Test the following improper integral for convergence:

The integral diverges.

*Proof.* First, we show that for all . To do this let

This derivative is 0 at and is less than 0 for and greater than 0 for . Hence, has a minimum at . But, (since implies ). So, has a minimum at and is positive there; thus, it is positive for all , or

So, since we know

But then, consider the limit

Therefore, by the limit comparison test (Theorem 10.25), the convergence of would imply the convergence of (for any ), but we know by Example 5 that this integral diverges. Hence, we must also have the divergence of

Split in two, and then compare the negative lower integral with 1/x^(3/2), which converges. The upper one diverges, as can be compared with 1/x. A substitution t = 1/x could help.

RoRi, dude, log x<0 for 0<x<1 that said it isn't possible to 1/[sqrt(x) log(x)] to be greater than 1/x

You are wrong about theorem 7.11. The hypothesis of theorem 7.11 is with x is going to infinity, not to zero.

Rafael, the book says that the theorem is easily exploted to other situations, such as the x approaching 0.