# Determine expected value for continuous random variable [duplicate]

This question already has an answer here: I know that $$E[X^n]$$ is found by $$\displaystyle\int_{0}^\infty{x^nf_x(x)dx}$$
I simplified this to $$\displaystyle\int_{0}^\infty{ \frac{x^{\frac{v}{2}-1+n}e^{\frac{-x}{2}}}{\displaystyle\int_{0}^\infty{{x^{\frac{v}{2}-1}e^{\frac{-x}{2}}}}dx} dx }$$

But I don't know how to proceed, since i shouldn't solve the denominator.

## marked as duplicate by whuber♦ probability StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Feb 24 at 16:43

\begin{align}E[X^n]& = \frac1{c_V}\int_0^\infty x^{n+v/2-1}\exp(-\frac{x}2) \, dx\\ &=\frac1{c_V}\left[\left.-2x^{n+v/2-1}\exp\left(-\frac{x}2 \right)\right|_0^\infty\right.\\&\left.+2(n+v/2-1)\int_0^\infty x^{(n-1)+v/2-1}\exp(-\frac{x}2) \, dx\right]\\ &=\frac{2(n+\frac{v}{2}-1)}{c_v}\int_0^\infty x^{(n-1)+v/2-1}\exp(-\frac{x}2) \, dx\\ &=2\left(n+\frac{v}2-1\right)E[X^{n-1}]\end{align}