1. Introduction: Why Improper Integrals?

The Riemann integral $\int_a^b f(x)\,dx$ requires two conditions:

The interval $[a, b]$ must be bounded (finite length)
The function $f$ must be bounded on $[a, b]$

But many important integrals in mathematics, physics, and probability violate these conditions:

Gaussian Integral

\int_{-\infty}^{+\infty} e^{-x^2}\,dx = \sqrt{\pi}

Infinite interval

Arc Length

\int_0^1 \frac{1}{\sqrt{x}}\,dx = 2

Unbounded integrand

Remark 6.1: Two Types of Improper Integrals

Type I (Infinite Limits): Integration interval extends to $\pm\infty$
Type II (Singular Points): Integrand becomes unbounded at some point(s)

Both types are defined using limits, extending the Newton-Leibniz formula.

2. Type I: Improper Integrals with Infinite Limits

Definition 6.1: Type I Improper Integral (Upper Limit)

Let $f$ be defined on $[a, +\infty)$ and integrable on every $[a, b]$ for $b > a$ .

If the limit

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

exists and is finite, we say the improper integral $\int_a^{+\infty} f(x)\,dx$ converges, and define:

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

If the limit does not exist or is infinite, the integral diverges.

Definition 6.2: Type I Improper Integral (Lower Limit)

Similarly, for $f$ defined on $(-\infty, b]$ :

\int_{-\infty}^b f(x)\,dx = \lim_{a \to -\infty} \int_a^b f(x)\,dx

Definition 6.3: Type I Improper Integral (Both Limits)

For $f$ defined on $(-\infty, +\infty)$ , we split at any point $c$ :

\int_{-\infty}^{+\infty} f(x)\,dx = \int_{-\infty}^c f(x)\,dx + \int_c^{+\infty} f(x)\,dx

The integral converges if and only if both parts converge independently.

Remark 6.2: Independence of Splitting Point

The choice of $c$ does not affect convergence or the value. Any $c \in \mathbb{R}$ works.

Example 6.1: Basic Type I Integral

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

Solution:

\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 \to +\infty} \left(-\frac{1}{b} + 1\right) = 1

The integral converges with value $1$ .

Example 6.2: Divergent Type I Integral

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

Solution:

\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.

Example 6.3: Both Limits Infinite

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

Solution:

Split at $c = 0$ :

\int_{-\infty}^0 \frac{1}{1+x^2}\,dx = \lim_{a \to -\infty} [\arctan x]_a^0 = 0 - (-\frac{\pi}{2}) = \frac{\pi}{2}

\int_0^{+\infty} \frac{1}{1+x^2}\,dx = \lim_{b \to +\infty} [\arctan x]_0^b = \frac{\pi}{2} - 0 = \frac{\pi}{2}

Total: $\frac{\pi}{2} + \frac{\pi}{2} = \pi$

Remark 6.3: Warning: Independent Limits

Common Mistake:

For $\int_{-\infty}^{+\infty} x\,dx$ , one might think:

\lim_{R \to +\infty} \int_{-R}^R x\,dx = \lim_{R \to +\infty} 0 = 0 \quad \text{(WRONG!)}

This is the Cauchy Principal Value, not the improper integral. The correct evaluation:

\int_{-\infty}^0 x\,dx = -\infty, \quad \int_0^{+\infty} x\,dx = +\infty

Both diverge, so $\int_{-\infty}^{+\infty} x\,dx$ diverges.

3. Type II: Improper Integrals with Singular Points

Definition 6.4: Singular Point

A point $c$ is a singular point (or singularity) of $f$ if $f(x) \to \pm\infty$ as $x \to c$ .

Definition 6.5: Type II Improper Integral (Singularity at Left Endpoint)

Let $f$ be defined on $(a, b]$ with $f$ unbounded near $x = a$ .

