# $\left(-\frac{1}{2}\right)! = \sqrt{\pi}?$ [duplicate]

I recently learned that $\left(-\frac{1}{2}\right)! = \sqrt{\pi}$ but I don't understand how that makes sense. Can someone please explain how this is possible?

Thanks!

-

## marked as duplicate by Start wearing purple, Ayman Hourieh, Amzoti, Joe, sdcvvcJun 16 '13 at 22:52

Look up the gamma function on Wikipedia. – Ted Shifrin Jun 16 '13 at 20:31
That's because the factorial that you're familiar with is only for positive numbers; it's an abuse of notation. – Chris Gerig Jun 16 '13 at 20:41

In order to extend the factorial function to any real number, we introduce the Gamma Function, which is a strange object defined as follows:

$$\Gamma(s)=\int_0^\infty t^{s-1}e^{-t} \, dt$$

The gamma function comes with the special property that $n!=\Gamma(n+1)$ for natural numbers $n$, so to evaluate $(-1/2)!$, (which by itself is not technically defined) we define it to be $\Gamma(1/2)$ and hence we evaluate the integral

$$(-1/2)!:=\Gamma(1/2)=\int_0^\infty\frac{e^{-t}}{\sqrt{t}} \,dt$$ To evaluate this integral, we make the substitution $u=\sqrt{t}$, which results in the well known Gaussian integral:

$$\int_0^\infty \frac{e^{-t}}{\sqrt{t}}dt=2\int_0^\infty \frac{e^{-u^2}}{u}u \, du=\int_{-\infty}^\infty e^{-u^2} \, du=\sqrt{\pi}$$

-

I don't know where you've seen this notation. One thing you surely know is the socalled Gamma Function $\Gamma (z)$. This function is a complex function with a host of properties. One of its properties is that if evaluated on $\mathbb{N}$ it coincides with the factorial function, this is, $\Gamma (n+1) = n!$ if $n\in \mathbb{N}$. Also, one can find that $\Gamma (1/2) = \sqrt{\pi}$. So maybe by using a notation I don't know about someone could write $(-1/2)!$ instead of $\Gamma (1/2)$.

-

I posted this very question and an answer to it: Why is $\Gamma(1/2)=\sqrt{\pi}$?

What follows is the answer I posted there. Several others also posted good answers.

If there's any justice in the universe, someone must have asked here how to show that $$\int_{-\infty}^\infty e^{-x^2/2}\,dx = \sqrt{2\pi}.$$ Let's suppose that has been answered here. Let (capital) $X$ be a random variable whose probability distribution is $$\frac{e^{-x^2/2}}{\sqrt{2\pi}}\,dx.$$ Consider the problem of finding $\mathbb E(X^2)$. It is $$\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty x^2 e^{-x^2/2}\,dx = \text{(by symmetry)} \frac{2}{\sqrt{2\pi}} \int_0^\infty x^2 e^{-x^2/2}\,dx$$ $$\sqrt{\frac2\pi}\int_0^\infty xe^{-x^2/2}\Big(x\,dx\Big) = \sqrt{\frac2\pi}\int_0^\infty \sqrt{u}\ e^{-u}\,du = \sqrt{\frac2\pi}\ \Gamma\left(\frac32\right) = \frac12\sqrt{\frac2\pi} \Gamma\left(\frac12\right).$$ So it is enough to show that this expected value is $1$. That is true if the sum of two independent copies of it has expected value $2$. So: $$\Pr\left(X^2+Y^2<w\right) = \frac{1}{2\pi}\iint\limits_\mathrm{disk} e^{-(x^2+y^2)/2}\,dx\,dy$$ where the disk has radius $\sqrt{w}$. This equals $$\frac{1}{2\pi}\int_0^{2\pi}\int_0^{\sqrt{w}} e^{-\rho^2/2} \,\rho\,d\rho\,d\theta = \int_0^{\sqrt{w}} e^{-\rho^2/2} \,\rho\,d\rho.$$ This last equality holds because we are integrating with respect to $\theta$ something not depending on $\theta$. Differentiating this with respect to $w$ gives the probability density function of the random variable $X^2+Y^2$: $$e^{-w/2}\sqrt{w}\frac{1}{2\sqrt{w}} = \frac{e^{-w/2}}{2}\text{ for }w>0.$$ So $$\mathbb E(X^2+Y^2) = \int_0^\infty w \frac{e^{-w/2}}{2}\,dw =2.$$

-

factorial of negative numbers are $defined$ in terms of Gamma function, i.e. $(x!) := \Gamma(x+1)\forall x\in\mathbb R$, except for $x$ to be negative integer. this is because the two of them agree on positive integers, and so this is just a convention.

-
Actually, $\Gamma(n+1)=n!$ for $n\in\mathbb{Z}$ and neither exist for $n\lt0$. – robjohn Jun 16 '13 at 21:33
@robjohn sorry for my mistake. I have edited it. Is it correct now? – mathmansujo Jun 16 '13 at 21:47
Better. However, this is a bit more than convention: $\Gamma$ is the only log-convex function so that $\Gamma(1)=1$ and $\Gamma(x+1)=x\Gamma(x)$. – robjohn Jun 16 '13 at 22:28