crazedbeat said:
Given that
F(x) = \int_{0}^{1} f(s) ds
and
F(\infty)=1
prove that for any
\alpha \geq 1,
\int_{0}^{\infty} x^(\alpha) dF(x) = \alpha \int_{0}^{\infty} x^(\alpha - 1)(1 - F(x))dx
where two sides either converge or diverge together.
note: some of it didn't come out correctly. on the left hand side it means x raised alpha, and on the left x raised to (alpha - 1).
I have NO idea how to even START the problem. I've been doing convergence/divergence by finding limits, ratio test, etc. But these are not even close-- these are integrals and they don't make any sense. Any guidance into what i should be looking at would be greatly appreciated.
From the problem statement, we have:
1: \ \ \ \ F(x) \ = \ \int_{0}^{x} f(s) \, ds \hspace{0.8cm} \mbox{where:} \ \ \lim_{x \to \infty} F(x) \ = \ \left(1\right) \ \ \mbox{and} \ \ \ F(0) \ = \ \left(0\right)
Define the following:
2: \ \ \ \ g\left(x\right) \ \, = \, \ x^{\displaystyle \alpha} F\left(x\right)
Thus:
3: \ \ \ \ \frac{dg(x)}{dx} \ \ \, = \, \ \ \alpha x^{\displaystyle (\alpha - 1)} F(x) \ \ \, + \ \, \ x^{\displaystyle \alpha} \left ( \frac{dF(x)}{dx}\right )
4: \ \ \ \ \Longrightarrow \ \ x^{\displaystyle \alpha} \left (\frac{dF(x)}{dx}\right) \ \ \, = \, \ \ \frac{dg(x)}{dx} \ \ \, - \ \, \ \alpha x^{\displaystyle (\alpha - 1)} F(x)
5: \ \ \ \ \Longrightarrow \ \ x^{\displaystyle \alpha} \left (\frac{dF(x)}{dx}\right ) \ \ \, = \, \ \ \frac{d}{dx} \left \{ x^{\displaystyle \alpha} F(x) \right \} \ \ \, - \ \, \ \alpha x^{\displaystyle (\alpha - 1)} F(x)
6: \ \ \ \ \Longrightarrow \ \ \int_{0}^{w} x^{\displaystyle \alpha} \left (\frac{dF(x)}{dx}\right ) \, dx \ \ \, = \, \ \ \color{blue}\int_{0}^{w} \frac{d}{dx} \left \{ x^{\displaystyle \alpha} F(x) \right \} \, dx \color{black} \ \ - \ \ \int_{0}^{w} \alpha x^{\displaystyle (\alpha - 1)} F(x) \, dx
7: \ \ \ \ \Longrightarrow \ \ \int_{0}^{w} x^{\displaystyle \alpha} \, dF(x)\right ) \ \ \, = \, \ \ \color{blue}\left [ x^{\displaystyle \alpha} F(x) \right ]_{0}^{w}\color{black} \ \ \, - \ \, \ \int_{0}^{w} \alpha x^{\displaystyle (\alpha - 1)} F(x) \, dx
8: \ \ \ \ \Longrightarrow \ \ \int_{0}^{w} x^{\displaystyle \alpha} \, dF(x)\right ) \ \ \, = \, \ \ \color{blue}\alpha \int_{0}^{w} x^{\displaystyle (\alpha - 1)} F(w) \, dx \color{black} \ \ \, - \ \, \ \alpha \int_{0}^{w} x^{\displaystyle (\alpha - 1)} F(x) \, dx
9: \ \ \ \ \Longrightarrow \ \ \int_{0}^{w} x^{\displaystyle \alpha} \, dF(x)\right ) \ \ \, = \, \ \ \alpha \int_{0}^{w} x^{\displaystyle (\alpha - 1)} \left \{ F(w) \, - \, F(x) \right \} \, dx
10: \ \ \ \Longrightarrow \ \ \lim_{w \rightarrow \infty} \, \int_{0}^{w} x^{\displaystyle \alpha} \, dF(x)\right ) \ \ \, = \, \ \ \lim_{w \rightarrow \infty} \, \left ( \alpha \int_{0}^{w} x^{\displaystyle (\alpha - 1)} \left \{ F(w) \, - \, F(x) \right \} \, dx \right )
11: \ \ \ \color{red} \Longrightarrow \ \ \int_{0}^{\infty} x^{\displaystyle \alpha} \, dF(x)\right ) \ \ \, = \, \ \ \alpha \int_{0}^{\infty} x^{\displaystyle (\alpha - 1)} \left \{ 1 \, - \, F(x) \right \} \, dx \hspace{1.5cm} \mbox{\LARGE \textbf{Q.E.D.}}
~~