Moment of a probability distribution

[dbb04]

when you calculate the Moment of the following equation



p(x)=\left\{\begin{array}{cc}2Axe^{-Ax^2},&\mbox{ if }
x\geq 0\\0, & \mbox{ if } x<0\end{array}\right.


We get


Mn =2A \int_0^\infty x^{n+1}e^{-Ax^2}


solving it by parts I am getting


Mn=(n+1)\int_0^\infty x^{n-1}e^{-Ax^2}


but, apparently, the right solution is


Mn=n\int_0^\infty x^{n-1}e^{-Ax^2}



What am I doing wrong? What is the proper way to solve it? Could you please do it step by step?

Thanks

[Tide]

It looks like you didn't differentate properly when you integrated by parts:

\frac {d x^n}{dx} = n x^{n-1}

[dbb04]

sorry, still not following you.
If we integrate by parts we have


Mn =2A \int_0^\infty x^{n+1}e^{-Ax^2}



\int_0^\infty u\frac{dv}{dx} dx=uv-\int_0^\infty v\frac{du}{dx} dx


where


u=x^{n+1} \ \ \ \ \ \frac{du}{dx}=(n+1)x^n


and


\frac{dv}{dx}=e^{-Ax^2} \ \ \ \ \ v= -\frac{e^{-Ax^2}}{2xA}


so


Mn=2A \ \{-x^{n+1} \ \frac{e^{-Ax^2}}{2xA} \ \ |_0^\infty +\int_0^\infty \frac{e^{-Ax^2}}{2xA}(n+1)x^n dx \ \}



Mn=0 +(n+1)\int_0^\infty x^{n-1} \ e^{-Ax^2} dx


So, How you get rid of the (n+1) term.
Thanks

[Tide]

This should get you where you want to go:

\int_0^{\infty}x^{n+1} e^{-Ax^2} dx = - \frac {1}{2A}
\int_0^{\infty}x^n \frac {d}{dx}e^{-Ax^2}dx

[dbb04]

Thanks Tide,

appreciate your patience