Given a continuous function $f\colon \mathbb{R}^n \to \mathbb{R}$ such that, for some M, R, p constant values:
$$\vert f(x) \vert \leq \frac{M}{\vert\vert x \vert\vert^p}$$
If $\vert\vert x \vert\vert\geq R$ and $p>n$. Then the integral over the entire space $\mathbb{R}^n$ exists.
Now, so far I've realized that for $R> 1, M>0$ the exponential function $g(x)=R^x$ is increasing. Therefore, given $p>n \implies \vert\vert x \vert\vert^p> \vert\vert x \vert\vert^n \implies \frac{M}{\vert\vert x \vert\vert^p}<\frac{M}{\vert\vert x \vert\vert^n}$
So $\vert f(x) \vert < \frac{M}{\vert\vert x \vert\vert^n}$. This tells me that the function is bounded, but I'm not sure where to go from here. Are there any theorems or lemmas that could help with the proof?
Thank you so much!
This is an improper Riemann integral.
To show that $\lim_{r \to \infty} \int_{\|x\| \le r} f$ exists, it is sufficient to show that $\lim_{r \to \infty} \int_{\|x\| \le r} f_+$ and $\lim_{r \to \infty} \int_{\|x\| \le r} f_-$ exist, where $f_+=\max(f,0), f_- = \max(-f,0)$. Since $f_+, f_-$ are continuous, and are bounded above by $|f|$, it is sufficient to show that $\lim_{r \to \infty} \int_{\|x\| \le r} |f|$ exists (and is finite).
Note that if $0 \le h \le g$ and $g$ is improperly integrable, then $h$ is improperly integrable.
In the above case, it is sufficient to show that $\int_{R \le \|x\| \le r} {1 \over \|x\|^p} dx$ exists (and is finite). For simplicity we can assume that $R$ is an integer.
Let $S$ be the volume of $B(0,1)$, note that the volume of $B(0,r)$ is $S^n$. Let $g$ be the function $g(x) = \sum_{k=R}^\infty 1_{B(0,k+1)\setminus B(0,k)}(x) {1 \over k^p}$, note that $g(x) = {1 \over \lfloor \|x\| \rfloor^p} \ge {1 \over \|x\|^p}$ for $\|x\| \ge R$. Note that the set of discontinuities of $g$ have measure zero, and hence $g$ is integrable on any closed ball of finite radius.
We have $\int_{\|x\| \le r} g \le \sum_{k=R}^\infty ((k+1)^n-k^n) S {1 \over k^p}$, and $p(x) = (x+1)^n -x^n$ is a polynomial of degree $n-1$ and since $p-(n-1) > 1$ we see that the series is summable and hence $\lim_{r \to \infty} \int_{\|x\| \le r} g$ exists and so $f$ is integrable.