How to evaluate the integral $\int_0^{\infty}[I_{(0,2)}(z)]\frac{(n-1)(y-z)^{n-2}}{y^{n-1}}dy$

Question

I am doing a statistical calculation from a statistical exercise but get stuck at the following integral.

$$\int_0^{\infty}[I_{(0,2)}(z)] \frac{(n-1)(y-z)^{n-2}}{y^{n-1}}dy$$ $0<z<y$

$I_{(0,2)}(z)$ is an indicator function.

Answer 1

Ok,I solved the problem, in fact, our book has a typo it should be $dz$,not $dy$. Thank you for all your comments.

Now the problem becomes easy.

The indicator function is just $1$ or $0$, if , $z\in (0,2) $ then the function is $1$ else is $0$. If it is $0$ integral will be $0$.

So when $z\in (0,2) $

$$\int_0^2 \frac{(n-1)(y-z)^{n-2}}{y^{n-1}}dz\\=\frac{n-1}{y^{n-1}}\int_0^{2} (y-z)^{n-2}dz\\=-\frac{(n-1)}{y^{n-1}}\int_0^2 (y-z)^{n-2}d(y-z)\\=-\frac{(n-1)}{y^{n-1}}\frac{(y-z)^{n-1}}{n-1}\mid_0^2\\=-\frac{(n-1)}{y^{n-1}}[\frac{(y-2)^{n-1}}{n-1}-\frac{(y-0)^{n-1}}{n-1}]\\=1-(\frac{y-2}{y})^{n-1}$$

Which is the solution.

When I try to solve the problem I find that the integral $$\int_0^{\infty}[I_{(0,2)}(z)] \frac{(n-1)(y-z)^{n-2}}{y^{n-3}}dy$$ is solvable just by a small modification.