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Earlier today I asked about this question: Improper integral (is it convergent?)

where the integral fortunately seems to be convergent. So we have that given $\alpha\in (-1/2,0)$ there is a $\gamma \in (1,2)$ such that $$\int_0^1 \int_0^{u} \frac{((1-v)^{\alpha}-(1-u)^{\alpha})^2}{(u-v)^{\gamma}}dvdu < \infty.$$

My question now is, if we take a middle value $a\in (0,1)$ is then the following integral also finite? $$\int_0^1 \int_0^{u} \frac{((a-v)^{\alpha}-(a-u)^{\alpha})^2}{(u-v)^{\gamma}}dvdu < \infty.$$

Of course a naiv start would be to split up the integral as follows:

\begin{align*} \int_0^1 \int_0^{u} \frac{((a-v)^{\alpha}-(a-u)^{\alpha})^2}{(u-v)^{\gamma}}dvdu =& \int_0^a \int_0^{u} \frac{((a-v)^{\alpha}-(a-u)^{\alpha})^2}{(u-v)^{\gamma}}dvdu\\ &+ \int_a^1 \int_0^{u} \frac{((a-v)^{\alpha}-(a-u)^{\alpha})^2}{(u-v)^{\gamma}}dvdu\\ &=: (A) + (B). \end{align*}

Here, $(A)$ is essentially the same as the one on top and therefore convergent. So the question is equivalent to proving that $(B)$ is either convergent or divergent.

What are your feelings? Should the integral above still be convergent for any values $a\in (0,1)$ and not only when $a=1$? Any impressions? or ideas on how to prove it?

Thanks a lot guys!

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  • $\begingroup$ You use $u$ as both a limit of integration and a variable of integration. Your integral cannot be evaluated as written. $\endgroup$ – user76844 Jan 19 '15 at 21:22
  • $\begingroup$ I don't see the problem there. First, one integrates w.r.t. $v$ and evaluate $u$ and then w.r.t. $u$ and evaluate from 0 to 1. It's just an integral w.r.t. the two dimensional simplex. $\endgroup$ – Martingalo Jan 20 '15 at 17:13
  • $\begingroup$ Ah, I see...you have dvdu, not dudv...my mistake. It didn't impede my analysis below. $\endgroup$ – user76844 Jan 20 '15 at 18:30
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    $\begingroup$ I see an issue here: how $(a-u)^{\alpha}$ is defined when $u>a$? $\endgroup$ – Jack D'Aurizio Jan 21 '15 at 14:26
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    $\begingroup$ @JackD'Aurizio exactly, and in the original question with $a=1$...nothing is stopping $u$ from going beyond $a$, resulting in undefined integrand. $\endgroup$ – user76844 Jan 21 '15 at 14:32
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Assuming the first integrate $(a=1)$ is convergent, the second integrate is not such different from the first type. By a simple variable conversion: $$a-u=1-x$$ $$a-v=1-y$$

The second integral will convert to:

$$ \int_a^{1-a} \int_{1-a}^x \frac{((1-x)^\alpha-(1-y)^\alpha)^2}{(y-x)^\gamma} dy dx$$

which is as same as the first one except for boundaries. If critical points would not make problem in the first integral, they must not make problem in the second one too.

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  • $\begingroup$ Aha! I see. Good observation. Yes, this is more or less the feeling I had, that essentially it is the same integral With different boundaries but of the same "type" of singularity. Then intuition was good here. Thanks a lot for your help! $\endgroup$ – Martingalo Jan 23 '15 at 15:25

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