Here's the informal argument. I'll leave it to you to make it rigorous.

The singularity occurs at t=1, so we need to think about what the function looks like when t is close to 1.

For t close to 1, is approximately 1–t, so we should look at the function In fact, that function is negative for 0<t<1, so it would be more convenient to look at its negative, namely

As x decreases to 0, |log x| increases to infinity, but more slowly than any negative power of x: for any Put x = 1–t to get Choosing , you see that The integral converges, hence (by the comparison test for improper integrals) so should the integral that you started with.