Detailed Study Notes on Improper Integrals
7 Techniques of Integration
7.8 Improper Integrals
1. Definition of Improper Integrals
Improper integrals extend the concept of definite integrals to cases where:
The interval is infinite.
The function has an infinite discontinuity within the interval \[a, b].
In these scenarios, the integral is classified as an improper integral.
2. Type 1: Infinite Intervals
2.1 Infinite Area Concept
Consider an infinite region S, defined as the area under the curve above the x-axis and to the right of the line (x = 1).
Initially, one might assume the area is infinite due to its extent. However, examining specific intervals (e.g., from (x = 1) to (x = t)) reveals otherwise.
2.2 Area Limitation
The area (A(t)) calculated as shaded in the region approaches 1 as (t o \infty). This suggests:
Therefore, despite the infinite extent, the area of region S is finite and equal to 1.
2.3 Definition of Improper Integral of Type 1
The integral of (f) over an infinite interval is defined as:
\int{a}^{\infty} f(x) \, dx = \lim{t \to \infty} \int_{a}^{t} f(x) \, dx
\int{-\infty}^{b} f(x) \, dx = \lim{t \to -\infty} \int_{t}^{b} f(x) \, dx
These limits must exist as finite numbers for the improper integrals to be defined.
3. Convergence and Divergence of Improper Integrals
3.1 Categories
Improper integrals are classified:
Convergent: If the limit exists as a finite number.
Divergent: If the limit does not exist.
3.2 Combined Convergence
If both integrals in Definition 1 are convergent, we define: \int{a}^{\infty} f(x) \, dx + \int{-\infty}^{b} f(x) \, dx
Any real number (a) is valid for this expression (reference to Exercise 88).
4. Area Interpretation of Improper Integrals
When (f) is a positive function, improper integrals can be interpreted as areas. Specifically:
Case (a): If (f(x) \geq 0) and the integral is convergent, it is defined as the area of the region.
Illustrated by the area under the graph of (f) from (a) to (t) as (t) approaches infinity.
5. Example 1: Convergence Determination
Problem: Determine if the integral (\int_{1}^{\infty} f(x) \, dx) is convergent or divergent.
Solution: From Definition 1(a), we find that the limit does not exist as a finite number, thus:
The improper integral is divergent.
6. Comparison of Results in Type 1 Integrals
Example comparisons reveal distinct results even when curves appear similar:
Although curves look akin for (x > 0), one has a finite area while the other diverges.
Type 2: Discontinuous Integrands
1. Definition and Context
Consider (f) as a positive continuous function defined on a finite interval but exhibits a vertical asymptote at (b).
The unbounded region S is then under the graph of (f) and above the x-axis between (a) and (b).
2. Area Concept for Type 2
The area between (a) and (t) approaches a definite number (A) as (t) approaches the vertical asymptote:
Denote this area mathematically as:
\lim_{t \to b^{-}} A(t) = A
3. Definition of Improper Integral of Type 2
For Type 2, the improper integral is defined as:
\int{a}^{b} f(x) \, dx = \lim{t \to b^{-}} \int_{a}^{t} f(x) \, dx
When (f) is discontinuous at (b).
\int{a}^{b} f(x) \, dx = \lim{t \to a^{+}} \int_{t}^{b} f(x) \, dx
When (f) is discontinuous at (a).
4. Convergence and Divergence for Type 2
Similar to Type 1, improper integrals are also classified:
Convergent: If the limit exists as a finite number.
Divergent: If it does not exist.
When discontinuities occur at a point (c) within (a < c < b), if both are convergent, the combined definition applies.
5. Example 5: Finding Convergence
Problem: Evaluate the improper integral near a vertical asymptote at (x = 2).
Determination: Since the discontinuity occurs at the left endpoint of the interval [2, 5], we apply part (b) of Definition 3 and find:
The integral is convergent, with the area being the value of the integral.
A Comparison Test for Improper Integrals
1. Need for Comparison Theorems
Difficulty may arise in evaluating the exact value of an improper integral; thus, knowing if it is convergent or divergent becomes significant.
Comparison Theorem establishes bounds:
Let (f) and (g) be continuous functions such that (f(x) \geq g(x) \geq 0) for (x \geq a).
2. Theorem Consequence
If:
(\int{a}^{\infty} g(x) \, dx) is convergent, then (\int{a}^{\infty} f(x) \, dx) is convergent.
If (\int{a}^{\infty} f(x) \, dx) is divergent, then (\int{a}^{\infty} g(x) \, dx) is also divergent.
It's important to note that these conditions are not reversible; convergence of one does not ensure the convergence of the other.
3. Example 9: Demonstrating Convergence
Task: Show that the integral (\int_{1}^{\infty} f(x) \, dx) is convergent.
Solution Analysis: Direct evaluation isn't possible due to non-elementary functions. Instead, we can express it in terms of known integrals:
Breaking it down, we connect it to an easier known definite integral.
Therefore, invoking results from the comparison, we conclude:
The integral with respect to infamous analytic and continuous behavior is convergent.