# Thread: Today's calculation of integral #3

1. ## Today's calculation of integral #3

Prove that $\displaystyle \left(\frac12\right)!=\frac{\sqrt\pi}2.$

2. ## IM stupid but?

Isnt this just (1/2)!=$\displaystyle \gamma$(-1/2)?=$\displaystyle \int_{0}{1}{-ln(x)^(1/2) \dx$?

3. Sorry I meant isnt it just gammafunction(-1/2)?

4. so it is $\displaystyle \int_{0}{1}{1/(-ln(x)^(1/2)) \ dx}[/$ which is pi^(1/2)*erfi(ln(x)^(1/2))*ln(x)^(1/2)/(-ln(x))^(1/2) evaluated from 0 to one?

5. Originally Posted by Krizalid
Prove that $\displaystyle \left(\frac12\right)!=\frac{\sqrt\pi}2.$
$\displaystyle \Gamma(1/2)=\int_{0}^{\infty}t^{-1/2}e^{-t}dt$

using the sub $\displaystyle u^2=t \mbox{ then } 2udu=dt$

$\displaystyle \Gamma(1/2)=\int_{0}^{\infty}u^{-1}e^{-u^2}(2udu)=2\int_{0}^{\infty}e^{-u^2}du$

we could do the same sub with $\displaystyle v^2=t$ and get

$\displaystyle \Gamma(1/2)=2\int_{0}^{\infty}e^{-v^2}dv$

so

$\displaystyle [\Gamma(1/2)]^2=(2\int_{0}^{\infty}e^{-u^2}du)(2\int_{0}^{\infty}e^{-v^2}dv)=4\int_{0}^{\infty}\int_{0}^{\infty}e^{-(u^2+v^2)dudv}$

Convert to polar coordinates the region of integration is the 1st quadrant so we get...

$\displaystyle 4 (\int_{0}^{\pi/2}d\theta)(\int_{0}^{\infty}re^{-r^2}dr)=4 \left( \frac{\pi}{2}\right) \left( \frac{1}{2}\right)=\pi$

so

$\displaystyle [\Gamma(1/2)]^2=\pi \iff \Gamma(1/2)=\sqrt{\pi}$

6. Yeah, that's the standard solution.

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By considering $\displaystyle \int_0^1x^n(1-x)^n\,dx$ for $\displaystyle n\ge0,$ a direct application of Beta function shows that $\displaystyle \int_0^1x^n(1-x)^n\,dx=\frac{(n!)^2}{(2n+1)!}.$

Eventually, for $\displaystyle n=\frac12$ we have $\displaystyle \int_0^1\sqrt{x(1-x)}=\bigg\{\left(\frac12\right)!\bigg\}^2\cdot\fra c12=\frac\pi8,$ this 'cause we're integrating a semicircle whose radius is $\displaystyle \frac12.$ The conclusion follows.