5.4 Improper integrals

(a) $\displaystyle{\frac{\pi}{2}}$ (b) 1(c) $\displaystyle{\frac{\pi^2}{16}}$ (d) 1(e) $\pi$ (f) $\displaystyle{\frac{\pi^2}{2}}$

Let $\displaystyle{t = x^{\frac{1}{2}}}$ . Then $\displaystyle{2tdt = dx}$ ．Thus，

$\displaystyle \Gamma(\frac{1}{2}) = \int_{0}^{\infty} x^{-\frac{1}{2}}e^{-x} dx = 2\int_{0}^{\infty} e^{-t^2} dt$

Let $\displaystyle{I = \int_{0}^{\infty} e^{-x^2} dx, I = \int_{0}^{\infty} e^{-y^2} dt}$ . Then

$\displaystyle I^2 = \iint_{0}^{\infty} e^{-(x^2 + y^2)} dx dy$

Note that by 5.5， $\displaystyle{I^2 = \frac{\pi}{4}}$ より $\displaystyle{\Gamma(\frac{1}{2}) = \sqrt{\pi}}$