How to prove this :$\left(\cfrac{1}{2}\right)! = \cfrac {\sqrt {\pi}}{2}$?

Question

$\left(\cfrac{1}{2}\right)! = \cfrac {\sqrt {\pi}}{2}$

I saw this in a math book. I tried to prove it but I'm unable to. Any hints or solutions would be most appreciated.

You have to use the Gamma function. Also there is a high possibility that is a duplicate. — vitamin d, Mar 01 '21 at 08:38
This is a duplicate of Why is $\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$?. As $\Gamma(3/2) = \Gamma(1/2)/2$, I would advise you to check the linked question here which has more ways to prove this. — Toby Mak, Mar 01 '21 at 08:40
If you are unfamiliar with the gamma function, check Wikipedia for the definition. — Toby Mak, Mar 01 '21 at 08:45

score 2 · Accepted Answer · answered Mar 01 '21 at 08:50

Please also keep in mind that this is nothing more than an analytic continuation of the factorial. In other words, when you take the factorial of a fraction, it loses its initial meaning of multiplying the number with a number lower until you reach $1$. The Gamma function is, in a sense, an interpolation between the integer values of the factorial that do make sense with the original meaning. So it wouldn't have been possible proving this with the ''ordinary'' definition of the factorial.

score 1 · Answer 2 · answered Mar 01 '21 at 08:43

1

Nothing to prove, we know $n!=\Gamma(n+1)$ where: $$\Gamma(z)=\int_{0}^{+\infty}t^{z-1}e^{-z}dt$$

Now we can "deduce" $\big(\frac{1}{2}\big)!=\frac{\sqrt\pi}{2}$

answered Mar 01 '21 at 08:43

Davide Trono

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How can you "deduce" $\Gamma(1/2) = \int_0^{+\infty}... = \sqrt{\pi}/2 $ ? – NN2 Mar 01 '21 at 08:50

How to prove this :$\left(\cfrac{1}{2}\right)! = \cfrac {\sqrt {\pi}}{2}$?

2 Answers2