Improper integral (is it convergent?)

Question

I would like to either prove or disapprove the following: Let $\alpha\in (-1/2,0)$ be given. Then we can find $\gamma \in (1,2)$ such that $$\int_0^1 \int_0^{u} \frac{((1-v)^{\alpha}-(1-u)^{\alpha})^2}{(u-v)^{\gamma}}dvdu < \infty.$$

I have tried a couple of things with no success.. Any ideas are welcome.

Thank you guys!

Answer 1

Let me substitute $x=1-u$ and $y=1-v$. This is not important but like $x^\alpha$ better than $(1-u)^\alpha$.

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In the first part we solve the original problem.

Lemma: If $0<x<y\le1$ and $-1\le \alpha\le0$ then $$ \left|\frac{y^\alpha-x^\alpha}{y-x}\right| < 2x^\alpha y^{-1}. $$

Proof: If $y\ge 2x$ then $$ \left|\frac{y^\alpha-x^\alpha}{y-x}\right| \le \frac{x^\alpha}{\frac12y} = 2 x^\alpha y^{-1}. $$

If $x<y<2x$ then apply Lagrange's midvalue theorem to the function $x^\alpha$. By the midvalue theorem, there is some $\xi\in(x,y)$ such that $\dfrac{y^\alpha-x^\alpha}{y-x} = \alpha\xi^{\alpha-1}$, so $$ \left|\frac{y^\alpha-x^\alpha}{y-x}\right| = |\alpha|\xi^{\alpha-1} < x^{\alpha-1} < 2 x^\alpha y^{-1}. $$ The Lemma has been proved.

Now for $1<\gamma\le2$ we get $$ \int_{u=0}^1 \int_{v=0}^u \frac{\big((1-v)^\alpha-(1-u)^\alpha\big)^2}{(u-v)^\gamma} dv du = \int_{x=0}^1 \int_{y=x}^1 \frac{\big(y^\alpha-x^\alpha\big)^2}{(y-x)^\gamma} dy dx = \\ = \int_{x=0}^1 \int_{y=x}^1 \left|\frac{y^\alpha-x^\alpha}{y-x}\right|^2 (y-x)^{2-\gamma} dy dx \le \int_{x=0}^1 \int_{y=x}^1 (2x^\alpha y^{-1})^2 y^{2-\gamma} dy dx \le \\ \le 4 \int_{x=0}^1 x^{2\alpha} \int_{y=x}^\infty y^{-\gamma} dy dx = 4 \int_{x=0}^1 x^{2\alpha}\cdot \frac{x^{1-\gamma}}{\gamma-1} dx. = \frac4{\gamma-1} \int_{x=0}^1 x^{1+2\alpha-\gamma} dx. $$

If the last exponent, ${1+2\alpha-\gamma}$ is higher than $-1$ then the integral will be finite. We can be achieve that if $1<\gamma<2(1+\alpha)$. Due to $\alpha>-\frac12$, this interval is not empty, so there exists a suitable $\gamma$.

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Now we prove that the integral is $\infty$ if $\alpha=-\frac12$ and $\gamma\ge1$.

If $\alpha=-\frac12$ the computation can be in a simpler way. For $0<x<y<1$ and and $\gamma\ge1$ we have $$ \frac{\big(y^{\alpha}-x^{\alpha}\big)^2}{(y-x)^\gamma} \ge \frac{\big(y^{-1/2}-x^{-1/2}\big)^2}{y-x} = \frac{y-x}{xy\big(\sqrt{y}+\sqrt{x}\big)^2} > \frac{y-x}{4xy^2}. $$ This is not very significant if $y-x$ is small. Narrow the integration domain to $0<x<\frac13$, $2x<y<1$; in that set we have $y-x>\frac12y$, so $$ \int_{x=0}^1 \int_{y=x}^1 \frac{\big(y^\alpha-x^\alpha\big)^2}{(y-x)^\gamma} dy dx \ge \int_{x=0}^{1/3} \int_{y=2x}^1 \frac{y-x}{4xy^2} dy dx \ge \int_{x=0}^{1/3} \int_{y=2x}^1 \frac1{8xy} dy dx =\\ = \frac18 \int_{x=0}^{1/3} \frac{\ln\frac1{2x}}{x} dx \ge \frac{\ln\frac32}8 \int_{x=0}^{1/3} \frac{dx}{x} =\infty. $$

Answer 2

I found the following way to (maybe) prove it. It would be really helpful if anyone could confirm its validity.

Make the following change of variables: $x= 1-v$ and $y= \frac{u-v}{x}$, the Jacobian is $x$. So we obtain

$$\int_0^1 \int_0^u \frac{((1-v)^{\alpha} - (1-u)^{\alpha})^2}{(u-v)^{\gamma}}dvdu=\int_0^1 x^{2\alpha-\gamma+1} \int_0^1 \frac{(1-(1-y)^{\alpha})^2}{y^{\gamma}}dydx.$$

Now, using comparison with $\frac{1}{y^{\gamma-2}}$ we have that $$C:=\int_0^1 \frac{(1-(1-y)^{\alpha})^2}{y^{\gamma}}dy<\infty \mbox{ iff } \int_0^1 \frac{1}{y^{\gamma-2}}dy <\infty,$$ i.e. $\gamma <3$. Good, so the inner integral in $y$ is a finite constant (call it C). Finally we have $$\int_0^1 \int_0^u \frac{((1-v)^{\alpha} - (1-u)^{\alpha})^2}{(u-v)^{\gamma}}dvdu=C\int_0^1 v^{2\alpha-\gamma+1}dv < \infty$$ if and only if $\gamma < 2\alpha + 2$, since $\alpha \in (-1/2,0)$ then $\gamma\in (1,2)$.