improper integrals

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The point of this page is to look at improper integrals, and how the limits to find them make sense.

Let's consider the function f(x) = 1/(x2 - 1). If we want to find either of the integrals

int_2^infty f(x) dx or int_1^5 f(x) dx

then, we have an improper integral: in the first case the y-range is unbounded, and in the second the x-range is unbounded.

graph of a f(x) for x in [1,infty)
figure 1: The graph of f(x)
graph of a f(x) for x in [1,b]
figure 2: The graph of f(x), with area between 2 and b.

Let's consider the second of these: we want to find the area shown in the figure to the right. However, because one of the dimensions of the graph is unbounded, we can't just apply our usual fundamental theorem of calculus methods: it doesn't make to say

int_2^infty f(x) dx = F(infty) - F(2)

for some antiderivative F(x) of f(x), because infinity isn't a point and we can't evaluate a function there.

So instead we consider a well defined finite domain by stopping the integral at some point b, as shown in figure 2 to the right. This integral we can do (using partial fractions or a table): we have

int_2^b f(x) dx = (1/2)(ln|x-1| - ln|x+1|)|_2^b
=(1/2)ln(|(x-1)/(x+1)|)|_2^b = (1/2)(ln|(b-1)/(b+1)| - ln(1/3))

Then, we know this is the same as the original integral in the limit as b goes to infinity. Thus

int_2^infty f(x) dx = lim(b to infty) (1/2)(ln|(b-1)/(b+1)| - ln(1/3)) = -ln(1/3) = ln(3).

Note that it is not true that the integral with the upper limit of infinity is the same as the integral with the upper limit of b: the two are equal only as b goes to infinity:

int_2^infty f(x) dx = lim(b to infty) int_2^b f(x) dx

So, if you're writing out the solution, either stop and work the integral with b in the limit (making a note in your solution that you're considering a different integral), and then say later that the integral you want is the limit of your result---or write limits the whole way through:

Solution 1:
To find
int_2^infty f(x) dx
we consider
int_2^b (1/(x^2-1)) dx = (1/2)ln|(x-1)/(x+1)||_2^b = (1/2)(ln|(b-1)/(b+1)| - ln(1/3))
Then
int_2^infty f(x) dx = lim(b to infty) int_2^b f(x) dx
so we want to take the limit (as b goes to infinity) of the evaluated form of the integral on the right-hand side. We know that as b goes to infinity, the argument of the ln|(b-1)/(b+1)| term approaches 1, so the term approaches ln|1| =0, and we therefore find
int_2^infty f(x) dx = lim(b to infty) (1/2)(ln|(b-1)/(b+1)| - ln(1/3)) = -ln(1/3) = ln(3).
Solution 2:
Find:
int_2^infty f(x) dx
We evalute the integral with limits:
int_2^infty (1/(x^2-1)) dx = lim(b to infty) int_2^b (1/(x^2-1)) dx = lim(b to infty) (1/2)(ln|x-1| - ln|x+1|)|_2^b
=lim(b to infty) (1/2)(ln(|(b-1)/(b+1)|) - ln(1/3)) = -(1/2)ln(1/3) = ln(3)
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Last modified: Tue Jan 29 15:55:29 EST 2008
Comments to:glarose(at)umich(dot)edu
©2008 Gavin LaRose, UM Math Dept.