If $\int_1^\infty x^{-p} dx\,$ exists, then does $\int_1^\infty x^{-q} dx\,$ exist, where $q > p$?

Question

If $\int_1^\infty x^{-p} dx\,$ exists, then does $\int_1^\infty x^{-q} dx\,$ exist, where $q > p$?

My initial assumption is that the answer was true. Because if p is an arbitrary number 5, then the area is $1/4$. So then q could be 2, giving 1 for an area.

These values for p and q both make the integral exist, however what if you use 1 for q. The answer is $\infty$, which is divergent, so it does not exist.

Since no values are given for p and q, how do I know whether the improper integral exists or not?

Answer 1

Hint: If $q>p $ and $|x|\geq 1$ then $|x|^q\geq |x|^p$. So $\frac{1}{|x|^p}\geq\frac{1}{|x|^q}$.

Answer 2

To show that $\int_1^\infty x^{-q}\,dx$ exists, we need to show that $$\lim_{B\to\infty}\int_1^B x^{-q}\,dx$$ exists. We are told that $\int_1^\infty x^{-p}\,dx$ exists. In particular, this implies that $p$ is positive.

We are also told that $p\lt q$. From this it follows that $x^{-q}\le x^{-p}$ for all $x\ge 1$, and therefore $$f(B)=\int_1^Bx^{-q}\,dx\le \int_1^\infty x^{-p}\,dx$$ for all $B$. Since the function $f(B)$ is increasing and bounded above, $\lim_{B\to\infty} f(B)$ exists, and therefore our improper integral exists.