Integral limit

**durrrrrrrr** · July 12th 2011, 04:22 AM

Does the integral in the attached image converge?

Can you give me a hint how to start solving this question. Thanks

**Opalg** · July 12th 2011, 11:36 AM

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 $(\log|\log t|)^7$ looks like when t is close to 1.

For t close to 1, $|\log t|$ is approximately 1–t, so we should look at the function $\bigl(\log(1-t)\bigr)^7.$ In fact, that function is negative for 0<t<1, so it would be more convenient to look at its negative, namely $\bigl|\log(1-t)\bigr|^7.$

As x decreases to 0, |log x| increases to infinity, but more slowly than any negative power of x: $|\log x|\leqslant x^{-\alpha}$ for any $\alpha>0.$ Put x = 1–t to get $\bigl|\log(1-t)\bigr|^7 \leqslant (1-t)^{-7\alpha}.$ Choosing $\alpha = 1/14$ , you see that $\bigl|\log(1-t)\bigr|^7 \leqslant (1-t)^{-1/2}.$ The integral $\int_{1/2}^1(1-t)^{-1/2}dt$ converges, hence (by the comparison test for improper integrals) so should the integral that you started with.