Let's consider the function f(x) = 1/(x2 - 1). If we want to find either of the integrals
then, we have an improper integral: in the first case the y-range is unbounded, and in the second the x-range is unbounded.
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
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
Then, we know this is the same as the original integral in the limit as b goes to infinity. Thus
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:
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: