Does the integral in the attached image converge?

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

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- Jul 12th 2011, 04:22 AMdurrrrrrrrIntegral limit
Does the integral in the attached image converge?

Can you give me a hint how to start solving this question. Thanks - Jul 12th 2011, 11:36 AMOpalgRe: Integral limit
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 $\displaystyle (\log|\log t|)^7$ looks like when t is close to 1.

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

As x decreases to 0, |log x| increases to infinity, but more slowly than any negative power of x: $\displaystyle |\log x|\leqslant x^{-\alpha}$ for any $\displaystyle \alpha>0.$ Put x = 1–t to get $\displaystyle \bigl|\log(1-t)\bigr|^7 \leqslant (1-t)^{-7\alpha}.$ Choosing $\displaystyle \alpha = 1/14$, you see that $\displaystyle \bigl|\log(1-t)\bigr|^7 \leqslant (1-t)^{-1/2}.$ The integral $\displaystyle \int_{1/2}^1(1-t)^{-1/2}dt$ converges, hence (by the comparison test for improper integrals) so should the integral that you started with. - Jul 13th 2011, 12:43 AMdurrrrrrrrRe: Integral limit
- Jul 13th 2011, 01:27 AMFernandoRevillaRe: Integral limit
Verify $\displaystyle \lim_{t\to 1^-}\frac{|\log t|}{1-t}=\ldots=1$ and $\displaystyle \lim_{x\to 0^+}\frac{|\log x|}{x^{-\alpha}}=\ldots=0\;\;(\alpha>0)$

- Jul 13th 2011, 01:40 AMdurrrrrrrrRe: Integral limit
- Jul 13th 2011, 01:47 AMFernandoRevillaRe: Integral limit
Use the substitution $\displaystyle u=1-t$ and the well known property: $\displaystyle \int_{0}^1\frac{du}{u^p}$ converges iff $\displaystyle p<1$ .