See $\int_0^{\infty}\frac{x^2}{(1+x)^{n-1}}$ as a Beta distribution

Question

Let $n>4, x>0$. I need help with this integral. Using Wolfram,I get that it is $\frac{2\Gamma(n-4)}{\Gamma(n-1)}$, as if it was a Beta distribution with parameters $(3,n-4)$, but I can’t see how to transform this expresion, I can only obtain that it is a Beta$(3,2-n)$.

Also, I do it in a more exhaustive way, using a change of variable and I got $\frac{n^2-6x+10}{(n-3)^2(n-4)}$. Can someone explain me how to get the result of Wolfram?

Answer 1

The result given by Wolfram Alpha is the right one.

Explanation: take a look at this very well done site ; you will see there that you shouldn't refer to a Beta distribution but to a "Beta prime" distribution with pdf:

$$f(x)=\frac{1}{B(a,b)}\frac{x^{a-1}}{(1+x)^{a+b}}$$

here with $a=3$ and $b=n-4$.

Just express that

$$\int_0^{\infty} f(x)=1$$

in order to get

$$\int_0^{\infty}\frac{x^{a-1}}{(1+x)^{a+b}}=B(a,b)=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}=\frac{2\Gamma(n-4)}{\Gamma(n-1)}$$

which is indeed the result given by WA.

Answer 2

Letting $y=\frac{1}{1+x}$ transforms the integral into a beta function as $$ \begin{aligned} I &=\int_1^0\left(\frac{1}{y}-1\right)^2 \cdot y^{n-1} \cdot\left(-\frac{d y}{y^2}\right) \\ &=\int_0^1 y^{n-5}(1-y)^2 d y \\ &=B(n-4,3) \\ &=\frac{[(n-4) \cdot \Gamma(3)}{\Gamma(n-1)} \\ &=\frac{2 \Gamma(n-4)}{\Gamma(n-1)} \end{aligned} $$