Integrability of $\frac{1}{(x^2+y^2+z^2)^a}$ on $E=\{(x,y,z)\in \mathbb{R}^3: z>1, \ z^2(x^2+y^2)<1 \}$ Let $E=\{(x,y,z)\in \mathbb{R}^3: z>1, \ z^2(x^2+y^2)<1 \}$ and $$f_{a}(x)=\frac{1}{(x^2+y^2+z^2)^a}$$
I need to find all $a\in \mathbb{R}$ such that $f_a\in L^1(E).$ 

I already know a solution to this problem, and I want to understand why mine is wrong.

My attempt
1) [Measurability check] 
First of all $E$ is open in $\mathbb{R}^3$ being a union over all $z>1$ of open disks of radius $\frac{1}{z}$: $E=\bigcup_{z>1} E_z=\bigcup_{z>1}\{(x,y): x^2+y^2<\frac{1}{z^2}\}.$ Next, since $z>1,$ $f_a$ looks continous on every point of $E,$ thus is measurable.
2) [Positivity] 
The function $f_a$ is strictly positive for every $a\in \mathbb{R},$ thus to check integrability we can reduce ourselves to compute 
$$\int_E f_a \ d(x,y,z)$$
and by positivity of the function we can apply Tonelli's theorem, rewriting the integral as 

$$\int_1^{+\infty}\int_{x^2+y^2<\frac{1}{z^2}} f_a \ \ d(x,y) \ d(z)$$

3)[Polar coordinates]
We now compute the inner integral using polar coordinates to obtain
$$\int_{x^2+y^2<\frac{1}{z^2}} f_a \ \ d(x,y)= \int_0^{2\pi} \int_0^{1/z} \frac{r}{(r^2+z^2)^a} \ dr d\theta= 2\pi \int_0^{1/z} \frac{r}{(r^2+z^2)^a} \ dr =...$$
changing variable $r\mapsto \ t= \phi(r)=r^2+z^2 $ 
$$...=\pi \int_{z^2}^{1/z^2+z^2} \frac{1}{t^a} \ dt= F(z,a) $$
4) [Computation by cases]
4.i)
For $a=1,$ we have $F(1,z)= \log(1/z+z^2)-\log(z^2)$ and $$\int_1^{+\infty}\log(1/z^2+z^2)-\log(z^2) \ dz= \int_1^{+\infty} \log(z^2(1/z^4+1))-\log(z^2) \ dz= \int_1^{+\infty}\log(1+1/z^4) \ dz < +\infty$$
and so $f_{a=1}\in L^1(E).$
4.ii)
Now we let $a \neq 0,$ and we have
$$\int_{z^2}^{1/z^2+z^2}t^{-a} \ dt=\frac{(1/z^2+z^2)^{1-a}}{1-a}-\frac{(z^2)^{1-a}}{1-a}$$
and we have to evaluate 
$$\int_1^{+\infty}\frac{(1/z^2+z^2)^{1-a}}{1-a}-\frac{(z^2)^{1-a}}{1-a} \ dz $$
But 
$$\int_1^{+\infty}\frac{(1/z^2+z^2)^{1-a}}{1-a}-\frac{(z^2)^{1-a}}{1-a}
 \ dz \geq \int_1^{+\infty}\frac{(1/z^2)^{1-a}}{1-a}+
 \frac{(z^2)^{1-a}}{1-a} -\frac{(z^2)^{1-a}}{1-a} \ dz=
 \int_1^{+\infty}\frac{(1/z^2)^{1-a}}{1-a} \ dz > +\infty$$
for $0<2-2a\leq 1 \iff 0<1-a \leq \frac{1}{2} \iff a \geq \frac{1}{2}$ so that $f_a \notin L^1(E)$ for $a \geq \frac{1}{2};$
on the other hand, we have 
$$\int_1^{+\infty}\frac{(1/z^2+z^2)^{1-a}}{1-a}-\frac{(z^2)^{1-a}}{1-a}
 \ dz \leq \int_1^{+\infty}\frac{(1/z^2+z^2)^{1-a}}{1-a} \ dz$$
and since in a ngbh of $+\infty$ $$1/z^2+z^2 \sim z^2 $$ the latter is finite if and only if $$\int_1^{+\infty}\frac{(1/z^2)^{1-a}}{1-a}<+\infty$$
if and only if 
$$2-2a >1 \iff a<\frac{1}{2}$$

Can someone proof-read and tell me if there are any mistakes?
The other solution I have, using spherical coordinates, has $f_a $ integrable $\iff$ $a=3/2$ or $a>-\frac{1}{2}, $ while my solution gives $f_a$ integrable iff $a=1$ or $a<\frac{1}{2}$. I want to understand what went wrong with my solution.

 A: Estimating $\int_{z^2}^{z^2+z^{-2}} \frac{dt}{t^a}$ correctly is how you would get the correct range for convergence. For $a\ge 0$, the integrand is decreasing, so we can estimate 
$$ \int_{z^2}^{z^2+z^{-2}} \frac{dt}{t^a} \le \frac1{(z^{2})^a}(z^2 + z^{-2} - z^2) = \frac1{z^{2a+2}}\in L^1_{\text{loc}}(1,\infty),$$
and this is integrable at infinity if $2a+ 2>1$, which gives $a>-1/2$, so $a\ge 0$. (Note that this estimate works better than just trying to use that $z^2+z^{-2}\sim z^2$ as a black box.)
For $a<0$, the integrand is increasing, we instead go
$$\int_{z^2}^{z^2+z^{-2}} \frac{dt}{t^a} \le \frac1{z^2+z^{-2}}(z^2 + z^{-2} - z^2) = \frac1{1+z^{2\alpha + 2}}$$
and the same analysis applies, giving the remaining cases $-1/2 < a < 0$.
As for the divergence part, I don't believe its true that
$$ (a+b)^s \ge a^s + b^s$$
where you used this for $s\in[0,1/2]$. Actually in this regime, you can use the concavity of $x\mapsto x^s$ to deduce that the opposite inequality holds. Also, the original 3D integral is integrable on the larger set $\{z>1\}\supset E$ when $2a>3$, so that's already a sign that you can't prove it diverges for the range you claimed.
For a correct proof, actually the above proof gives lower bounds as well, by using the other integration bound. So $\int_{z^2}^{z^2+z^{-2}} \frac{dt}{t^a} \sim \frac1{z^{2a+2}}$ at infinity, and this diverges if $a\le 1/2$.
