I'm trying to figure out how to prove $\theta y^{\theta} x^{-\theta -1}$ where $y>0, \theta>1, x\geq y$ is a PDF. So far I have $\int_{-\infty}^\infty \theta y^{\theta} x^{-\theta -1}dx=-\frac{y^\theta}{x^\theta}|_{-\infty}^\infty$ but don't know where to go from there. I know given the dependencies that the fraction will result in something at most 1, but don't how to make the integral equal to 1.
Edit: I think I figured it out by changing the bounds. Since $x\geq y >0$, then the bounds are from $y$ to $\infty$. Thus $-\frac{y^\theta}{x^\theta}|_{y}^\infty=1$. Any confirmation would be great. It is also trivial that the function is $\geq0$, so it satisfies both properties to be a PDF.
Being positive, you only need to show that this function integrates to $1$:
\begin{align*} \int_y^\infty \theta y^\theta x^{- \theta - 1}dx = \left[-\frac{y^\theta}{x^\theta} \right]_y^\infty = 1 \end{align*}
Notice that in your original solution attempt you didn't account for the support of your PDF being $x \geq y$.