We want to know when does the function $f:\Bbb R^n \to \Bbb R^n$ defined by $$f: x \mapsto \frac{1}{\lvert x \rvert^\beta}e^{-\frac{\lvert x \rvert^2 }{2}}$$ belong to $L^p(\Bbb R^n)$. I did an integration with polar coordinates and used Taylor expansion of exponential :
$$\int_{\Bbb R^n}\lvert f(x) \rvert ^p dx= \int_0^{+ \infty} \int_{\mathbb{S}^{n - 1} (r)} \frac{e^{-\frac{\lvert \sigma \rvert^2 p}{2}}}{ | {\sigma |^{\beta p}}} d \sigma d r \nonumber\\=\int_0^{+ \infty} \frac{e^{-\frac{r^2 }{2}}}{ {r^{\beta p}}} \left( \int_{\mathbb{S}^{n - 1} (r)} d \sigma \right) d r \nonumber = | \mathbb{S}^{n - 1} | \int_0^{+ \infty} \frac{r^{n - 1}}{r^{\beta p}(1 + r^2/2+r^4/96+\cdots)^{ p}} d r \nonumber$$ $$= | \mathbb{S}^{n - 1} | \int_0^{+ \infty} \frac{r^{n - 1 - \beta p}}{(1 + r^2/2+r^4/96+\cdots)^{ p}} d r. \nonumber$$
The integral from $0$ to $1$ is the one I should worry about right ? Because since there are infinitely many degrees in the denominator, the integral from $1$ to $\infty$ will always converge. The integral from $0$ to $1$ is well defined if $n-1-\beta p>1 \iff n-2>\beta p.$ However the answer I was expected to find is $n>\beta p$. Did I make any mistake ?
I don't see the need for Taylor when it was only used to obtain the imprecise statement about an "infinite degree polynomial". We have that $r^k\exp(-ar^2)\to 0$ for any $k$ and $a>0$ as $r\to\infty$. Near $0$ we have
$$r^{n-1-p\beta}\exp\left(-\frac{pr^2}{2}\right) \sim r^{n-1-p\beta}$$
The only condition needed to make this integral converge is
$$n-1-p\beta > -1 \implies n > p\beta$$
as you expected to find.