If the limit

\lim_{\epsilon \to 0^+} \int_{a+\epsilon}^b f(x)\,dx

exists and is finite, we say the integral converges and define:

\int_a^b f(x)\,dx = \lim_{\epsilon \to 0^+} \int_{a+\epsilon}^b f(x)\,dx

Definition 6.6: Type II Improper Integral (Singularity at Right Endpoint)

For $f$ unbounded near $x = b$ :

\int_a^b f(x)\,dx = \lim_{\epsilon \to 0^+} \int_a^{b-\epsilon} f(x)\,dx

Definition 6.7: Type II Improper Integral (Interior Singularity)

If $f$ has a singularity at $c \in (a, b)$ , split the integral:

\int_a^b f(x)\,dx = \int_a^c f(x)\,dx + \int_c^b f(x)\,dx

The integral converges if and only if both parts converge.

Example 6.4: Singularity at Left Endpoint

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

Solution:

The function $1/\sqrt{x} \to +\infty$ as $x \to 0^+$ .

\int_0^1 \frac{1}{\sqrt{x}}\,dx = \lim_{\epsilon \to 0^+} \int_\epsilon^1 x^{-1/2}\,dx

= \lim_{\epsilon \to 0^+} [2\sqrt{x}]_\epsilon^1 = \lim_{\epsilon \to 0^+} (2 - 2\sqrt{\epsilon}) = 2

The integral converges with value $2$ .

Example 6.5: Divergent Type II Integral

Problem: Evaluate $\int_0^1 \frac{1}{x}\,dx$

Solution:

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

The integral diverges.

Example 6.6: Interior Singularity

Problem: Evaluate $\int_0^2 \frac{1}{(x-1)^{2/3}}\,dx$

Solution:

Singularity at $x = 1$ . Split:

\int_0^1 \frac{1}{(x-1)^{2/3}}\,dx = \lim_{\epsilon \to 0^+} [3(x-1)^{1/3}]_0^{1-\epsilon}

= \lim_{\epsilon \to 0^+} (3(-\epsilon)^{1/3} - 3(-1)^{1/3}) = 0 - 3(-1) = 3

\int_1^2 \frac{1}{(x-1)^{2/3}}\,dx = \lim_{\epsilon \to 0^+} [3(x-1)^{1/3}]_{1+\epsilon}^2 = 3 - 0 = 3

Total: $3 + 3 = 6$

4. Mixed Improper Integrals

Definition 6.8: Mixed Improper Integral

An integral is mixed if it has both an infinite limit AND a singular point, or multiple singularities.

Strategy: Split the integral so each part has only one "problem" (one infinite limit or one singularity), then check each part independently.

Example 6.7: Mixed Type Integral

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

Solution:

Issues: singularity at $x = 0$ AND infinite limit. Split at $x = 1$ :

\int_0^{+\infty} = \int_0^1 + \int_1^{+\infty}

Near 0: $\frac{1}{\sqrt{x}(1+x)} \approx \frac{1}{\sqrt{x}}$ , and $\int_0^1 1/\sqrt{x}\,dx = 2$ converges.

Near ∞: $\frac{1}{\sqrt{x}(1+x)} \approx \frac{1}{x^{3/2}}$ , and $\int_1^{+\infty} 1/x^{3/2}\,dx = 2$ converges.

Both parts converge, so the integral converges.

Remark 6.4: Splitting Strategy

When splitting a mixed integral:

Choose a splitting point away from singularities
Each resulting integral should have only one "problem"
All parts must converge for the whole to converge
If any part diverges, the whole integral diverges

5. Basic Properties

Theorem 6.1: Linearity

If $\int_a^{+\infty} f(x)\,dx$ and $\int_a^{+\infty} g(x)\,dx$ both converge, then for constants $\alpha, \beta$ :

\int_a^{+\infty} [\alpha f(x) + \beta g(x)]\,dx = \alpha\int_a^{+\infty} f(x)\,dx + \beta\int_a^{+\infty} g(x)\,dx

Theorem 6.2: Interval Additivity

For any $c \in (a, +\infty)$ :

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

And when convergent: $\int_a^{+\infty} f = \int_a^c f + \int_c^{+\infty} f$

Remark 6.5: Convergence is a Tail Property

Whether an improper integral converges depends only on the behavior near the "problem" (∞ or the singularity). Changing the integrand on any bounded interval away from singularities does not affect convergence.

