I have the following probability density function :
$\ p(x)=p_{a,\lambda}(x)=\frac{a\lambda^a}{(x+\lambda)^{1+a}},$ $ \lambda>0, a>0$ and I need to:
1) Give the values $\beta$ ,$\beta>0$ for which $\mathbb{E}(X^\beta)$ is finite
2) Show that $\mathbb{E}(X^\beta)$ can be written as $\lambda^bc$, where $b$ and $c$ are expressions of $a$ and $\beta$.
To solve question 2 I calculated $\int_{0}^{+\infty} x^{\beta}\frac{a\lambda^a}{(x+\lambda)^{1+a}}dx$ by parts and found $a\lambda^a[\frac{x^\beta}{-a(x+\lambda)^a}]^{+\infty}_{0}+\beta\lambda^a \int_{0}^{+\infty}\frac{x^{\beta-1}}{(x+\lambda)^a}dx$, could someone give me a hint on how to proceed from here? And for question 1 I don't really have any clue. Thanks a lot
You can rewrite the integral as a Beta function with $x=\lambda\tan^2t,\,y=\sin^2t$. The boundary term is $0$ provided $\beta<\alpha$; you can see this from the $x\to\infty$ behaviour of $\frac{x^\beta}{(x+\lambda)^\alpha}$.