For what value of $k$ is $f(x)=\frac{k}{(1+x^3)}$a distrib. function and what is its variance?

Question

Let $X$ be a random variable with the folowing distribution function: $f(x)=\frac{k}{(1+x^3)}$ for all $x>0$.

1. Find a value for the constant $k$ for which $f$ will be a distribution >function.

2. Calculate the variance of $X$.

In 1, I assumed that in order for $f$ to be a distribution function, I should demand $$\int_0^\infty f(x) = 1$$ Since $$\int_0^\infty \frac{k}{(1+x^3)} = (-\frac{ln(x^2-x+1)}{6}+\frac{ln(x+1)}{3}+\frac{arctan {\frac{(2x-1)}{\sqrt{3}}}}{\sqrt{3}})|_0^\infty = \frac{2\pi}{\sqrt{27}}$$ I concluded $$k = \frac{\sqrt{27}}{2\pi}$$ in 2, Since $Var(X) = E[(x-\mu)^2]$, (and since $\mu = \int xf(x) = 1$) I thought that $Var(X)$ should be $$ \int_0^\infty \frac{(x-1)^2}{1+(x-1)^6}dx$$

But, I think that something is wrong. These integrals get too difficult for a probabiity course.

Any ideas?

Thank you!

Answer 1

The integration to find $k$ was indeed not pleasant. I have not checked it.

The rest is easy! Note that the variance of $X$ is $E(X^2)-(E(X))^2$. The expectation of $X$ exists. Let us not compute it.

For $E(X^2)$, we need $\int_0^\infty \frac{kx^2}{1+x^3}\,dx$. This integral diverges.