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It should be easy but I'm not sure... For which $\alpha \in \mathbb{R}$ the following integral is convergent:

$$\int_0^1 \int_0^1 \frac{1}{|y-x|^\alpha}dxdy \ \ ?$$

I get for all $\alpha \neq 1,2$ but I think it is wrong.

Thanks for any suggestion!

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I would say $\alpha <1$ –  Emanuele Paolini Mar 1 '13 at 13:21

2 Answers 2

Make a change of variables: $t=x-y$, $s=x+y$ so you integral becomes something like (up to constants): $$ \iint \frac {1}{|t|^\alpha} \, dt\, ds $$ which has the same convergence as $$ \int_0^1 \frac 1 {t^\alpha} \, dt = C\left[t^{1-\alpha}\right]_0^1 < +\infty $$ if $\alpha<1$.

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Yes! Now I see that, one gets $\alpha < 1$ and $\alpha < 2$ when evaluating the limits.. so $\alpha < 1$ indeed.. The change of variables is smart as well. Thanks a lot! –  Dan Mar 1 '13 at 13:28

Make the change of variable $u=x$ and $v=y-x$.

This yields $$ \int_{u=0}^1\int_{v=-u}^{1-u}\frac{1}{|v|^\alpha}dvdu. $$

The inner integral converges if and only if $\alpha<1$, in which case it is equal to $$ \int_{-u}^0\frac{1}{(-v)^\alpha}dv+\int_0^{1-u}\frac{1}{v^\alpha}dv=\frac{u^{1-\alpha}}{1-\alpha}+\frac{(1-u)^{1-\alpha}}{1-\alpha}. $$

Then the remaining $u$ integral is improper at $0$ and $1$, where it converges if and only if $\alpha -1<1$, ie $\alpha<2$.

So your integral converges if and only if $$ \alpha<1. $$

Note you can even compute it.

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