Improper integral and volume integral

Question

I want to calculate the integral:

$$\displaystyle\iiint\limits_{\mathbb{R}^3}{\frac{5\cdot e^ {- 4\cdot \left(z^2+y^2+x^2\right) }}{\sqrt{z^2+y^2+x^2}}}\,\mathrm{d}V$$

I know a spherical coordinate system $(x,y,z)\mapsto(\rho\cos\theta\sin\phi,\rho\sin\theta\sin\phi,\rho\cos\phi)$ where $\rho^2 = x^2+y^2+z^2$ and the special case $\rho\to\infty$

and

$$\iiint\limits_{\mathbb{R}^3}{\frac{5\cdot e^ {-4\rho^2}}{\sqrt{\rho^2}}}\,\mathrm{d}V= \iiint\limits_{\mathbb{R}^3}{\frac{5\cdot e^ {-4\rho^2}}{\rho^2}}\,\mathrm{d}V$$

$\theta\to[0,\pi]$

$\phi\to[0,2\pi]$

But what are the limit values of the integral?

Answer 1

Since the surface area of $x^2+y^2+z^2=\rho^2$ is $4\pi\rho^2$ your integral equals

$$ 5\int_{0}^{+\infty}4\pi\rho^2 \frac{e^{-4\rho^2}}{\rho}\,d\rho = \frac{5\pi}{2}\left[-e^{-4\rho^2}\right]_{0}^{+\infty}=\frac{5\pi}{2}.$$

Answer 2

By spherical coordinates we have that $dV=\rho^2 \sin\theta\, d\rho \,d\theta\, d\phi$ and therefore

$$\iiint\limits_{\mathbb{R}^3}{\frac{5\cdot e^ {-4\rho^2}}{\sqrt{\rho^2}}}\,\mathrm{d}V= \int_0^\pi d\theta\int_0^{2\pi}d\phi\int_0^\infty{\frac{5\cdot e^ {-4\rho^2}}{\rho}}\rho^2 \sin\theta\,d\rho$$