 Since the integral test always gives us an answer, we can use it in both ways. If the integral converges or diverges, the series does the same thing, that's how we typically use the integral test, but if the series converges or diverges, the integral does the same thing. And just to dot all the p's and cross all the q's, our ability to use the integral test this way is an example of a proof by contradiction, also known as an indirect proof, and it works as follows. So I suppose we're in a situation where we can use the integral test, namely that f is continuous, positive, and decreasing over the infinite interval. Suppose you know that the infinite series converges. Now there's a corresponding integral, and maybe this integral could converge or diverge. But if the integral diverged, we could then use the integral test, and that would tell us that the series diverged as well. But that's a contradiction, and so we know the integral can't diverge, it must converge. And we get a similar result if we know that the series diverges. So for example, let's determine the convergence or divergence of this particularly important integral. And because it's so very important, we'll actually answer this question in three ways. Well, so let's see if we can actually find the value of this integral. So it's an improper integral, and so we'll let the upper limit go to infinity, and all we need to do is find an antiderivative for e to minus x squared, except we can't. No problem. Fortunately, we can take multivariable calculus, where we'll determine the value of the integral. And so our first solution is to take another course and find the value of the integral directly. Now the only problem with that is our solution in multivariable calculus only works for this one integral. For anything else, we have to do the equivalent of original research in mathematics. So maybe we should find another way. And so we might begin with the idea that if x is greater than 1, e to power minus x squared is less than e to minus x. And so this means that the integrals have the same relationship. And the good news is we can figure out the thing on the right-hand side pretty easily. And the right-hand side is bounded. The integral is continuous positive and increasing. And so if we take a course in real analysis, we can then prove that the limit exists. Now you'll need to take a couple more courses before you can take real analysis. So let's take a few more courses and then prove that our limit exists. The advantage here is this process works more generally for other functions besides e to minus x squared. What's that? You haven't taken multivariable calculus or real analysis yet? Because this is only second semester calculus? Well, that's okay. We can still solve the problem. Now, the earlier observation is still useful. And this time we note that for all n greater than or equal to 1, we have e to minus n squared less than e to power minus n. And while the actual names of the variables don't matter, this is set up to consider these as series. Now, the series with terms e to minus n, well, that's a geometric series with ratio 1 divided by e, which is strictly less than 1. So this series converges. And since e to minus n squared is smaller, our series of smaller terms also converges by the comparison test. But wait, that means that the integral test also has to tell us that the series converges. And that can only happen if the integral itself converges. And so our integral must converge because what? Oh, wait, the integral we want starts at zero. And we only know about the integral from one. Oh, wait a minute, the integral from zero to one is a finite interval, and our function is continuous over the interval. So the integral also exists. And if they include that value, we get the infinite integral also must converge by the integral test. So if you don't want to take other math courses, why wouldn't you though? But if you don't have time to take other math courses before having to answer this question, we can answer it using the integral test. So the series whose terms are e to minus n converges. So the series whose terms are e to minus n squared also converges. That's by our comparison test. So the integral over the infinite interval for one to infinity also converges by the integral test. And since the first portion is also finite, the integral over the infinite interval from zero to infinity also converges by the integral test. Now admittedly, that was a hard problem because you had the function and you could have found the actual value of the integral. So let's make the problem easier by eliminating the function. Now, of course, you have to know something about the function. So let's suppose it's positive, continuous and decreasing over our infinite interval. And maybe we do know that the infinite integral converges. Does this other improper integral converge or diverge? Well, to answer this question, we'll take multivariable calculus and real enough. No, wait, we don't want to do that. We'll use the integral test. Now we know that the integral converges, so the corresponding infinite series must also converge. Our comparison test tells us that a series that looks a lot like this second integrand will also converge, which tells us that the corresponding integral must also converge.