Calculus 7 : return of $\pi$

$\boxed{1.-}$ $J = \int_{-\infty}^{\infty} e^{- x^2} \, dx=\int_{-\infty}^0 e^{- x^2} \, dx + \int_0^\infty e^{- x^2} \, dx =$ Due to $f(x) = e^{-x^2}$ is and even function $= 2\int_0^\infty e^{- x^2} \, dx = 2I$ Now, we are going to make the change $x = r \cos\theta, y = r\sin\theta \Rightarrow |\det J(x,y))| = r$ where $J(x,y)$ is the Jacobian matrix... $I = \int_0^\infty e^{- x^2} \, dx \Rightarrow I^2 = \int_0^\infty \int_0^\infty e^{-x^2} e^{-y^2} \, dy dx = \int_0^{\pi/2} \int_0^\infty r e^{-r^2} \, dr d\theta = \int_0^{\pi/2} \frac{1}{2} \,d\theta = \frac{\pi}{4} \Rightarrow$ $I = \frac{\sqrt{\pi}}{2} \Rightarrow J = \sqrt{\pi} = \pi^{1/2} \Rightarrow a + b = 1 + 2 = 3$ $\boxed{2.-}$ $\Gamma (s)=\int_0^{\infty} t^{s-1} e^{-t} dt,$ . Make on $I = \int_0^\infty e^{- x^2} \, dx$ the change $x^2 = t \rightarrow dt = 2x \space dx \Rightarrow$ $I = \int_0^\infty e^{-t} \frac{1}{2\sqrt{t}} \, dt = \frac{1}{2} \Gamma[1/2] = \frac{\sqrt{\pi}}{2}$ ...

$\boxed{3.-}$

Because of 1 and 2, $\Gamma[1/2] = \sqrt{\pi} = 2I$ . The betta function $B[x,y] = \int_0^1 t^{x-1}(1-t)^{y-1} \, dt$ fulfills $B[1/2, 1/2] = \frac{\Gamma[1/2] \cdot \Gamma[1/2]}{\Gamma[1]} = 4I^{2} = \Gamma[1/2]^{2} = \pi \Rightarrow I = \frac{\sqrt{\pi}}{2}$ $B[1/2, 1/2] = \int_0^1 \frac{1}{\sqrt{t} \sqrt{1 - t}} \, dt =$ Making the change $t = \sin^2 (x) \rightarrow dt = 2\sin (x) \cos(x) \space dx$ $B[1/2, 1/2] = \int_0^{\pi/2} 2 dx = \pi...$