- Topics At a Glance
- Indefinite Integrals Introduction
- Integration by Substitution: Indefinite Integrals
- Legrange (Prime) Notation
- Leibniz (Fraction) Notation
- Integration by Substitution: Definite Integrals
- Integration by Parts: Indefinite Integrals
- Some Tricks
- Integration by Parts: Definite Integrals
- Integration by Partial Fractions
- Integrating Definite Integrals
- Choosing an Integration Method
- Integration by Substitution
- Integration by Parts
- Integration by Partial Fractions
- Thinking Backwards
- Improper Integrals
- Badly Behaved Limits
- Badly Behaved Functions
- Badly Behaved Everything
- Comparing Improper Integrals
- The p-Test
- Finite and Infinite Areas
- Comparison with Formulas
- In the Real World
- I Like Abstract Stuff; Why Should I Care?
- How to Solve a Math Problem
Introduction to :
Improper integrals with badly-behaved functions are deceptive.
They look like normal definite integrals
,
but somewhere in the interval from a to b, possibly at one of the endpoints, there will be a vertical asymptote where the function f tends towards ∞ or -∞.
In fact, the function may have more than one vertical asymptote in the interval of integration.
Because of all these possibilities for bad behavior, we recommend graphing functions and looking at their formulas before trying to integrate. Be on the lookout for bad behavior! You want to catch all the places where the function is badly-behaved.
To evaluate this type of improper integral, the first thing we have to do is figure out where the function is badly behaved. Then we can write the improper integral as the limit it really is, work out the integral inside the limit, and finally evaluate the limit.
Badly-Behaved at One Endpoint
We'll start with improper integrals where the function is badly behaved at one limit of integration, but nowhere else.
Every other improper integral with a badly-behaved function can be broken down into integrals like this.
Convergence and Divergence
There's weird stuff going on again. If we take p > 1, the integral
diverges. If we take p < 1, the integral converges. This is the opposite of what happened with the other type of improper integral.
To take an example, on the interval [0,1] the area between 
and the x-axis is infinite but the area between 
and the x-axis is finite. The graphs don't even look that different. It's perfectly normal if this makes your brain bend a little.
When talking about things "converging" or "diverging," make sure you say whether you're talking about a function or an integral. The function
diverges as x approaches 0. However, the integral
converges. This means we can have a function diverge but a closely associated improper integral diverge. It's also possible for both the function and an improper integral to diverge. The function 
diverges as x approaches 0, and so does its improper integral
Badly-Behaved in the Middle
If a function is badly-behaved in between its limits of integration, we split the improper integral into two pieces. We do this by making a new endpoint at the value of x where the function is badly-behaved. Suppose that
is an improper integral because f has a vertical asymptote at x = b:
Then we split the integral into two new improper integrals:
The function f is badly behaved at just one limit of integration for each of these improper integrals.
When we do this, it's important to remember that
are both improper integrals, and thus are both limits. Both of these limits must exist in order for
to exist. Said another way, if either
diverges, then
diverges also.
There's both good news and bad news here. The good news is that if we find one of the new improper integrals diverges, we're done with the problem. The bad news is that if both of the new improper integrals converge, we have to work out both of them! This is the same good news / bad news we got for the other type of integral.
Practice:
Determine whether
converges or diverges. If it converges, find its value. Graph the region whose area is represented by this integral. |
is badly behaved at the lower limit of integration, x = 0. This means when we rewrite the integral as a limit we need to change the lower limit of integration.
Determine whether
converges or diverges. If it converges, find its value. Graph the region whose weighted area is represented by this integral. |
is badly behaved at the upper limit of integration.
This means
when we rewrite the integral as a limit we need to change the upper limit of integration, and use a one-sided limit with b approaching 0 from the left. 
Determine whether
converges or diverges. |
Determine whether the integral converges or diverges. Indicate on a graph the region whose weighted area is given by the integral. If the integral converges, find its value.
approaches ∞ as b approaches 0. That means the area of this region is infinite.Determine whether the integral converges or diverges. Indicate on a graph the region whose weighted area is given by the integral. If the integral converges, find its value.
.
Determine whether the integral converges or diverges. Indicate on a graph the region whose weighted area is given by the integral. If the integral converges, find its value.
Determine whether the integral converges or diverges. Indicate on a graph the region whose weighted area is given by the integral. If the integral converges, find its value.
for p > 1
in another problem.
Determine whether the integral converges or diverges. Indicate on a graph the region whose weighted area is given by the integral. If the integral converges, find its value.
for p < 1
is positive, which is reassuring since the area between 
and the x-axis on (0,1] is all above the x-axis.Determine whether each statement is true or false. Explain your answers.
If 
does not exist, then
must diverge.
on the interval [-1,0). This graph is the graph of 
reflected over the y-axis.
Then 
does not exist and 
If 
does not exist, then
must converge.
on the interval [-1,0), then 
does not exist and 
Determine whether the integral converges or diverges, and find its value if it converges.
approaches ∞, which means e1/b approaches ∞ also. This integral diverges, which means the original integral diverges. We don't have to bother with the other integral.Determine whether the integral converges or diverges, and find its value if it converges.