Theorem 6.3: Newton-Leibniz for Improper Integrals

If $F'(x) = f(x)$ on the relevant interval:

Type I:

\int_a^{+\infty} f(x)\,dx = \lim_{x \to +\infty} F(x) - F(a) = F(+\infty) - F(a)

Type II (singularity at a):

\int_a^b f(x)\,dx = F(b) - \lim_{x \to a^+} F(x) = F(b) - F(a^+)

6. The p-Integral: A Fundamental Example

Theorem 6.4: p-Integral (Type I)

For the integral with infinite upper limit:

\int_1^{+\infty} \frac{1}{x^p}\,dx \begin{cases} \text{converges} = \frac{1}{p-1} & \text{if } p > 1 \\ \text{diverges} & \text{if } p \leq 1 \end{cases}

Proof:

Case p ≠ 1:

\int_1^b x^{-p}\,dx = \left[\frac{x^{1-p}}{1-p}\right]_1^b = \frac{b^{1-p} - 1}{1-p}

As $b \to +\infty$ :

If $p > 1$ : $1-p < 0$ , so $b^{1-p} \to 0$ . Limit = $\frac{1}{p-1}$
If $p < 1$ : $1-p > 0$ , so $b^{1-p} \to +\infty$ . Diverges.

Case p = 1: $\int_1^b 1/x\,dx = \ln b \to +\infty$ . Diverges.

∎

Theorem 6.5: p-Integral (Type II)

For the integral with singularity at 0:

\int_0^1 \frac{1}{x^p}\,dx \begin{cases} \text{converges} = \frac{1}{1-p} & \text{if } p < 1 \\ \text{diverges} & \text{if } p \geq 1 \end{cases}

Remark 6.6: Opposite Conditions!

Critical Observation:

Type I ( $\int_1^{+\infty} 1/x^p$ ): converges when $p > 1$
Type II ( $\int_0^1 1/x^p$ ): converges when $p < 1$

The conditions are opposite! This is a common source of errors.

Example 6.8: Why ∫₀^∞ 1/x² Diverges

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

Solution:

Split at $x = 1$ :

$\int_1^{+\infty} 1/x^2\,dx$ : converges (p = 2 > 1) ✓
$\int_0^1 1/x^2\,dx$ : diverges (p = 2 ≥ 1) ✗

Since one part diverges, the whole integral diverges.

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 Solution

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

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

= \frac{1}{2}\left[-\frac{1}{u}\right]_1^{+\infty} = \frac{1}{2}(0 - (-1)) = \boxed{\frac{1}{2}}

The integral converges with value $\frac{1}{2}$ .

Challenge 2

For what values of α does the following converge?

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

Show Solution

Solution: Check singularities at x = 0 and x = 1.

Near x = 1: By L'Hôpital, $\lim_{x \to 1} \frac{x^\alpha - 1}{\ln x} = \lim_{x \to 1} \frac{\alpha x^{\alpha-1}}{1/x} = \alpha$

So the integrand is bounded near x = 1 (no singularity).

Near x = 0: $\frac{x^\alpha - 1}{\ln x} \approx \frac{-1}{\ln x}$ for $\alpha > 0$ , which is bounded.

For $\alpha \leq 0$ : $x^\alpha \to +\infty$ as $x \to 0^+$ , need more careful analysis.

Answer: The integral converges for all $\boxed{\alpha > -1}$ .

Challenge 3

Evaluate the following integral:

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

Show Solution

Solution: Split at x = 1:

I = \int_0^1 \frac{\ln x}{1+x^2}\,dx + \int_1^{+\infty} \frac{\ln x}{1+x^2}\,dx

For the second integral, substitute $u = 1/x$ , so $dx = -du/u^2$ :

\int_1^{+\infty} \frac{\ln x}{1+x^2}\,dx = \int_1^0 \frac{-\ln u}{1+1/u^2} \cdot \frac{-du}{u^2} = \int_0^1 \frac{-\ln u}{1+u^2}\,du

Therefore:

I = \int_0^1 \frac{\ln x}{1+x^2}\,dx - \int_0^1 \frac{\ln x}{1+x^2}\,dx = \boxed{0}

Challenge 4

Prove that the following integral converges:

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

Show Solution

Solution: Identify singularities in [0, 1]: x = 0 and x = 1.

