Suppose that $X$ is a non-negative rv with CDF $F$. Prove that for any $p \geq 1$, we have: $$\int_0^\infty x^p dF(x) = \int_0^\infty px^{p-1} (1-F(x)) dx$$
I tried to do this using integration by parts on the right integral, which gave me the result: $$ \lim_{x\rightarrow\infty} x^p(1-F(x)) + \int_0^\infty x^p dF(x) $$ but I am struggling to show that $\lim_{x\rightarrow\infty} x^p(1-F(x)) = 0$. Could someone please provide some hints?
Also just wondering, how does the result change if $X$ is no longer non-negative?
Note that $$ x^pI(X>x)\leq X^{p}I(X>x) $$ whence taking expectations of both sides $$ x^pP(X>x)\leq E(X^{p}I(X>x))\to 0 $$ as $x\to \infty$ by the dominated convergence theorem assuming that $EX^p<\infty.$
Alternatively
Let $X$ be have distribution function $F$. Note that $$ \begin{align*} \int_0^\infty px^{p-1} (1-F(x)) dx&=\int_0^\infty px^{p-1}\int_{\Omega} I(X>x)\, dP \, dx\\ &=\int_{\Omega}\int_0^Xpx^{p-1}\, dx\, dP\tag{0}\\ &=\int_{\Omega} X^{P} dP\\ &=EX^{p}\\ &=\int_{0}^\infty x^p\, dF(x) \end{align*} $$ where in (0) we used Tonelli's theorem.