Are these functions in $L^{1}(\mathbb{R}^{d})$?

Question

Are the functions $f:\mathbb{R}^{d}\to\mathbb{R}$ with

$$f(x):=\frac{1}{(1+\vert x\vert)^{n}}$$

in $L^{1}(\mathbb{R}^{d})$ if $n>d$?

These functions are obvious measurable, so we only need to show

$$\Vert f\Vert_{L^{1}}=\int_{\mathbb{R}^{d}}\vert f(x)\vert\,\mathrm{d}x<\infty$$

My first try was to bound $\frac{1}{(1+\vert x\vert)^{n}}\leq \frac{1}{\vert x\vert^{n}}$, but this obviously not works, because this bound diverges for $\vert x\vert=0$.

Just by looking at the functions it is quite clear that the integral is finite, because it is everywhere finite and we have $f(x)\to 0$ for $\vert x\vert\to\infty$, but I am not sure how to prove it rigorously...

Answer 1

Note that, polar coordinates in $\mathbb R^d$: $$ \int_{\mathbb R^d}f(x)\,dx=\int_0^\infty \left(\int_{S^{d-1}}f(rw)dw\right)\,r^{d-1}\,dr $$ where $S^{d-1}$ is the unit sphere in $\mathbb R^d$.

So $$ \int_{\mathbb R^d}\frac{dx}{(1+|x|^n)}=\int_0^\infty \left(\int_{S^{d-1}}\frac{dw}{(1+r^n)}\right)\,r^{d-1}\,dr =\omega_{d-1}\int_0^\infty \frac{r^{d-1}\,dr}{(1+r^n)}<\infty, $$ if $d<n$, where $\omega_{d-1}$ is the $(d-1)-$dimensional area of $S^{d-1}$.