I have a doubt about this passage:
$\ I_{\alpha} =\lim_{\lambda \to 0^+} \int_{\mathbb{R}^n/B_{1}} \frac{e^{-\lambda|x|}}{|x|^\alpha} dx $
$\ I_{\alpha} = \lim_{\lambda \to 0^+} \omega_n \int_{1}^{+\infty} \frac{e^{-\lambda\rho}}{\rho^\alpha} \rho^{n-1} d\rho $
I realize it concerns polar coordinates substitution but I don't understand how this change is done and where $\ \omega_n $ comes from. Can someone explain in details? Because it's not the first exercise that I see in which this has been done, thanks.
Let's start by writing the integrals for $n=1,2,3$. I will skip writing the limit, since it's not relevant to the problem. For $n=1$, $\mathbb R^n/B_1=(-\infty,-1)\cup(1,\infty)$. Then $$I_{\alpha,1}=\int_{-\infty}^{-1}\frac{e^{-\lambda|x|}}{|x|^\alpha}dx+\int_1^{\infty}\frac{e^{-\lambda|x|}}{|x|^\alpha}dx=2\int_1^{\infty}\frac{e^{-\lambda|x|}}{|x|^\alpha}dx$$We can now rename $|x|=r$, so that means that $\omega_1=2$.
Now go for $n=2$. We are going to use the components of $\textbf x$ as $x=r\cos\theta$ and $y=r\sin\theta$. Then $|\textbf x|=r$ and $d\textbf x=dx\ dy=r\ dr\ d\theta$. The integral limits in polar coordinates are $1$ to $\infty$ for $r$, and $0$ to $2\pi$ for $\theta$. You can immediately see that the integrand does not depend on $\theta$, so $\int_0^{2\pi}d\theta=2\pi$. Plugging it back into the integral, we get $$I_{\alpha,2}=2\pi\int_1^{\infty}\frac{e^{-\lambda r}}{r^\alpha}rdr=2\pi\int_1^{\infty}\frac{e^{-\lambda r}}{r^\alpha}r^{2-1}dr$$ From here $\omega_2=2\pi$.
We apply the same procedure for $n=3$. $\textbf x=(r\sin\theta\cos\phi,r\sin\theta\sin\phi,r\cos\theta)$. $d\textbf x=r^2\sin\theta dr d\theta d\phi$. The limits of integration are $1$ to $\infty$, $0$ to $\pi$, and $0$ to $2\pi$. Once again, the integrand does not depend on the angular part, so $$I_{\alpha,3}=4\pi\int_1^{\infty}\frac{e^{-\lambda r}}{r^\alpha}r^{3-1}dr$$ Therefore $\omega_3=4\pi$.
You can continue for higher dimension. The value of $\omega_n$ is the "surface area" of the $B_1$ in $\mathbb R^n$. You can calculate the hyper-surface area of the hypersphere.