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

This is a p-integral with p = 1/3 < 1, so it converges.

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

This is also p = 1/3 < 1, so it converges.

Since both singularities are integrable, the integral converges. ∎

13. Historical Context

Euler (1707-1783)

Leonhard Euler was the first to systematically study integrals with infinite limits, computing famous results like $\int_0^{+\infty} e^{-x^2}dx = \frac{\sqrt{\pi}}{2}$ using ingenious techniques.

Cauchy (1789-1857)

Augustin-Louis Cauchy provided rigorous definitions using limits, establishing the foundation for modern analysis. He introduced the concept of principal value for certain divergent integrals.

Riemann (1826-1866)

Bernhard Riemann extended integration theory to handle more general functions and contributed to our understanding of convergence for improper integrals.

14. Connections to Other Topics

Infinite Series

Improper integrals and infinite series are deeply connected. The integral test states:

\sum_{n=1}^{\infty} f(n) \text{ and } \int_1^{+\infty} f(x)\,dx

converge or diverge together (for positive, decreasing f).

Laplace Transform

The Laplace transform is an improper integral:

\mathcal{L}\{f(t)\} = \int_0^{+\infty} f(t)e^{-st}\,dt

Used extensively in differential equations and engineering.

Probability Theory

Continuous probability distributions require improper integrals:

\int_{-\infty}^{+\infty} f(x)\,dx = 1

The normal distribution uses the Gaussian integral.

Fourier Transform

The Fourier transform is another improper integral:

\hat{f}(\omega) = \int_{-\infty}^{+\infty} f(t)e^{-i\omega t}\,dt

Fundamental in signal processing and physics.

15. More Practice Examples

Example 6.18: Exponential with Power

Problem: Evaluate $\int_0^{+\infty} x^3 e^{-2x}\,dx$

Solution:

Use repeated integration by parts or substitute $u = 2x$ :

\int_0^{+\infty} x^3 e^{-2x}\,dx = \frac{1}{16}\int_0^{+\infty} u^3 e^{-u}\,du = \frac{\Gamma(4)}{16} = \frac{3!}{16} = \frac{6}{16} = \frac{3}{8}

Example 6.19: Trigonometric Substitution

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

Solution:

This is Type II with singularity at $x = 1$ .

\int_0^1 \frac{1}{\sqrt{1-x^2}}\,dx = [\arcsin x]_0^1 = \frac{\pi}{2} - 0 = \frac{\pi}{2}

Example 6.20: Rational Function at Infinity

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

Solution:

Substitute $x = \tan\theta$ :

\int_0^{+\infty} \frac{x^2}{(1+x^2)^2}\,dx = \int_0^{\pi/2} \frac{\tan^2\theta}{\sec^4\theta} \cdot \sec^2\theta\,d\theta

= \int_0^{\pi/2} \sin^2\theta\,d\theta = \frac{\pi}{4}

Example 6.21: Logarithm and Power

Problem: Evaluate $\int_0^1 x^2 \ln x\,dx$

Solution:

Type II at x = 0 (ln x → -∞). Integration by parts:

\int_0^1 x^2 \ln x\,dx = \left[\frac{x^3}{3}\ln x\right]_0^1 - \int_0^1 \frac{x^2}{3}\,dx

The boundary term at 0: $\lim_{x \to 0^+} \frac{x^3 \ln x}{3} = 0$

= 0 - 0 - \frac{1}{9} = -\frac{1}{9}

Example 6.22: Gaussian Type

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

Solution:

Substitute $u = x/\sqrt{2}$ , so $dx = \sqrt{2}\,du$ :

\int_0^{+\infty} e^{-x^2/2}\,dx = \sqrt{2}\int_0^{+\infty} e^{-u^2}\,du = \sqrt{2} \cdot \frac{\sqrt{\pi}}{2} = \sqrt{\frac{\pi}{2}}

16. Decision Flowchart

Step 1: Check for infinite limits → Type I

Step 2: Check for singularities in [a, b] → Type II

Step 3: If both, split into separate parts (mixed)

Step 4: Find antiderivative F(x)

Step 5: Evaluate limits:

