I want to show that that for $d\mu(x) = x^\alpha dx, d\nu = x^{\alpha + p} dx$
We have
$$\int_0^\infty \left(\int_x^\infty f(t)dt\right)^p d\mu(x) \leq c \int_0^\infty f^p(x) d\nu(x) \quad \alpha < - 1$$
$$\int_0^\infty \left(\int_0^x f(t)dt\right)^p d\mu(x) \leq c \int_0^\infty f^p(x) d\nu(x) \quad \alpha > - 1$$
For the first case, I used Holders inequality, which gives ($\int_0^x f(t)dt)^p \leq cx^{p-n-1}\int_0^\infty f^p(t)t^n dt$ for $p - n -1 > 0 $, then multiplying both sides by $x^\alpha$, integrating over $(0,\infty)$, and applying a change of order of integration which gave me
$$\int_0^\infty \left(\int_0^x f(t)dt\right)^p d\mu(x) \leq \frac{c}{\alpha + p -n} \int_0^\infty f^p(x) d\nu(x)$$
Which worries me for $\alpha = n - p$, however I think my process was correct.
I'm having problems with how to go about the second inequality for $\alpha > -1$, any ideas would be greatly appreciated.