As it is said in the title, I need to check the integrability of $$f(x,y)=(1-x)^{a}$$ for $a\in \mathbb{R}$ on the open disk $D= \{0<x^2+y^2<1\}.$
My attempt
- Let $a \geq 0,$ then since $D \subset [-1,1]\times[-1,1]$ is bounded and $f$ is bounded on $D$ we have $f_{a\geq 0}\in L^1(D).$
- Let $a<0.$ Then since $x \neq 0$ on $D,$ $f_a$ is continous on $D$ and thus measurable. But it is also unbounded for $(x,y)\to (1,0).$ I use polar coordinates and evaluate $$\int_0^{2\pi}d\theta\int_0^1(1-\rho \cos\theta)^a \rho \ d\rho$$ Changing variable $\rho \mapsto \rho\cos\theta=t,$ $\ t'=\cos\theta$, we have $$\int_0^{2\pi}\frac{1}{cos^2\theta}d\theta \int_0^{\cos\theta}(1-t)^at \ dt$$ Then I can evaluate $$\int_0^{cos\theta}t \ dt-\int_0^{cos\theta}t^{a+1} \ dt=\frac{1}{2}\cos^2\theta-\int_0^{cos\theta}t^{a+1} \ dt$$ and $$\int_0^{cos\theta}t^{a+1} \ dt < +\infty \iff -1<a+1 \iff -2<a$$ in which case it remains to check that $$\int_0^{2\pi}\frac{1}{2}-\frac{(\cos\theta)}{2}^{a+2}=\pi-\int_0^{2\pi}\frac{(\cos\theta)}{a+2}^{a+2}\ d\theta<+\infty$$
So, to conclude, $f_a \in L^1(D) \iff -2<a $.
Do you agree? Is my reasoning sound?
I'm not sure why you leave out $(0,0)$ in the integral. Your function $f$ is continuous on $\{x^2+y^2<1\}$ no matter what $a$ is.
The integral can be written
$$\int_{-1}^1 \int_{-\sqrt {1-x^2}}^{\sqrt {1-x^2}}\,dy\,(1-x)^a\,dx = \int_{-1}^1 2\sqrt {1-x^2}(1-x)^a\,dx$$ $$=\int_{-1}^1 2\sqrt {1+x}\sqrt {1-x}(1-x)^a\,dx =\int_{-1}^1 2\sqrt {1+x}(1-x)^{a+1/2}\,dx.$$
Now the integral from $-1$ to $0 $ converges no matter what $a$ is. Let's forget about it. And on $[0,1],$ $2\sqrt {1+x}$ lies between $2$ and $2\sqrt 2,$ so it has no impact on convergence/divergence. We're left thinking about
$$\int_{0}^1 (1-x)^{a+1/2}\,dx.$$
That converges iff $a+1/2>-1,$ or $a>-3/2.$