So our teacher doesnt use the same demonstration as most other sites use for proving that gamma of a half is the square root of pi. I dont understand the demonstration from the first step because he uses the Wallis product but first he represents $Γ(1/2)$ as :

$$Γ(n + 1/2) = 2^{-n}Γ(1/2)\prod_{k=1..n}(2k-1)$$

This is just the first step and i dont undderstand how they get that.. I understand the gamme function and that when you integrate it you get $Γ(x+1) = xΓ(x)$ and i know i need to somehow use this identity but i dunno how.

  • 6
    $\begingroup$ $$\Gamma(1+\tfrac{1}{2}) = \tfrac{1}{2}\Gamma(\tfrac{1}{2}) = 2^{-1}(2\cdot 1-1)\Gamma(\tfrac{1}{2});\; \Gamma(2+\tfrac{1}{2}) = (1+\tfrac{1}{2})\Gamma(1+\tfrac{1}{2}) = 2^{-1}(2\cdot 2 - 1)\Gamma(1+\tfrac{1}{2}) = 2^{-2}(2\cdot 2-1)(2\cdot 1-1)\Gamma(\tfrac{1}{2})$$ And so on. $\endgroup$ Feb 18, 2015 at 11:08
  • $\begingroup$ A pity the above is just a comment and not an answer, as it is the clearest one, imo, to the OP's post. $\endgroup$
    – Timbuc
    Feb 18, 2015 at 12:08
  • $\begingroup$ @Timbuc The OP was looking for ideas as to why and how the lecturer used the associated formula on line 4, Daniel's comment is fine as a comment, but doesn't explore how it potentially arose. $\endgroup$
    – Autolatry
    Feb 18, 2015 at 12:14
  • $\begingroup$ @user135688 could you please say why don't you accept any of the solutions below? This could help the authors below to refine their derivations or that you get a yet more tailored answer ... $\endgroup$
    – Math-fun
    Feb 18, 2015 at 12:28
  • $\begingroup$ @Autolatry "how it potentially arose"? Anyway, I think it definitely does as he's using exactly what the OP mentiones at the end of its question: $\;\Gamma(1+x)=x\Gamma (x)\;$ . This, together with an easy inductive argument on the above is all that's needed. $\endgroup$
    – Timbuc
    Feb 18, 2015 at 13:00

4 Answers 4


Here why $\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$ :$$\Gamma\left(\frac{1}{2}\right)=\intop_{t=0}^{+\infty}t^{\frac{1}{2}-1}e^{-t}dt=\intop_{t=0}^{+\infty}\frac{e^{-t}}{\sqrt{t}}dt,$$ and with $y=\sqrt{t}$, $dy=\frac{dt}{2\sqrt{t}}$, we get$$\Gamma\left(\frac{1}{2}\right)=2\intop_{y=0}^{+\infty}e^{-y^{2}}dy=\intop_{y=-\infty}^{+\infty}e^{-y^{2}}dy=\sqrt{\pi}.$$

  • 2
    $\begingroup$ The OP knew this result, but wanted clarification on his/her lecturer's demonstration. $\endgroup$
    – Autolatry
    Feb 18, 2015 at 11:44
  • $\begingroup$ I was not sure of this because of the title... but it can still be usefull for anyone who wants to see the calculus. $\endgroup$
    – Nicolas
    Feb 18, 2015 at 11:48
  • $\begingroup$ Absolutely! Your post is perfectly correct. $\endgroup$
    – Autolatry
    Feb 18, 2015 at 11:49
  • $\begingroup$ For some reason, I was not able to think of this despite of having the formula XD. Thanks so much @Nicolas! $\endgroup$
    – HarshDarji
    Jan 19 at 17:38

From the Legendre duplication formula; \begin{equation} \Gamma(z) \; \Gamma\left(z + \frac{1}{2}\right) = 2^{1-2z} \; \sqrt{\pi} \; \Gamma(2z). \end{equation} We can re-arrange by dividing by $\Gamma(z)$ and considering the quotient \begin{eqnarray} \frac{\Gamma(2n)}{\Gamma(n)} &=& \frac{1}{\Gamma(n)} \left(2n(2n-1)(2n-2)(2n-3)(2n-4)(2n-5)\ldots\right)\\ &=& \frac{1}{\Gamma(n)} \left(2^{n+1}n(2n-1)(n-1)(2n-3)(n-2)(2n-5)(n-3)\ldots\right)\\ &=& \frac{1}{\Gamma(n)} \left(2^{n+1}n! \prod_{k=1}^{n}(2n-(2k+1))\right)\\ &=& 2^{n+1} \prod_{k=1}^{n}(2n-(2k+1)) \end{eqnarray} Which means that \begin{eqnarray} \Gamma\left(z + \frac{1}{2}\right)&=& 2^{1-2n} \sqrt{\pi} 2^{n+1} \prod_{k=1}^{n}(2n-(2k+1)) \\ &=& \sqrt{\pi} 2^{-n} \prod_{k=1}^{n}(2n-(2k+1)) \end{eqnarray} In general for non-integer $n$,

\begin{eqnarray} \Gamma\left(\frac{1}{2}+n\right) &=& {(2n)! \over 4^n n!} \sqrt{\pi} \\ &=& \frac{(2n-1)!!}{2^n} \sqrt{\pi} \\ &=& \sqrt{\pi} \left[ {n-\frac{1}{2}\choose n} n! \right] \end{eqnarray}

  • $\begingroup$ I found three mistakes in this derivation. 2nd equation: Γ(2n)≠(2n)! but instead, Γ(2n)=(2n−1)!. Obviously the same for Γ(n). Also, one gets $2^{n−1}$ instead of $2^{n+1}$. Finally, the product in the third equation should be $...(2n-(2k-1))$. If you agree, edit your answer please $\endgroup$
    – Mencia
    Apr 13, 2016 at 12:44

Through $\Gamma(x+1)=x\,\Gamma(x)$ we have: $$\begin{eqnarray*}\Gamma\left(n+\frac{1}{2}\right) &=& \left(n-\frac{1}{2}\right)\Gamma\left(n-\frac{1}{2}\right) = \left(n-\frac{1}{2}\right)\left(n-\frac{3}{2}\right)\Gamma\left(n-\frac{1}{3}\right)\\&=&\ldots\;=\color{red}{\frac{(2n-1)!!}{2^n}\,\Gamma\left(\frac{1}{2}\right)}\end{eqnarray*}$$ as wanted.


We use Euler's Reflection Formula $\Gamma(1-z)\Gamma(z)= \frac\pi{\sin(\pi z)}$ with $z=\frac12$. We then have $\Gamma(1/2)^2=\frac\pi{\sin( \frac\pi2)}$. Since $\sin( \frac\pi2)=1$ we now have $\Gamma(1/2)^2=\pi$. Finally therefore $\Gamma(1/2)=\sqrt\pi$. Not $-\sqrt\pi$ because we know $\Gamma( 1/2) > 0$.


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