Question

For what values of p makes the integral converge?

Answer 1

`\displaystyle\int_1^\infty x^(-p)\,\mathrm dx=(x^(1-p))/(1-p)\bigg|_(x=1)^(x\to\infty)`

Note that this assume
`p\\eq1`; 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)(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
`\frac1{1-p}`. So the integral only converges for
`p>1`.