# How to find the normalizing constant for a distribution of unbounded support?

The probability density of a random variable is $$f(x) = ax^2 e^{-kx} ;k\gt0,0\le x\le \infty$$

What is the value of $a$?

I understand that first we'll have to take the integral of the function with limits $0$ and $\infty$ and then substitute them in the integral obtained. This is where I'm stuck because substituting $0$ and $\infty$ yields an undefined number. I need to know where I'm going wrong and how to proceed with it. Thank you !

• By definition, $\int_0^\infty f(x) dx = \lim_{a\to\infty}\int_0^a f(x) dx$. This avoids having to deal with undefined numbers. – whuber Apr 20 '15 at 16:04

If $X$ is a random variable with $\text{Gamma}(\text{shape}=\alpha,\text{rate}=\beta)$, then it's density is $$f_X(x)=\frac{\beta^\alpha}{\Gamma(\alpha)}x^{\alpha-1}e^{-\beta x}$$ for $x\geq 0$, and $f_X(x)=0$ for $x<0$.
Remember that $\int_{-\infty}^\infty f_X(x)\,dx=\Pr(X\in\mathbb{R})=1$.
You want to find $$a = \frac{1}{\int_0^\infty x^2e^{-kx}}.$$ Notice that $$\int_0^\infty x^2e^{-kx} = \frac{\Gamma(\alpha)}{\beta^\alpha}\int_0^\infty \frac{\beta^\alpha}{\Gamma(\alpha)}x^{\alpha-1}e^{-\beta x},$$ in which $\alpha=3$ and $\beta=k$.
This is all the information you need to find $a$ by simple inspection. You don't need to "do" any integrals.