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I know it's $\frac{\sqrt{\pi}}{2}$ but how can this be evaluated by hand? (Or can it not?) For quick reference: $$n!=\Gamma(n+1)$$ $$\Gamma(n)=(n-1)!$$ $$\Gamma(n)=\int_0^\infty x^{n-1} e^{-x}\,dx$$ Thanks. And if it's not too much trouble, maybe some guidelines in solving general Gamma functions.


marked as duplicate by Jack D'Aurizio, Jonas Meyer, Michael Hardy, Adriano, Grigory M Apr 6 '15 at 9:29

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    $\begingroup$ $\Gamma(\frac12)=\sqrt{\pi}$ is a classical result, and the fuctional equation $x\Gamma(x)=\Gamma(x+1)$ gives you the result. Is this the sort of thing you are looking for? $\endgroup$ – Olivier Bégassat Apr 6 '15 at 0:57
  • $\begingroup$ Thanks, but not really. I'm not understanding where $\sqrt{\pi}$ or $\frac{\sqrt{\pi}}{2}$ come from. $\endgroup$ – Jacob Wheeler Apr 6 '15 at 1:05

$$\Gamma(3/2) = \Gamma(1/2)/2 = \frac{1}{2}\int_0^{\infty} x^{-1/2} e^{-x} \, dx $$ Change variables to $y=x^{1/2}$, so $dy=dx/(2x^{1/2})$, and we have the integral $$ I = \int_0^{\infty} e^{-y^2} \, dy. $$

You may or may not have met this one before. Basically the easiest idea for evaluating this is to change coordinates to polars after squaring it. $$ I^2 = \int_0^{\infty} \int_0^{\infty} e^{-(x^2+y^2)} \, dx \, dy $$ Set $x=r\cos{\theta}$, $y=r\sin{\theta}$, then $ dx \, dy = r \, dr \, d\theta $. The integral is over the first quadrant, so $0<\theta<\pi/2$. $$ I^2 = \int_0^{\infty} \int_0^{\pi/2} re^{-r^2} \, d\theta \, dr = \frac{\pi}{2} \int_0^{\infty} re^{-r^2} \, dr $$ The remaining integral has an elementary antiderivative $e^{-r^2}/2$, so the whole thing evaluates to $$ I^2 = \frac{\pi}{4}, $$ and you take the positive root for obvious reasons.


By definition of the Gamma function, we have $$\Gamma\left(\frac{1}{2}\right)=\intop_{0}^{\infty}\frac{e^{-t}}{\sqrt{t}}\ dt.$$

Now let $u=\sqrt{t}$, so that $du=\frac{1}{2\sqrt{t}}\ dt$. Then we have


This last integral is the Gaussian integral, whose value can be found many different ways. Several are included in the link.


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