By the definition of the gamma function: $$\Gamma(n)=\int_0^{\infty}{e^{-x}x^{n-1}dx}$$ Let $n = \frac{1}{2}$ and we get: $$\Gamma(\frac{1}{2})=\int_0^{\infty}{e^{-x}x^{-\frac{1}{2}}dx}$$ I tried to use integration by parts but treating $x^{-\frac{1}{2}}$ as $u$ or as $v$ both makes it needlessly complicated and doesn't seem to get me anywhere. I then tried the subtitution $u=e^{-x}$, where $x=-\ln(u)$ and $dx=-\frac{1}{u}du$ gives me this new integral: $$\int_1^0{\frac{u}{\sqrt{\ln(u)}}\cdot\frac{1}{u}du}=\int_1^0{-\frac{1}{\sqrt{-\ln(u)}}du}$$ However, I've not studied enough analysis to know if this integral is equivalent to $\int_0^1{-\frac{1}{\sqrt{-\ln(u)}}du}$ where I just swap the bounds like that, and even if I try to integrate it without particularly caring for the limits I get stuck. WolframAlpha gives the answer to this $u$ integral in terms of the imaginary error function, and it gives the answer to my original integral in terms of the very gamma function I started with! I know that $\Gamma(\frac{1}{2})=\sqrt{\pi}$ but I wanna figure out how we derive it!
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1Hint: Use $x=\frac{1}{2}t^2$. Then you get something that looks like the normal distribution. For a proof of the integral of the normal distribution, see here – Reinhard Meier Dec 09 '22 at 12:49
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2$$\int_0^\infty e^{-x}x^{-1/2}d x=2\int_0^\infty e^{-t^2}d t=\sqrt{\pi}$$ – Craig Thone Dec 09 '22 at 14:04
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Yes, thank you @ReinhardMeier for the substitution I couldn't think of. – tai Dec 11 '22 at 20:22