How to show function raised to $-\alpha$ is in $L^p(\mathbb{R}^d)$ when $p\alpha < d$?

Question

Let $x \in \mathbb{R}^d$ and $|x|=(x_1^2 + \ldots + x_d^2)^{\frac{1}{2}}$.

Let $f_u(x)=|x|^{-\alpha}$ if $|x| <1$, $f_u(x)=0$ for $|x|\geq1$.

Let $f_k(x)=|x|^{-\alpha}$ if $|x| \geq 1$, $f_k(x)=0$ for $|x| <1$.

Show that:

(1) $f_u \in L^p(\mathbb{R}^d)$ iff $p \alpha < d$ and

(2) $f_k \in L^p(\mathbb{R}^d)$ iff $d < p \alpha$.

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My attempt:

I think what I need to do in order to show that $f \in L^p(\mathbb{R}^d)$ is to show that $\int_{\mathbb{R}^d} |f(x)| < \infty.$

So $f_u(x) = (x_1^p + \ldots + x_d^p)^{1/{p\alpha}}$.

Since $p\alpha<d$, $p/d < 1/\alpha$.

Then $(x_1^p + \ldots + x_d^p)^{1/{p\alpha}} > (x_1^p + \ldots + x_d^p)^{1/{d}}$

... not sure what now.

Answer 1

Remember that we can integrate radial functions as $$\int_{\Bbb R^n}f(|x|)\mathrm d^n x=A_{n-1}\int_0^\infty f(r)r^{n-1}\mathrm dr$$

With $$A_{n}=\frac{2\pi^{\frac{n+1}{2}}}{\Gamma\left(\frac{n+1}{2}\right)}$$

Hence $$\int_{\Bbb R^n}f_u(x)\mathrm d^n x=A_{n-1}\int_0^1 r^{-\alpha}r^{n-1}\mathrm dr \\ \int_{\Bbb R^n}f_k(x)\mathrm d^nx=A_{n-1}\int_1^\infty r^{-\alpha}r^{n-1}\mathrm dr$$ Can you proceed?