \leq 2$
$1/\sqrt{x}$ is monotonically decreasing to 0

By Dirichlet test, the integral converges.

4. The Abel Test

Theorem 6.17: Abel Test for Improper Integrals

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

$\int_a^{+\infty} f(x)\,dx$ converges
$g(x)$ is monotonic (increasing or decreasing)
$g(x)$ is bounded

Remark 6.17: Dirichlet vs Abel

Dirichlet Test

Requires: bounded partial integrals of f, g → 0 monotonically

Abel Test

Requires: ∫f converges, g bounded and monotonic

Example 6.27: Abel Test Application

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

Solution:

Let $f(x) = \sin x / x$ and $g(x) = 1/(1 + 1/x) = x/(x+1)$ .

$\int_1^{+\infty} \sin x / x\,dx$ converges (by Dirichlet)
$g(x) = x/(x+1)$ is increasing and bounded: $1/2 \leq g(x) \leq 1$

By Abel test, the integral converges.

5. Deep Dive: The sin(x)/x Integral

The integral $\int_1^{+\infty} \frac{\sin x}{x}\,dx$ is the canonical example of conditional convergence. Let's prove both parts: convergence and non-absolute convergence.

Theorem 6.18: Convergence of ∫sin(x)/x

The integral $\int_1^{+\infty} \frac{\sin x}{x}\,dx$ converges.

Proof:

Apply the Dirichlet test with $f(x) = \sin x$ and $g(x) = 1/x$ :

$\left|\int_1^b \sin x\,dx\right| = |{-\cos b + \cos 1}| \leq 2$ (bounded)
$1/x$ is monotonically decreasing
$\lim_{x \to +\infty} 1/x = 0$

All conditions satisfied. ∎

∎

Theorem 6.19: Divergence of ∫|sin(x)/x|

The integral $\int_1^{+\infty} \frac{|\sin x|}{x}\,dx$ diverges.

Proof:

Use the inequality $|\sin x| \geq \sin^2 x = \frac{1 - \cos(2x)}{2}$ :

\int_1^b \frac{|\sin x|}{x}\,dx \geq \int_1^b \frac{\sin^2 x}{x}\,dx = \frac{1}{2}\int_1^b \frac{1}{x}\,dx - \frac{1}{2}\int_1^b \frac{\cos(2x)}{x}\,dx

The first integral $\frac{1}{2}\ln b \to +\infty$ .

The second integral converges (by Dirichlet test).

So $\int |\sin x|/x\,dx \to +\infty$ . ∎

∎

Corollary 6.2: Conditional Convergence

Therefore, $\int_1^{+\infty} \frac{\sin x}{x}\,dx$ converges conditionally.

Remark 6.18: The Threshold Exponent

For $\int_1^{+\infty} \frac{\sin x}{x^p}\,dx$ :

$p > 1$ : Absolutely convergent (since $|\sin x|/x^p \leq 1/x^p$ )
$0 < p \leq 1$ : Conditionally convergent (converges by Dirichlet, but $\int |\sin x|/x^p$ diverges)
$p \leq 0$ : Divergent

6. Properties and Implications

Theorem 6.20: Linearity for Absolutely Convergent Integrals

If $\int |f|$ and $\int |g|$ both converge, then $\int (\alpha f + \beta g)$ converges absolutely for any constants $\alpha, \beta$ .

Remark 6.19: Conditionally Convergent Integrals are Fragile

Warning:

Unlike absolutely convergent integrals, conditionally convergent ones:

May not preserve convergence under rearrangement (for series analogs)
Cannot be split arbitrarily
Require careful handling in applications

Example 6.28: Determining Convergence Type

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

Solution:

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

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

This is a Fresnel-type integral. Since $\int \sin u / u\,du$ converges conditionally:

The original integral converges conditionally.

Example 6.29: Product of Oscillating Functions

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

Solution:

Use identity: $\sin x \cos x = \frac{1}{2}\sin(2x)$

\int_1^{+\infty} \frac{\sin x \cos x}{x}\,dx = \frac{1}{2}\int_1^{+\infty} \frac{\sin(2x)}{x}\,dx

Substitute $u = 2x$ :

= \frac{1}{2}\int_2^{+\infty} \frac{\sin u}{u/2} \cdot \frac{du}{2} = \frac{1}{2}\int_2^{+\infty} \frac{\sin u}{u}\,du

This converges conditionally (same as sin(x)/x).

