Интеграл Эйлера-Пуассона

$$ \int_{-\infty}^{+\infty} e^{-x^2} dx = \sqrt{\pi} $$

$$ \Gamma\left(\frac{1}{2}\right) = \int_{0}^{\infty} t^{-\frac{1}{2}} e^{-t} dt = 2 \int_{0}^{\infty} e^{-t} d\left(t^{\frac{1}{2}}\right) = 2 \int_{0}^{\infty} e^{-x^2} dx $$ $$ B\left(\frac{1}{2}, \frac{1}{2}\right) = \frac{\Gamma\left(\frac{1}{2}\right) \Gamma\left(\frac{1}{2}\right)}{\Gamma(1)} \Rightarrow \Gamma\left(\frac{1}{2}\right) = \sqrt{B\left(\frac{1}{2}, \frac{1}{2}\right)} $$ $$ B\left(\frac{1}{2}, \frac{1}{2}\right) = \int_{0}^{1} t^{-\frac{1}{2}} (1-t)^{-\frac{1}{2}} dt = [t = \sin^2 x] = \int_{0}^{\frac{\pi}{2}} \frac{2 \sin x \cos x}{\sin x \cos x} dx = \pi $$