MathIsimple
Course 12

Improper Integrals

Section 1: Type I Improper Integrals

Definition 1.1: Type I Improper Integral

If ff is integrable on [a,b][a, b] for all b>ab > a, then:

∫a+∞f(x) dx=lim⁡b→+∞∫abf(x) dx\int_a^{+\infty} f(x)\,dx = \lim_{b \to +\infty} \int_a^b f(x)\,dx

If the limit exists and is finite, the integral converges; otherwise it diverges.

Example 1.1: Convergent Type I Integral

Evaluate: ∫1+∞1x2 dx\int_1^{+\infty} \frac{1}{x^2}\,dx

Solution:

∫1+∞1x2 dx=lim⁡b→+∞∫1b1x2 dx=lim⁡b→+∞[−1x]1b\int_1^{+\infty} \frac{1}{x^2}\,dx = \lim_{b \to +\infty} \int_1^b \frac{1}{x^2}\,dx = \lim_{b \to +\infty} \left[-\frac{1}{x}\right]_1^b
=lim⁡b→+∞(−1b+1)=1= \lim_{b \to +\infty} \left(-\frac{1}{b} + 1\right) = 1

The integral converges to 1. ∎

Example 1.2: Divergent Type I Integral

Evaluate: ∫1+∞1x dx\int_1^{+\infty} \frac{1}{x}\,dx

Solution:

∫1+∞1x dx=lim⁡b→+∞[ln⁡x]1b=lim⁡b→+∞ln⁡b=+∞\int_1^{+\infty} \frac{1}{x}\,dx = \lim_{b \to +\infty} [\ln x]_1^b = \lim_{b \to +\infty} \ln b = +\infty

The integral diverges. ∎

Section 2: Type II Improper Integrals

Definition 2.1: Type II Improper Integral

If ff has a singularity at aa and is integrable on [a+ε,b][a+\varepsilon, b] for all ε>0\varepsilon > 0, then:

∫abf(x) dx=lim⁡ε→0+∫a+εbf(x) dx\int_a^b f(x)\,dx = \lim_{\varepsilon \to 0^+} \int_{a+\varepsilon}^b f(x)\,dx

If the limit exists and is finite, the integral converges; otherwise it diverges.

Example 2.1: Convergent Type II Integral

Evaluate: ∫011x dx\int_0^1 \frac{1}{\sqrt{x}}\,dx

Solution:

∫011x dx=lim⁡ε→0+∫ε1x−1/2 dx=lim⁡ε→0+[2x]ε1\int_0^1 \frac{1}{\sqrt{x}}\,dx = \lim_{\varepsilon \to 0^+} \int_\varepsilon^1 x^{-1/2}\,dx = \lim_{\varepsilon \to 0^+} [2\sqrt{x}]_\varepsilon^1
=lim⁡ε→0+(2−2ε)=2= \lim_{\varepsilon \to 0^+} (2 - 2\sqrt{\varepsilon}) = 2

The integral converges to 2. ∎

Section 3: Convergence Tests

Theorem 3.1: p-Integral Test (Type I)

The integral ∫1+∞1xp dx\int_1^{+\infty} \frac{1}{x^p}\,dx:

  • Converges if p>1p > 1
  • Diverges if p≤1p \leq 1
Theorem 3.2: p-Integral Test (Type II)

The integral ∫011xp dx\int_0^1 \frac{1}{x^p}\,dx:

  • Converges if p<1p < 1
  • Diverges if p≥1p \geq 1
Theorem 3.3: Comparison Test

If 0≤f(x)≤g(x)0 \leq f(x) \leq g(x) for x≥ax \geq a:

  • If ∫a+∞g(x) dx\int_a^{+\infty} g(x)\,dx converges, then ∫a+∞f(x) dx\int_a^{+\infty} f(x)\,dx converges
  • If ∫a+∞f(x) dx\int_a^{+\infty} f(x)\,dx diverges, then ∫a+∞g(x) dx\int_a^{+\infty} g(x)\,dx diverges
Example 3.1: Comparison Test Application

Show: ∫1+∞1x2+1 dx\int_1^{+\infty} \frac{1}{x^2+1}\,dx converges

Solution: Since 1x2+1<1x2\frac{1}{x^2+1} < \frac{1}{x^2} for x≥1x \geq 1:

And ∫1+∞1x2 dx\int_1^{+\infty} \frac{1}{x^2}\,dx converges (p = 2 > 1).

By comparison test, ∫1+∞1x2+1 dx\int_1^{+\infty} \frac{1}{x^2+1}\,dx converges. ∎

Section 4: Absolute Convergence

Definition 4.1: Absolute Convergence

An improper integral ∫a+∞f(x) dx\int_a^{+\infty} f(x)\,dx is absolutely convergent if:

∫a+∞∣f(x)∣ dx\int_a^{+\infty} |f(x)|\,dx

converges.

Theorem 4.1: Absolute Convergence Implies Convergence

If ∫a+∞∣f(x)∣ dx\int_a^{+\infty} |f(x)|\,dx converges, then ∫a+∞f(x) dx\int_a^{+\infty} f(x)\,dx converges.

