 So far we've seen indefinite integrals and definite integrals. Now let's take a look at improper integrals. How can we have improper integrals? This emerges from a very useful property of the definite integral. Suppose f of x is continuous over an interval between a and b. Then the definite integral from a to b of f of x dx is guaranteed to exist. You might contrast this with the other two features of calculus, the derivative and the limit. When we looked for these things we always had to qualify them with a provided that they exist. With the definite integral we can actually go a little bit further. It's guaranteed to exist as long as our function is continuous. This means that for most functions and for most intervals the definite integral will exist. So this definite integral, which we can find with a little bit of effort, has a definite value and the effort is worth making. This definite integral, on the other hand, we can't evaluate by finding the anti-derivative. No simple function has derivative e to the x cosine x squared. Nevertheless, because this is a continuous function over the interval between minus 5 and 5, the value of this integral exists. So we will be able to approximate it by looking at the Riemann sums. And similarly for this function, again, no function has derivative e to minus x squared over 3, but nevertheless, because it's continuous over the interval from 0 to 1, we know the value of the definite integral exists and so we can approximate it by using the Riemann sums. And unfortunately one of the bizarre things about being a mathematician is once a problem is solved, it's not very interesting. As soon as we know that a definite integral exists, we know it can be found. And we really don't care how it's found. And so our question moves on to something else. Can we have improper integrals? And there are two problematic cases. First, what if the interval isn't finite? And second, what if the function isn't continuous? And so this leads to two slightly different types of improper integrals. An integral is an improper integral if, first, the interval of integration extends to infinity or minus infinity, we sometimes call this a type 1 improper integral, or the integrand itself is discontinuous at some point in the interval, and this is a type 2 improper integral. It's possible to have both things happen. We may have an interval that extends to infinity where there is at least one discontinuity. And I suppose you could call this a modified type 1 or a modified type 2 or a type 3 integral or whatever. Well, mathematicians aren't good with coming up with names, we just call this an improper integral. So can we extend the notion of the definite integral to these cases? And so we'll define their value in the following way. For a type 1 improper integral where the limit of integration extends to infinity, we're going to define the integral from a to infinity of f of x to be the limit as b approaches infinity of the definite integral from a to b. Informally, we might say that the value of this improper integral is the value of the definite integral as the upper limit goes to infinity. And we'll make a similar definition if our lower limit is minus infinity, it'll be the limit as a goes to minus infinity of our definite integral. And because these are now defined as limits, we must contend with the possibility that these limits don't exist. So if these limits exist, we'll define the integral as having the value of the limit and we say that the integral converges to the limiting value. Otherwise, we'll say that the integral diverges. For example, suppose we want to evaluate the limit from 1 to infinity of 1 over x squared. So since our upper limit is going to infinity, this is a type 1 improper integral, not that we really care. What we do care about is that if it has a value, that value is going to be defined as the limit as b goes to infinity of the definite integral from 1 to b of 1 over x squared dx. Well, the last thing we do here is we worry about the limit. So let's take care of everything else. We'll evaluate the definite integral. And now I get an expression as b goes to infinity, this goes to 1. And so this does have a limit and so the improper integral will have the value of the limit. In other words, the improper integral will be equal to 1. Our definition for improper integrals of the first type required that either the upper or the lower limit go to infinity. But what happens if they both go to infinity? To answer that question, we'll fall back on a useful property of the definite integral. We can split the integral. And so suppose the integral from minus infinity to c and from c to infinity both converge. Then the integral from minus infinity to infinity is going to be the sum of these two integrals. For example, let's see if we can find the integral from minus infinity to infinity of 1 over 1 plus x squared. So our definition says that if the integral from c to infinity and the integral from minus infinity to c both exists, then their sum will be the integral from minus infinity to positive infinity. And so, wait a minute, what's c? This seems to be a major flaw in our definition. It doesn't tell us what c is. Well, if you don't play, you can't win. We'll use a time-honored academic tradition. Procrastination. We won't worry about c until we have to. So let's just find the integral from c to infinity and we do have a way of expressing that. We'll evaluate the definite integral and then we'll take the limit, which will exist regardless of the value of c. Similarly, we can find the integral from minus infinity up to c. First, we'll express it as a limit. Evaluate the integral. Then take the limit. And again, this limit will exist regardless of the value of c. And since both integrals converge, then our integral from minus infinity to infinity is going to be the sum of these two. And it turns out we don't actually need to know the value of c to find this integral.