Given density function finding constant

Question

Given X, a continuous random variable and the density:

$$ f(x) = \begin{cases} x^2, & \text{if $x$ $\in$ [-1,1] ,} \\ c \cdot \frac{1}{|x|^k}, & \text{else} \end{cases} $$ Where $c\in \mathbb{R}$ and $k\in \mathbb{N}\geq2$. We need to identify the value of $c$ so that $f$ is truely a density function.

We tried to solve $$\int_{-\infty}^{-1}c\cdot\frac{1}{|x|^k} + \int_{-1}^{1}x^2 + \int_{1}^{\infty}c\cdot\frac{1}{|x|^k} = 1$$ but failed miserably and dont know any other way we could determine $c$. Any hints would be much apreciated. Thanks in advance.

Answer 1

Guide:

note that $f$ is an even function:

$$\int_0^1 x^2 \, dx + c \int_1^\infty \frac{1}{x^k}\, dx =\frac12$$

Integrating $x^2$ and $\frac{1}{x^k}$ (note that $k \ge 2$), then solve for $c$ in terms of $k$.

Answer 2

I don't think I have any alternative to offer... that's how you determine $c,$ so it's just a matter of doing the integral.

Hopefully you can do the integral $A =\int_{-1}^1 x^2dx,$ which is just a number.

Then using the power rule (and the fact that $k\ge 2)$, we have $$ c\int_1^\infty \frac{1}{x^k} dx = -c \frac{1}{k-1}\left.\frac{1}{x^{k-1}}\right|_1^\infty = \frac{c}{k-1}.$$ Symmetry says $c\int_{-\infty}^{-1} \frac{1}{|x|^k}dx =\frac{c}{k-1}$ as well. So you have $$ A + \frac{2c}{k-1} = 1$$ and need to solve for $c.$