Question

For what values of p makes the integral converge?

Answer 1

$\displaystyle\int_1^\infty x^{-p}\,\mathrm dx=\frac{x^{1-p}}{1-p}\bigg|_{x=1}^{x\to\infty}$

Note that this assume $p\neq1$; if that were the case, we'd end up with a logarithm as the antiderivative, and it's easy to show that in that case the integral diverges.

$=\displaystyle\lim_{x\to\infty}\frac{x^{1-p}}{1-p}-\frac1{1-p}$

We have two cases remaining. If $p<1$, then the numerator is of degree larger than 0, and as $x\to\infty$ we have $x^{1-p}\to\infty$.

If $p>1$, then the degree of the numerator will be negative, which would mean the numerator approaches 0 as $x\to\infty$, leaving us with just $\dfrac1{1-p}$. So the integral only converges for $p>1$.