Example 4.1: Absolute Convergence

Show: ∫1+∞sin⁡xx2 dx\int_1^{+\infty} \frac{\sin x}{x^2}\,dx converges

Solution: Since ∣sin⁡xx2∣≤1x2\left|\frac{\sin x}{x^2}\right| \leq \frac{1}{x^2}:

And ∫1+∞1x2 dx\int_1^{+\infty} \frac{1}{x^2}\,dx converges, so:

∫1+∞∣sin⁡xx2∣ dx\int_1^{+\infty} \left|\frac{\sin x}{x^2}\right|\,dx converges.

Therefore the original integral converges absolutely. ∎

Section 5: Limit Comparison Test

Theorem 5.1: Limit Comparison Test

If f,g≥0f, g \geq 0 and lim⁡x→+∞f(x)g(x)=L\lim_{x \to +\infty} \frac{f(x)}{g(x)} = L where 0<L<∞0 < L < \infty, then:

∫a+∞f(x) dx converges ⇔∫a+∞g(x) dx converges\int_a^{+\infty} f(x)\,dx \text{ converges } \Leftrightarrow \int_a^{+\infty} g(x)\,dx \text{ converges}
Example 5.1: Application

Determine convergence: ∫1+∞x+1x3+2x dx\int_1^{+\infty} \frac{x+1}{x^3+2x}\,dx

As x→∞x \to \infty: x+1x3+2x∼xx3=1x2\frac{x+1}{x^3+2x} \sim \frac{x}{x^3} = \frac{1}{x^2}

Since ∫1+∞1x2 dx\int_1^{+\infty} \frac{1}{x^2}\,dx converges (p=2>1), the original converges.

Section 6: Cauchy Criterion

Theorem 6.1: Cauchy Criterion for Improper Integrals

∫a+∞f(x) dx\int_a^{+\infty} f(x)\,dx converges if and only if:

∀ε>0,∃M>a:∀b,c>M,∣∫bcf(x) dx∣<ε\forall \varepsilon > 0, \exists M > a: \forall b, c > M, \left|\int_b^c f(x)\,dx\right| < \varepsilon

Section 7: Mixed Improper Integrals

Definition 7.1: Mixed Improper Integral

An integral is mixed if it has both infinite limits and singularities. Split at a convenient point and analyze each part separately.

Example 7.1: Mixed Integral

Evaluate: ∫0+∞1x2 dx\int_0^{+\infty} \frac{1}{x^2}\,dx

This has a singularity at 0 and infinite limit. Split at x=1:

∫0+∞1x2 dx=∫011x2 dx+∫1+∞1x2 dx\int_0^{+\infty} \frac{1}{x^2}\,dx = \int_0^1 \frac{1}{x^2}\,dx + \int_1^{+\infty} \frac{1}{x^2}\,dx

Both parts diverge, so the integral diverges.

Work These Ideas by Hand — Then Check Yourself

Practice Quiz: Improper Integrals
10
Questions
0
Correct
0%
Accuracy
1
The integral ∫1+∞1x2 dx\int_1^{+\infty} \frac{1}{x^2}\,dx is:
Easy
Not attempted
2
The integral ∫011x dx\int_0^1 \frac{1}{\sqrt{x}}\,dx is:
Easy
Not attempted
3
The integral ∫1+∞1x dx\int_1^{+\infty} \frac{1}{x}\,dx is:
Easy
Not attempted
4
To show ∫1+∞1x2+1dx\int_1^{+\infty} \frac{1}{x^2+1}dx converges, we compare with:
Medium
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5
If lim⁡x→+∞f(x)g(x)=L\lim_{x \to +\infty} \frac{f(x)}{g(x)} = L where 0<L<∞0 < L < \infty, then:
Medium
Not attempted
6
For 0≤f(x)≤g(x)0 \leq f(x) \leq g(x) and ∫a+∞g(x)dx\int_a^{+\infty} g(x)dx converges, then:
Easy
Not attempted
7
The integral ∫011x0.5dx\int_0^1 \frac{1}{x^{0.5}}dx:
Easy
Not attempted
8
For Type II integrals with singularity at x=a, the p-test says convergence occurs when:
Easy
Not attempted
9
The integral ∫1+∞x+1x3+2xdx\int_1^{+\infty} \frac{x+1}{x^3+2x}dx:
Medium
Not attempted
10
For ∫1+∞sin⁡xx2dx\int_1^{+\infty} \frac{\sin x}{x^2}dx, which test is most appropriate?
Medium
Not attempted
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Frequently Asked Questions

What makes an integral 'improper'?

An integral is improper if either (1) the integration interval is infinite (Type I), or (2) the integrand becomes unbounded at some point in the interval (Type II). Both types require limits to define.

How do I know if an integral is Type I or Type II?

Type I: Look for ±∞ in the limits. Type II: Check if the function has any singularities (points where it goes to ±∞) within [a,b]. An integral can be both types (mixed).

What is the p-integral test?

For Type I: ∫₁^∞ 1/xᵖ dx converges if p > 1, diverges if p ≤ 1. For Type II: ∫₀¹ 1/xᵖ dx converges if p < 1, diverges if p ≥ 1.

What is the comparison test?

If 0 ≤ f(x) ≤ g(x) and ∫g converges, then ∫f converges. If f(x) ≥ g(x) ≥ 0 and ∫g diverges, then ∫f diverges.

What is absolute convergence?

An improper integral ∫f is absolutely convergent if ∫|f| converges. Absolute convergence implies convergence, but not vice versa.