Why does the following equality (from a book I'm reading) hold? $$\int_0^\infty\mathbb{P}\big\{|X|^{2q}>x\big\}dx=\int_0^\infty 2qx^{2q-1}\mathbb{P}\{|X|>x\}dx$$
I know that $\int_0^\infty\mathbb{P}\big\{|X|^{2q}>x\big\}dx=\mathbb{E}[x^{2q}]$ (this is the starting point that led to this equation), and I notice that $\tfrac{\partial}{\partial x}x^{2q}=2qx^{2q-1}$.
I though it might be a specialization of some general case, something like - $$\int_0^\infty\mathbb{P}\Big\{f\big(g(x)\big)>x\Big\}dx=\int_0^\infty f'(x)\mathbb{P}\{g(x)>x\}dx$$ but the above doesn't hold in general.
It seems as if it should be trivial. I guess I'm missing something obvious.
If $y=x^{1/p}$, then $dx=py^{p-1}dy$. Consequently, for $p>0$, \begin{align} \int_0^\infty\mathsf{P}\!\big(|X|^{p}>x\big)dx&=\int_0^\infty\mathsf{P}\big(|X|>x^{1/p}\big)dx \\ &=\int_0^\infty py^{p-1}\mathsf{P}\big(|X|>y\big)dy, \end{align} where the first equality holds because $x\mapsto x^{p}$ is increasing (on $[0,\infty)$).