I have been asked by a lot of students how to tell what to do when faced with a comparison theory question and they can’t pick the comparison function.

The comparison theorem is used to tell whether a given integral is divergent or convergent if you can’t actually solve the integral itself.

Let’s say we have an integral which is an improper integral of the first kind. eg. the limit of integration goes up to \infty. If the function has no divergences at any point, then we know that the only behaviour we have to be concerned about is how the function goes as x\rightarrow \infty. In general we are concerned as to whether the function falls off fast enough in this limit. Clearly if the function goes to a constant, then it’s not going to converge, and if it falls off any slower than \frac{1}{x} then it’s also not going to converge. Let’s look at some simple examples.

Asked about whether:

\int_{2}^\infty \frac{x}{x^3+1}dx

converges we can see that for any x>-1 this is bounded. So we only have to ask how this goes as x\rightarrow \infty. The denominator in the function is x^3+1. As x gets larger and larger, we can see that this is well approximated by x^3. For x=1000 this is 1000000001 which is pretty close to x^3. Basically as we take the limit, we can ignore the extra 1. So we can see that in this limit \frac{x}{x^3+1}\sim \frac{x}{x^3}. However, we can also see that the left is actually less than the right. ie. \frac{x}{x^3+1}<\frac{x}{x^3}=\frac{1}{x^2} for our range of integration. Thus, using the comparison theorem, if the right hand expression converges, then so does the left hand expression.

Indeed the integral does converge for \frac{1}{x^2} so we’re good – our integral does indeed converge.

Here we used the limiting properties of our integrand to find an appropriate comparison function, and then, by working out which was greater and smaller, applied the comparison theorem to our problem.

Let’s look at a slightly more contrived example which turns out to be the same idea entirely:

\int_{2}^\infty \frac{x-\sin x-\sqrt{x}-32}{x^3+1+x^2}dx

We can make exactly the same arguments here, both looking at the numerator and the denominator (and making sure that the only improper nature of this integral is in the way the upper limit is \infty – ie. there are no divergences anywhere else in the integrand).

As x\rightarrow \infty we see that the numerator tends to x because all the other terms are dominated by the linear in x term, and on the bottom the dominant behaviour is x^3. We are left then with exactly the same comparison function as before: \frac{1}{x^2}. Again we can show that this is greater than our complicated function (at least for large values of x). Thus, because this function converges in our range of integration, then so does the integral in question.

The rule about finding a function to compare as a test function is to understand the behaviour of the numerator and denominator in the limit of interest – what is the tendency of the function as x\rightarrow \infty (for example).

If you have specific examples which you’ve struggled with, put them in the comments and I’ll add some explanations.

How clear is this post?