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

Note that this assume [tex]p\neq1[/tex]; 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.

[tex]=\displaystyle\lim_{x\to\infty}\frac{x^{1-p}}{1-p}-\frac1{1-p}[/tex]

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

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

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