Calculating $\int_{\mathbb{R}^5}\frac{e^{-x^2-y^2-z^2}}{1+w^2+s^2}\,dx\,dy\,dz\,dw\,ds$

95 Views Asked by At

Calculate the following:
$$\int_{\mathbb{R}^5}\frac{e^{-x^2-y^2-z^2}}{1+w^2+s^2}\,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z\,\mathrm{d}w\,\mathrm{d}s$$

I tried to do the following (based on the suggestion below):

\begin{align} \int_{\mathbb{R}^5}\frac{e^{-x^2-y^2-z^2}}{1+w^2+s^2}\,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z\,\mathrm{d}w\,\mathrm{d}s &= \int_{\mathbb{R}^3}{e^{-x^2-y^2-z^2}}\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z\int_{\mathbb{R}^2}\frac{1}{1+w^2+s^2}\,\mathrm{d}w\,\mathrm{d}s \\&=\pi^{3/2}\int_{0}^{2\pi}\int_{0}^{\infty}\frac{1}{1+r^2}r\,\mathrm{d}r\,\mathrm{d}\theta \\&=\pi^{3/2}\cdot2\pi\left(\frac{1}{2}\ln(1+r^2)\right)\bigg|_{0}^{\infty} \\&=\infty \end{align}

but I'm not sure about it, would appreciate your help:)

1

There are 1 best solutions below

0
On BEST ANSWER

I would choose the product of a sphere in $x,y,z$ with a circle in $w,s$. So a 5-D version of a cylinder.
Then break up into $\iiint dxdydz\iint dwds$