# Improper integral (is it convergent?) (v 2.0)

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!

• You use $u$ as both a limit of integration and a variable of integration. Your integral cannot be evaluated as written. – user76844 Jan 19 '15 at 21:22
• 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. – Martingalo Jan 20 '15 at 17:13
• Ah, I see...you have dvdu, not dudv...my mistake. It didn't impede my analysis below. – user76844 Jan 20 '15 at 18:30
• I see an issue here: how $(a-u)^{\alpha}$ is defined when $u>a$? – Jack D'Aurizio Jan 21 '15 at 14:26
• @JackD'Aurizio exactly, and in the original question with $a=1$...nothing is stopping $u$ from going beyond $a$, resulting in undefined integrand. – user76844 Jan 21 '15 at 14:32

Assuming the first integrate $(a=1)$ is convergent, the second integrate is not such different from the first type. By a simple variable conversion: $$a-u=1-x$$ $$a-v=1-y$$
$$\int_a^{1-a} \int_{1-a}^x \frac{((1-x)^\alpha-(1-y)^\alpha)^2}{(y-x)^\gamma} dy dx$$