 A discussion of improper integrals can't be complete without addressing one important issue. Sometimes it's not possible to evaluate the integral. However, because the integral is a sum, we might be able to determine convergence or divergence without computing the actual value. And we might be able to proceed this way. Suppose 0 is less than or equal to f of x, is less than or equal to g of x, for x greater than or equal to sum value k, then the integral from k to infinity of f of x should be less than or equal to the integral from k to infinity of g of x. Since the integral from a to infinity of f of x is the integral from a to k of f of x, plus the integral from k to infinity of f of x dx, and the integral from a to k is finite, Then the convergence or divergence of the integral from a to infinity depends only on the convergence or divergence of the integral from k to infinity. And a similar result holds for the integral from a to infinity of g of x. What this suggests is the following theorem. Suppose 0 is less than or equal to gf of x is less than or equal to g of x for x greater than or equal to some value k. Then if the integral from a to infinity of f of x diverges, so does the integral from a to infinity of g of x. Likewise if the integral from a to infinity of g of x converges, so does the integral from a to infinity of f of x. We could look at it another way. If you're larger than a divergent function, then the integral diverges. If you're smaller than a convergent function, the integral converges. For example, suppose we want to determine if the integral from 0 to infinity of e to power minus x squared dx diverges or converges. Now there is no anti-derivative of e to the minus x squared among these simple functions, so we can't evaluate this integral directly. Instead we need to be able to compare it to some function whose convergence or divergence over the same interval is known. This is unfortunately the hardest part of the problem, and there is no algorithm, no formula, no process that will allow you to find what that function is. However in order to gain some insight we might proceed as follows. e to minus x squared is really 1 over e to the x squared. So one possible comparison function is 1 over something. Now we've determined the integral from 1 to infinity of 1 over x squared converges, so maybe we'll try and compare this to 1 over x squared. Since we don't know whether this is less than or greater than, we'll leave that empty until the end of our analysis. Since both e to power x squared and x squared are positive numbers, if I cross multiply I won't change the direction of the inequality. So now the problem is comparing x squared to e to the x squared. One way to do this is to look at the ratio between the two expressions. Since we really hope that x squared is the smaller expression, let's look at the ratio of x squared over e to power x squared. So we'll take the limit as x goes to infinity of this ratio, and this is good for a L'Hopital's visit. We find that the limit as x goes to infinity is 0. Consequently, x squared less than e to power x squared is eventually true, and so their reciprocals will have the same inequality, and that gives us e to minus x squared must be eventually less than 1 over x squared. So we know that the integral from k to infinity of e to minus x squared dx is less than or equal to the integral from k to infinity of 1 over x squared for some value k. This work is only useful if this larger integral converges, so let's check. The integral from k to infinity of 1 over x squared is an improper integral, so it's going to be the limit as b goes to infinity of k to b of 1 over x squared. We'll find the integral and take the limit, which is finite, so our integral converges. So our smaller integral converges, as another example, let's consider the integral from 1 to infinity of secant squared of x over x. So again, we need a suitable comparison function, and here the problem is the secant function. Since secant is 1 over cosine, we know that secant squared is 1 over cosine squared, and so secant squared is always going to be greater than or equal to 1. So secant squared over x is always going to be greater than or equal to 1 over x. So the integral from k to infinity of secant squared over x is going to be greater than or equal to the integral from k to infinity of 1 over x. But the integral from k to infinity of 1 over x diverges, and so does our integral.