One Step in Proving the Gamma of $1 \over 2$

Question

Good Afternoon All. There is one step in the proof that I never quite understood.

Let $I = \int_{0}^{\infty} e^{{-u}^2} du$. Then $$I^2 = \int_{0}^{\infty} e^{{-u}^2} du \int_{0}^{\infty} e^{{-v}^2} dv$$ Now, if I say that since $\int_{0}^{\infty} e^{{-u}^2} du$ is not a function of $v$, I can stack the integral as such $$I^2 = \int_{0}^{\infty} \int_{0}^{\infty} e^{{-u}^2} du (e^{{-v}^2}) dv \space (*)$$ Then because $e^{{-v}^2}$ is not a function of $u$, $$I^2 = \int_{0}^{\infty} \int_{0}^{\infty} e^{{-v}^2}e^{{-u}^2} du dv$$ However, to make such a argument in $(*)$, do I need to first know that $\int_{0}^{\infty} e^{{-u}^2} du$ converges?

Answer 1

Yes, you do need to know it converges. But $\exp (-u^2) \lt \frac 1{1+u^2}$ for large $u$ is enough to show it.

Answer 2

Another approach: if you don't need to stick to the rather nice double integral thing, you could use Euler's Reflection Formula :

$$\Gamma(1-z)\Gamma(z)=\frac\pi{\sin\pi z}\;,\;\;\text{ so}\;\;z=\frac12\implies \Gamma\left(\frac12\right)^2=\frac\pi{\sin\frac\pi2}=\pi\;\ldots\ldots$$