- Calculate integrals over infinite intervals
- Find integrals when there’s an infinite discontinuity inside your interval
- Use the comparison theorem to determine if an improper integral converges
Integrating over an Infinite Interval
The Main Idea
Improper integrals handle two situations that break the rules of regular integration: infinite limits of integration and functions with discontinuities. We solve these by replacing the problematic parts with limits, transforming impossible calculations into manageable ones.
The Three Types of Infinite Limits:
Type 1: [latex]\int_a^{+\infty} f(x) , dx = \lim_{t \to +\infty} \int_a^t f(x) , dx[/latex]
Type 2: [latex]\int_{-\infty}^b f(x) , dx = \lim_{t \to -\infty} \int_t^b f(x) , dx[/latex]
Type 3: [latex]\int_{-\infty}^{+\infty} f(x) , dx = \int_{-\infty}^c f(x) , dx + \int_c^{+\infty} f(x) , dx[/latex] (split at any point [latex]c[/latex])
Convergence vs. Divergence:
- Converges: The limit exists and equals a finite number
- Diverges: The limit doesn’t exist or approaches [latex]\pm\infty[/latex]
For Type 3, both pieces must converge for the whole integral to converge.
Problem -Solving Strategy:
- Set up the limit by replacing infinity with a variable
- Integrate normally over the finite interval
- Evaluate the limit as the variable approaches infinity
- Use L’Hôpital’s rule if needed for indeterminate forms
Evaluate [latex]{\displaystyle\int }_{\text{-}\infty }^{0}\frac{1}{{x}^{2}+4}dx[/latex]. State whether the improper integral converges or diverges.
Evaluate [latex]{\displaystyle\int }_{-3}^{+\infty }{e}^{\text{-}x}dx[/latex]. State whether the improper integral converges or diverges.
For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.
You can view the transcript for this segmented clip of “3.7 Improper Integrals” here (opens in new window).
Integrating a Discontinuous Integrand
The Main Idea
Sometimes the problem isn’t an infinite interval—it’s that your function has a vertical asymptote or other discontinuity right in the middle of where you want to integrate. These improper integrals require the same limit approach, but now you’re approaching the troublesome point from one or both sides.
The Three Discontinuity Cases:
Type 1 – Right Endpoint Problem: [latex]\int_a^b f(x) , dx = \lim_{t \to b^-} \int_a^t f(x) , dx[/latex] (Function continuous on [latex][a,b)[/latex], discontinuous at [latex]b[/latex])
Type 2 – Left Endpoint Problem: [latex]\int_a^b f(x) , dx = \lim_{t \to a^+} \int_t^b f(x) , dx[/latex] (Function continuous on [latex](a,b][/latex], discontinuous at [latex]a[/latex])
Type 3 – Interior Discontinuity: [latex]\int_a^b f(x) , dx = \int_a^c f(x) , dx + \int_c^b f(x) , dx[/latex] (Function has discontinuity at some point [latex]c[/latex] inside [latex][a,b][/latex])
Critical Rule for Type 3: Both pieces must converge for the whole integral to converge. If either piece diverges, the entire integral diverges.
Problem -Solving Strategy:
- Identify the discontinuity – where does the function blow up or have a jump?
- Set up the appropriate limit based on where the discontinuity occurs
- Integrate normally over the continuous parts
- Evaluate the limit as you approach the problematic point
- Use L’Hôpital’s rule if you encounter indeterminate forms
Evaluate [latex]{\displaystyle\int }_{0}^{2}\frac{1}{x}dx[/latex]. State whether the integral converges or diverges.
For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.
You can view the transcript for this segmented clip of “3.7 Improper Integrals” here (opens in new window).
A Comparison Theorem
The Main Idea
When you can’t evaluate an improper integral directly, the comparison test lets you determine convergence or divergence by comparing your function to a simpler one whose behavior you already understand. It’s like using a known yardstick to measure an unknown quantity.
If [latex]0 \leq f(x) \leq g(x)[/latex] for [latex]x \geq a[/latex], then:
- Larger function dominates: If the smaller function [latex]f(x)[/latex] diverges, then the larger function [latex]g(x)[/latex] must also diverge
- Smaller function is bounded: If the larger function [latex]g(x)[/latex] converges, then the smaller function [latex]f(x)[/latex] must also converge
The Two Comparison Cases:
Case 1 – Proving Divergence: Find a smaller function that you know diverges If [latex]\int_a^{+\infty} f(x) , dx = +\infty[/latex] and [latex]f(x) \leq g(x)[/latex], then [latex]\int_a^{+\infty} g(x) , dx = +\infty[/latex]
Case 2 – Proving Convergence: Find a larger function that you know converges
If [latex]\int_a^{+\infty} g(x) , dx[/latex] converges and [latex]f(x) \leq g(x)[/latex], then [latex]\int_a^{+\infty} f(x) , dx[/latex] converges
Helpful Tips:
- Common comparison functions: [latex]\frac{1}{x^p}[/latex] (diverges for [latex]p \leq 1[/latex], converges for [latex]p > 1[/latex]), [latex]e^{-x}[/latex] (converges)
- Reciprocal rule: If [latex]0 < f(x) \leq g(x)[/latex], then [latex]\frac{1}{f(x)} \geq \frac{1}{g(x)}[/latex]
- Focus on the tail: For large [latex]x[/latex], identify the dominant behavior of your function
For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.
You can view the transcript for this segmented clip of “3.7 Improper Integrals” here (opens in new window).