Calculate $E(Y)$ where $Y=X^{1.5}$

Question

Let $X$ be a rv which is $Exp(\lambda=2)$.

The pdf of $X$ is given by $f_X(x)=2e^{-2x}, x\geq 0$ (and $0$ otherwise).

We define $Y=X^{1.5}$ and ask $E(Y)$.

It does not look like the moment generating function can help me here. So, instead I evaluate the integral:

$$2\int^{\infty}_{0}e^{-2x}x^{1.5}dx$$

I know the Gamma function is given by: $$\int_{0}^{\infty}x^{\alpha-1}e^{-x}dx$$ but I have $e^{-2x}$ in my original integral. So I don't think this can help me either.

My question is the following:

1) If I were to evaluate the integral, it seems very messy by parts. Is there a simple way to evaluate the integral?

2) Is there a simpler way to solve the question that I am missing?

Answer 1

Just make a change of variables $u = 2x$, so that ${\rm d}x = {\rm d} u / 2$ and

$$ 2\int_0^{+\infty}x^{3/2}e^{-2x}{\rm d}x = 2^{-3/2} \int_0^{+\infty}u^{3/2}e^{-u}{\rm d}u = 2^{-3/2} \int_0^{+\infty}u^{5/2 - 1}e^{-u}{\rm d}u = 2^{-3/2}\Gamma(5/2) $$

You can further simplify this by using $\Gamma(5/2) = 3\pi^{1/2}/4$

$$ \mathbb{E}[X^{3/2}] = 3\cdot2^{-7/2}\pi^{1/2} $$