Poisson Kernel on the generalized upper half plane We have the function $P(x)=C_n\frac{1}{(1+||x||^2)^{(n+1)/2}}$ on $\mathbb{R}^n$, where $C_n=\pi^{-(n+1)}\Gamma\big(\frac{n+1}{2}\big)$. And we define the Poisson kernel on the generalized upper half plane $\{(x,y)\big|x\in\mathbb{R}^n,y>0\}.$ We also know that $P_y(x)$ is the Fourier transform of the function $e^{-2\pi  y||x||}$ for $y>0$. Now want to show that :

If $f\in L^p(\mathbb{R}^n),1\leq p<\infty,$ then $f*P_y\rightarrow f$ in $L^p(\mathbb{R}^n)$ as $y\rightarrow 0$.

My question is how to show that the Poisson Kernel is in $L^1$? And can we do it without integrating it explicitly? I am not so confident about integrating on whole of $\mathbb{R}^n$. May be one can use that it's a Fourier Transform of some function? Also, does the function $e^{-2\pi  y||x||}$ lie in any $L^p?$
 A: ΗΙΝΤ
$P(x),e^{-2\pi y||x||} \in L^p,\forall p \in [1,+\infty]$ since $y>0$.
You can see that by using the polar coordinates formula: $$
\int_{\Bbb{R}^n} f(x) dx = \int_0^\infty \left( \int_{S^{n-1}}  f(re)d\sigma(e) \right) r^{n-1} dr.
$$
where $\sigma(E)$ is the surface borel measure on the unit sphere.
In this case your function is radial so the integration won't depend on $e \in S^{n-1}$ thus things are easy especialy since your function is non-negative.
A: By using the equivalent* of spherical coordinates $r=|x|$ in $n$ dimensions, one has by the radial symmetry of $P$, 
$$\int_{\mathbb R^n} P(x) dx = C'_n \int_0^\infty\frac{1}{(1+r^2)^{(n+1)/2}}r^{n-1}dr$$
the constant $C'_n$ includes the surface area of the unit $n$-sphere. Now split into two integrals, one for $r<1$ and one for $r>1$. For $r<1$,
$$ \int_0^1 \frac{1}{(1+r^2)^{(n+1)/2}}r^{n-1}dr \le \int_0^1 \frac1{(1+0)^{(n+1)/2}} dr = 1.$$
For $r>1$, we have
\begin{align}
\int_1^\infty  \frac{1}{(1+r^2)^{(n+1)/2}}r^{n-1}dr 
&\le \int_1^\infty \frac{r^{n-1} dr}{(0 + r^2)^{(n+1)/2}} \\
&=  \int_1^\infty \frac{r^{n-1}}{r^{n+1}} dr \\
&= \int_1^\infty \frac{dr}{r^2} = 1\end{align}
In summary, $\int_0^\infty\frac{1}{(1+r^2)^{(n+1)/2}}r^{n-1}dr\le 2<\infty$, and therefore 
$$\|P\|_{L^1(\mathbb R^n)} = \int_{\mathbb R^n} |P(x)|dx =  \int_{\mathbb R^n} P(x)dx \le 2C'_n  < \infty. $$

*: The general formula is
$$ \int_{B(0,r)}f(x) dx = \int_0^r \int_{\partial B(0,1)} f(r\omega) \ d\sigma(\omega)\ r^{n-1} dr.$$ $\Omega$ represents an element of the unit sphere, and the $d\sigma$ is the surface measure on the unit sphere. If the function is radial, then $f(r\omega) = \tilde f(r)$, so $\int_{\partial B(0,1)} d\sigma$ is a constant, leaving you with a 1D integral. 
You should compare with the 2D polar coordinates you know, and also the 3D one. 
A quick google led me to this PDF. If you want to search more, here are some Google/wikipedia keywords: polar coordinates, higher dimensions, spherical, co-area formula for sphere...
These coordinates are extremely useful for radial functions and worth learning. Some other examples:
For what values of $ p \in (0, \infty] $ do we have $ f \in L^p (\mathbb{R}^3) $?
Can I still use polar coordinates when the function is defined on $\mathbb{R}^n$?
