For which $a$ does this integral converge?

Question

This is an exercise given during my course in Lebesgue integrals. I thought about it for two weeks but I couldn't find a solution. Do you have any hint? The exercise asks to find for which values of $a\in\mathbb{R}$ the function $\frac{1}{n^2+x^{a}+2y^{a}+3z^{a}}$ is $L^1([0,+\infty)^3)$, for all $n\in\mathbb{N}\cup\{0\}$. Thank you very much!

Answer 1

Hint: You'll want to use Fubini. Note that

$$\int_0^\infty \frac{dt}{1+t^a}$$

converges iff $a>1.$ Note further that if $c>0,$ then

$$\int_0^\infty \frac{dt}{c+t^a} = c^{1/a-1}\int_0^\infty \frac{ds}{1+s^a}.$$

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Added later Let's deal with just $1/(x^a+y^a+y^a).$ Using what I previously mentioned, the $x$ integral converges iff $a>1.$ In that case we get $C(y^a+z^a)^{1/a-1}.$ If we go to polar coordinates for the $y,z$ integral, we are then looking at

$$\left(\int_0^{\pi/2}(\cos^a t+ \sin^a t)^{1/a-1}\,dt \right )\left ( \int_0^\infty r^{2-a} \,dr \right ).$$

Note the $t$-integeral is nice and convergent for any $a.$