Calculate $ \iiint_{D} e^{-(x^2+y^2+z^2)^p} d x d y d z $

Question

Calculate $$ I = \iiint_{D} e^{-(x^2+y^2+z^2)^p} d x d y d z, D = \{x,y,z: x>0, y>0, z>0 \}. $$ I used spherical coordinates and get an answer $\frac{\pi}{4p} \Gamma(\frac32 p) $. My solution: $$ I = \int\limits_0^{\pi/2} \int\limits_0^{\pi/2}\int\limits_0^{+\infty} d r d \varphi d \theta e^{-r^{2p}} r^2 \sin \theta = \frac{\pi}{2} \int\limits_0^{\infty} e^{-r^{2p}} r^2 d r $$ Then I make change of variables $$ t = r^{2p}, r = t^{1/2p}, d r = \frac{t^{\frac{1-2p}{2p}}}{2p} d t $$ So we have $$ I = \frac {\pi}{2} \int_0^{\infty} \frac{e^{-t} t^{1/p} t^{\frac{1-2p}{2p}}}{2p} dt = \frac{\pi}{4p} \int_0^{\infty} e^{-t} t^{\frac32 p - 1} d t = \frac{\pi}{4p} \Gamma(\frac32 p). $$ Is that correct? Seems a little bit dubious.

Answer 1

Result using Wolfram for $Re(p)>0$: $$\int^{+\infty}_{0}\int^{+\infty}_{0}\int^{+\infty}_{0}e^{-(x^{2}+y^{2}+z^{2})^{p}}dxdydz=\dfrac{1}{6}\pi \Gamma \left( 1+\dfrac{3}{2p}\right)$$

The $p$ is in the denominator. $$\dfrac{\pi}{4p}\int^{+\infty}_{0}e^{-t}t^{-1+\frac{3}{2p}}dt=\dfrac{\pi}{4p}\Gamma\left( \dfrac{3}{2p}\right)=$$Using $\Gamma(x)=\dfrac{1}{x}\Gamma (x+1)$ $$=\dfrac{\pi}{4p}\cdot \dfrac{2p}{3}\Gamma\left( 1+\dfrac{3}{2p}\right)=\dfrac{\pi}{6}\Gamma \left( 1+\dfrac{3}{2p}\right)$$