\lim_{b \to \infty} F(b)

or

\lim_{\epsilon \to 0^+} F(a+\epsilon)

Step 6: If limit exists and finite → converges; otherwise → diverges

17. Summary Tables

Common Improper Integrals

Integral	Condition	Value
$\int_0^{+\infty} e^{-ax}\,dx$	a > 0	1/a
$\int_0^{+\infty} x^n e^{-ax}\,dx$	a > 0, n ≥ 0	$n!/a^{n+1}$
$\int_{-\infty}^{+\infty} e^{-x^2}\,dx$	—	√π
$\int_0^{+\infty} \frac{1}{1+x^2}\,dx$	—	π/2
$\int_0^1 x^{\alpha-1}(1-x)^{\beta-1}\,dx$	α, β > 0	B(α,β)
$\int_0^{+\infty} x^{\alpha-1}e^{-x}\,dx$	α > 0	Γ(α)

p-Integral Summary

Integral	Converges when	Diverges when
$\int_1^{+\infty} \frac{1}{x^p}\,dx$	p > 1	p ≤ 1
$\int_0^1 \frac{1}{x^p}\,dx$	p < 1	p ≥ 1
$\int_e^{+\infty} \frac{1}{x(\ln x)^p}\,dx$	p > 1	p ≤ 1
$\int_1^e \frac{1}{x\|\ln x\|^p}\,dx$	p < 1	p ≥ 1

18. Final Remarks

Improper integrals extend the power of integration to handle infinite intervals and unbounded functions. The key insights from this section are:

Improper integrals are defined as limits of proper integrals
Type I handles infinite limits; Type II handles singularities
The p-test provides quick convergence criteria with opposite conditions
Mixed integrals require splitting at convenient points
Newton-Leibniz formula extends naturally with appropriate limits

In the next section, we'll develop powerful convergence tests (comparison, limit comparison) that help analyze integrals without computing them explicitly.

19. Verification Checklist

Before concluding that an improper integral converges or diverges, verify the following:

Identified all infinite limits (Type I)

Identified all singular points (Type II)

Split mixed integrals correctly

Used correct p-test conditions for each type

Evaluated limits properly (not symmetric cancellation)

Checked interior singularities within [a, b]

Verified antiderivative is correct

Confirmed all sub-integrals converge independently

Definition & Properties Quiz

5

Questions

0

Correct

0%

Accuracy

1

The integral

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

is:

Easy

Not attempted

2

The integral

\int_0^1 \frac{1}{\sqrt{x}}\,dx

is:

Easy

Not attempted

3

The integral

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

is:

Easy

Not attempted

4

For

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

, the correct approach is:

Medium

Not attempted

5

If

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

and

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

both converge, then:

Medium

Not attempted

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).

Can I use the Newton-Leibniz formula for improper integrals?

Yes, but with limits! For Type I: ∫_a^{+∞} f(x)dx = lim_{b→+∞} F(b) - F(a). For Type II with singularity at a: ∫_a^b f(x)dx = F(b) - lim_{ε→0^+} F(a+ε).

What if both limits are infinite?

Split the integral at any convenient point c: ∫_{-∞}^{+∞} f(x)dx = ∫_{-∞}^c f(x)dx + ∫_c^{+∞} f(x)dx. Both parts must converge independently.

Definition and Basic Properties

1. Introduction: Why Improper Integrals?

2. Type I: Improper Integrals with Infinite Limits

3. Type II: Improper Integrals with Singular Points

4. Mixed Improper Integrals

5. Basic Properties

6. The p-Integral: A Fundamental Example

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. Historical Context

Euler (1707-1783)

Cauchy (1789-1857)

Riemann (1826-1866)

14. Connections to Other Topics

Infinite Series

Laplace Transform

Probability Theory

Fourier Transform

15. More Practice Examples

16. Decision Flowchart

17. Summary Tables

Common Improper Integrals

p-Integral Summary

18. Final Remarks

19. Verification Checklist

Frequently Asked Questions

What makes an integral 'improper'?

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

Can I use the Newton-Leibniz formula for improper integrals?

What if both limits are infinite?

Type I (Infinite Limits)

Type II (Singular)

p-Test (Type I)

p-Test (Type II)

Remember