Page 46 - Calculus for the Clueless, Calc II

Example 3—

This integral does not go to infinity. Yet it still diverges since, as a goes to infinity, sin (a) takes on every value
between -1 and 1. The integral diverges because it doesn't go to one finite value.

If we have , we break it up into two pieces, , where, for convenience, c is often
0 but certainly does not have to be.

Example 4—

Whenever you have infinity at both ends, you should try to do the piece that diverges first. If you choose the
piece that diverges, you do not have to do the other piece, since the integral diverges (the whole integral
diverges). If the first piece converges, then you still must do the other piece. You sharp-eyed readers have
probably spotted the fact that the negative infinity piece diverges since, roughly speaking, goes to infinity.
Note is 0.

Let us take a look at the other infinity kind of improper integral.

Example 5—

The improper part, f(x) = (x - 1) -1/2 , is infinite at x= 1.

This integral converges to 4.

Note

For this kind of improper integral, if the exponent in the denominator is less than 1, the integral converges; if
the exponent is greater than or equal to 1, it diverges.

Example 6—

First, note that this is an improper integral since f(3) is undefined. Second, most of the time, if
one piece diverges, both diverge, so that it is not important which piece to choose first.