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 $\pi$ .

Example 6.18: Oscillating with Faster Decay

Problem: Classify $\int_1^{+\infty} \frac{\cos(2x)}{x^3}\,dx$

Solution:

Check absolute convergence:

\left|\frac{\cos(2x)}{x^3}\right| \leq \frac{1}{x^3}

Since p = 3 > 1, $\int 1/x^3$ converges.

The integral converges absolutely.

Example 6.19: Product of Trig and Power

Problem: Classify $\int_1^{+\infty} \frac{\sin^2 x}{x}\,dx$

Solution:

Use identity: $\sin^2 x = \frac{1 - \cos(2x)}{2}$

\int_1^{+\infty} \frac{\sin^2 x}{x}\,dx = \frac{1}{2}\int_1^{+\infty} \frac{1}{x}\,dx - \frac{1}{2}\int_1^{+\infty} \frac{\cos(2x)}{x}\,dx

First integral diverges (p=1). Second converges (Dirichlet).

Since one part diverges, the integral diverges.

Example 6.20: Exponential Times Oscillation

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

Solution:

Check absolute convergence:

|e^{-x^2}\cos x| \leq e^{-x^2}

The Gaussian integral $\int_0^{+\infty} e^{-x^2}\,dx = \sqrt{\pi}/2$ converges.

The integral converges absolutely.

Example 6.21: Alternating Near Singularity

Problem: Classify $\int_0^1 \frac{\sin(1/x)}{\sqrt{x}}\,dx$

Solution:

Check absolute convergence:

\left|\frac{\sin(1/x)}{\sqrt{x}}\right| \leq \frac{1}{\sqrt{x}}

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

The integral converges absolutely.

Example 6.22: Dirichlet with Non-Standard g

Problem: Classify $\int_1^{+\infty} \frac{\sin x}{\ln x}\,dx$

Solution:

Let $f = \sin x$ , $g = 1/\ln x$ .

Check Dirichlet conditions:

Partial integrals of sin x are bounded ✓
$1/\ln x$ is monotonically decreasing for x > 1 ✓
$\lim_{x \to \infty} 1/\ln x = 0$ ✓

By Dirichlet test, the integral converges conditionally.

8. Quick Reference: When to Use Each Test

Use Comparison/p-Test When:

Integrand is non-negative
Checking for absolute convergence
Function behaves like a power function

Use Dirichlet Test When:

Product of oscillating and decaying functions
f has bounded partial integrals (sin, cos)
g → 0 monotonically

Use Abel Test When:

∫f already known to converge
Multiplied by bounded monotonic g
g doesn't need to go to 0

Classification Strategy:

First: Check if ∫|f| converges
If yes → Absolute convergence
If no → Try Dirichlet/Abel for ∫f
If ∫f converges → Conditional

8. Common Mistakes to Avoid

Mistake 1: Assuming Convergence Implies Absolute Convergence

Wrong: Thinking that if $\int f$ converges, then $\int |f|$ must also converge.

Right: Convergence does NOT imply absolute convergence. Example: $\int \sin x/x$ converges but $\int |\sin x/x|$ diverges.

Mistake 2: Using Comparison Test for Oscillating Integrands

Wrong: Trying to apply comparison test directly to $\sin x / x$ .

Right: Comparison test requires non-negative functions. For oscillating integrands, use Dirichlet or Abel test.

Mistake 3: Misapplying Dirichlet Conditions

Wrong: Applying Dirichlet test when partial integrals are not bounded.

Right: Verify ALL conditions: (1) bounded partial integrals of f, (2) g monotonically decreasing, (3) g → 0.

Mistake 4: Confusing Dirichlet and Abel Tests

Wrong: Thinking Abel test requires g → 0.

Right: Abel requires $\int f$ to converge + g bounded monotonic. Dirichlet requires bounded partials of f + g → 0 monotonically.

Mistake 5: Forgetting the Threshold Exponent

Wrong: Assuming $\sin x / x^{0.5}$ converges absolutely.

Right: For $\sin x / x^p$ : absolute convergence only when p > 1. Here p = 0.5 < 1, so it's conditionally convergent.

Mistake 6: Not Checking Both Conditions

Wrong: Concluding conditional convergence after showing $\int f$ converges without checking $\int |f|$ .

Right: You must verify BOTH that $\int f$ converges AND that $\int |f|$ diverges to conclude conditional convergence.

Mistake 7: Treating sin²x Like |sin x|

Wrong: Thinking $\sin^2 x = |\sin x|$ .

Right: $\sin^2 x \leq |\sin x|$ , but they're not equal. Use $\sin^2 x = (1-\cos 2x)/2$ for analysis.

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 Classification Summary

Integral	Classification	Reason
$\int_1^\infty \frac{\sin x}{x}\,dx$	Conditional	Dirichlet test; $\int \|\sin x\|/x$ diverges
$\int_1^\infty \frac{\cos x}{x^2}\,dx$	Absolute	$\|\cos x\|/x^2 \leq 1/x^2$ , p-test
$\int_1^\infty \frac{\sin x}{\sqrt{x}}\,dx$	Conditional	Dirichlet test; p = 1/2 < 1
$\int_1^\infty \frac{\sin x}{x^{1.5}}\,dx$	Absolute	p = 1.5 > 1, comparison test
$\int_1^\infty e^{-x}\sin x\,dx$	Absolute	$\|e^{-x}\sin x\| \leq e^{-x}$
$\int_0^\infty \frac{\sin(x^2)}{x}\,dx$	Conditional	Fresnel-type, substitution u=x²

14. Decision Tree for Classification

1

Check if ∫|f| converges

Use comparison, p-test, or other standard tests on |f(x)|

2

If ∫|f| converges → Absolutely Convergent

Done! The integral converges and is well-behaved.

3

If ∫|f| diverges, check if ∫f converges

Use Dirichlet or Abel test for oscillating integrands

4

If ∫f converges but ∫|f| diverges → Conditionally Convergent

The convergence depends on cancellation effects.

5

If both diverge → Divergent

The integral does not converge.

15. More Classification Examples

Example 6.30: Borderline Case

Problem: Classify $\int_1^{+\infty} \frac{\sin x}{x \ln x}\,dx$

Solution:

Step 1: Check absolute convergence.

$|\sin x/(x\ln x)| \leq 1/(x\ln x)$

Since $\int 1/(x\ln x)\,dx$ diverges (logarithmic p-test with p=1):

Step 2: Apply Dirichlet test.

Let $f = \sin x$ (bounded partials), $g = 1/(x\ln x)$ (decreasing to 0).

The integral converges conditionally.

Example 6.31: Product with Exponential

Problem: Classify $\int_0^{+\infty} e^{-x}\sin(x^2)\,dx$

Solution:

Check absolute convergence:

|e^{-x}\sin(x^2)| \leq e^{-x}

Since $\int_0^{+\infty} e^{-x}\,dx = 1$ converges:

The integral converges absolutely.

Example 6.32: Alternating Power

Problem: Classify $\int_1^{+\infty} \frac{\sin(\ln x)}{x}\,dx$

Solution:

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

\int_1^{+\infty} \frac{\sin(\ln x)}{x}\,dx = \int_0^{+\infty} \sin u\,du

This integral diverges (oscillates without decay).

Absolute & Conditional Convergence Quiz

12

Questions

0

Correct

0%

Accuracy

1

The integral

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

is:

Medium

Not attempted

2

If

\int_a^{+\infty} |f(x)|\,dx

converges, then

\int_a^{+\infty} f(x)\,dx

:

Easy

Not attempted

3

The Dirichlet test requires g(x) to be:

Medium

Not attempted

4

The integral

\int_1^{+\infty} \frac{\cos x}{x^2}\,dx

is:

Medium

Not attempted

5

A conditionally convergent integral can have its value changed by:

Hard

Not attempted

6

The integral

\int_1^{+\infty} \frac{\sin x}{x^{1.5}}\,dx

is:

Medium

Not attempted

7

For the Abel test, which condition is NOT required?

Hard

Not attempted

8

The integral

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

:

Hard

Not attempted

9

If

\int f

converges conditionally and

\int g

converges absolutely, then

\int(f+g)

:

Hard

Not attempted

10

For Dirichlet test, which function f(x) has bounded partial integrals?

Easy

Not attempted

11

The integral

\int_1^{+\infty} \frac{\sin(1/x)}{x}\,dx

:

Hard

Not attempted

12

Which statement about absolute convergence is FALSE?

Medium

Not attempted

Frequently Asked Questions

What is the difference between absolute and conditional convergence?

Absolute: ∫|f| converges. Conditional: ∫f converges but ∫|f| diverges. Absolute convergence is stronger and implies regular convergence.

When should I use the Dirichlet test?

Use Dirichlet when you have a product f(x)g(x) where g(x) → 0 monotonically and the partial integrals of f are bounded (like sin x or cos x).

What is the Abel test?

If ∫f converges and g is bounded and monotonic, then ∫fg converges. It's useful when g doesn't go to zero but is well-behaved.

Why does ∫sin(x)/x dx converge but ∫|sin(x)/x| dx diverge?

The oscillations of sin(x) cause cancellation that makes the integral converge. But |sin(x)| ≥ sin²(x), and ∫sin²(x)/x dx ~ ∫1/(2x) dx which diverges.

How do I determine if an integral is absolutely or conditionally convergent?

First check if ∫|f| converges using comparison or p-tests. If yes, it's absolutely convergent. If ∫|f| diverges but ∫f converges (often by Dirichlet/Abel), it's conditionally convergent.

What functions have bounded partial integrals?

Oscillating functions like sin(x), cos(x), sin(ax+b), cos(ax+b), and bounded periodic functions. The key is that positive and negative areas cancel out.

Can I use comparison test for conditionally convergent integrals?

Not directly. Comparison test requires non-negative functions. For oscillating integrands, use Dirichlet or Abel tests instead.

What is the threshold exponent for ∫sin(x)/x^p dx?

For p > 1: absolutely convergent. For 0 < p ≤ 1: conditionally convergent. For p ≤ 0: divergent. The threshold is p = 1.

Why is absolute convergence 'better' than conditional?

Absolutely convergent integrals can be rearranged, split arbitrarily, and composed more freely. Conditionally convergent integrals are 'fragile' - their value can change with rearrangement.

How do Fresnel integrals fit in?

Fresnel integrals ∫sin(x²)dx and ∫cos(x²)dx converge conditionally. They arise in optics and are related to ∫sin(x)/x through substitution.

Absolute and Conditional Convergence

1. Introduction: Why Two Types of Convergence?

2. Definitions of Absolute and Conditional Convergence

3. The Dirichlet Test

4. The Abel Test

5. Deep Dive: The sin(x)/x Integral

6. Properties and Implications

7. More Worked Examples

8. Quick Reference: When to Use Each Test

Use Comparison/p-Test When:

Use Dirichlet Test When:

Use Abel Test When:

Classification Strategy:

8. Common Mistakes to Avoid

Mistake 1: Assuming Convergence Implies Absolute Convergence

Mistake 2: Using Comparison Test for Oscillating Integrands

Mistake 3: Misapplying Dirichlet Conditions

Mistake 4: Confusing Dirichlet and Abel Tests

Mistake 5: Forgetting the Threshold Exponent

Mistake 6: Not Checking Both Conditions

Mistake 7: Treating sin²x Like |sin x|

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 Classification Summary

14. Decision Tree for Classification

15. More Classification Examples

Frequently Asked Questions

What is the difference between absolute and conditional convergence?

When should I use the Dirichlet test?

What is the Abel test?

Why does ∫sin(x)/x dx converge but ∫|sin(x)/x| dx diverge?

How do I determine if an integral is absolutely or conditionally convergent?

What functions have bounded partial integrals?

Can I use comparison test for conditionally convergent integrals?

What is the threshold exponent for ∫sin(x)/x^p dx?

Why is absolute convergence 'better' than conditional?

How do Fresnel integrals fit in?

Absolute Convergence

Conditional Convergence

Dirichlet Test

Abel Test

The Threshold Exponent

Remember