The issue of treating an inverse Fourier transform in terms of a tempered distribution. Consider the wave equation
$$
u_{tt}=\Delta{u} \quad u(x,0)=f(x) \quad u_t(x,0)=g(x) \tag{*}
$$
A solution to this equation is given by
$$
u(.,t)=f*\partial_t\Phi_t+g*\Phi_t \tag{**}
$$
where $\Phi_t$ is the inverse Fourier transform of the function $\Psi_t(\xi)=\frac{\text{sin}2\pi |\xi|t}{2\pi |\xi|}$. Now for $n=3$, how do we show that $\Phi_t$ as a tempered distribution is given by 
$$
\displaystyle  \langle \Phi_t, h \rangle := \frac{t}{4\pi} \int_{S^2} h(t\omega)\ d\omega \tag{***}
$$
That is
$$
\displaystyle  \Phi_t (h) := \frac{t}{4\pi} \int_{S^2} h(t\omega)\ d\omega
$$
for all Schwartz function $h$?
Note that, I am using the following definition of Fourier transform:
$$
F[f]=\hat{f}(\xi)=\!\int\limits_{\mathbb{R}^3}\!f(x)e^{-2\pi ix\cdot\xi}dx
$$
and the Fourier transform of a tempered distribution is defined as:
$$
\bigl\langle F[\lambda],\varphi\bigr\rangle
=\bigl\langle \lambda,\check{\varphi}\bigr\rangle                           
\quad\forall\,\varphi\in \mathcal{S}
$$
OR:
How to verify that $(**)$ is a solution to $(*)$ where $\Phi_{t}$ is a tempered distribution given by $(***)$?
 A: Representation $\Psi_t(\xi)=\frac{\sin{(2\pi |\xi|t)}}{2\pi |\xi|}$ corresponds to Fourier transform written in the form 
$$
F[f]=\hat{f}(\xi)=\!\int\limits_{\mathbb{R}^3}\!f(x)e^{-2\pi ix\cdot\xi}dx
\quad\Rightarrow\quad F[\partial_x^{\alpha}f]=(2\pi i\xi)^{\alpha}\hat{f}(\xi)\tag{1}
$$
generally preferred in Harmonic Analysis.  In Partial Differential Equations, another form of Fourier transform is generally preferred
$$
F[f]=\hat{f}(\xi)=\!\int\limits_{\mathbb{R}^3}\!f(x)e^{-ix\cdot\xi}dx
\quad\Rightarrow\quad F[\partial_x^{\alpha}f]=(i\xi)^{\alpha}\hat{f}(\xi)\tag{2}
$$
which looks more habitual in the section tagged pde, with the inverse transform
written in the form 
$$
F^{-1}[g]=\check{g}(x)=\frac{1}{(2\pi)^3}\!\int\limits_{\mathbb{R}^3}\!g(\xi)e^{ix\cdot\xi}d\xi.
$$
In version $(1)$,  a pde $\,u_{tt}=\Delta u\,$ transforms to $\,\hat{u}_{tt}=
-4\pi^2|\xi|^2\hat{u}\,$, while it transforms to $\,\hat{u}_{tt}=-|\xi|^2\hat{u}\,$ in version $(2)$.   In fact, this extra $2\pi$-factor looks just like an absolutely useless stumbling block in Partial Differential Equations, where exactly for this reason version $(1)$ is overwhelmingly disliked.  So, in version $(1)$, solution of the Cauchy problem really looks like
$$
\hat{u}(\xi,t)=\hat{f}(\xi)\cos{(2\pi|\xi|t)}+
\hat{g}(\xi)\frac{\sin{(2\pi|\xi|t)}}{2\pi|\xi|},
$$
while it must look like
$$
\hat{u}(\xi,t)=\hat{f}(\xi)\cos{(|\xi|t)}+
\hat{g}(\xi)\frac{\sin{(|\xi|t)}}{|\xi|}
$$
in version $(2)$.  Namely thereby, in version $(2)$, it must be  $\Psi_t(\xi)=\frac{\sin{(|\xi|t)}}{|\xi|}$.
Denote by $\mathcal{S}'$ the Schwartz space of tempered distributions dual
to the Schwartz space of rapidly decreasing functions $\mathcal{S}=\mathcal{S}
(\mathbb{R}^3)$. Denote by $\delta_R\in \mathcal{S}'$ a delta function supported on a sphere $S_R=\{x\in\mathbb{R}^3\colon |x|=R\}$ of radius $R>0$, otherwise called a spherical delta, defined by the identity
$$
\langle \delta_R,\varphi\rangle=\int\limits_{S_R}\varphi(x)\,ds_x\quad
\forall\,\varphi\in \mathcal{S}. 
$$ 
To find Fourier transform of $\delta_{S_R}$ in $\mathcal{S}'$, recall that
a Fourier transform $\hat{f}\in\mathcal{S}'$ of some distribution $f\in\mathcal{S}'$
is traditionally defined as the linear functional $\hat{f}$ subject to the rule
$$
\langle\hat{f},\varphi\rangle=\langle f,\hat{\varphi}\rangle\quad\forall\,
\varphi\in\mathcal{S}.\tag{3}
$$
Note that there can be no definition of the distribution Fourier transform other than $(3)$ according to the fundamental principle underlying the Standard Theory of Distributions. This principle reads something like: A distribution definition must not contradict to its classical prototype. Say, for a function $f\in L^1$, with its classical Fourier transform $\hat{f}$ defined by $(2)$, the distribution Fourier transform $\hat{f}$ is to be defined exactly by the identity $(3)$, since $f$ is to be embedded into $\mathcal{S}'$ as a regular distribution, i.e., as the linear functional defined by the identity
$$
\langle f,\varphi\rangle=\int\limits_{\mathbb{R}^3}f(x)\varphi(x)\,dx\quad
\forall\,\varphi\in\mathcal{S},\tag{4}
$$
whence readily follows 
$$
\langle\hat{f},\varphi\rangle=
\!\int\limits_{\mathbb{R}^3}\!\hat{f}(\xi)\varphi(\xi)\,d\xi=
\!\int\limits_{\mathbb{R}^3}\!\varphi(\xi)\,d\xi
\!\int\limits_{\mathbb{R}^3}\!f(x)e^{-i x\cdot\xi}dx=
\!\int\limits_{\mathbb{R}^3}\!f(x)\,dx
\!\int\limits_{\mathbb{R}^3}\!\varphi(\xi) e^{-i x\cdot\xi}\,d\xi=
\langle{f},\hat{\varphi}\rangle
$$
which implies definition $(3)$.  Note that there is no alternative to $(4)$ for embedding $\,f\in L^1\,$ into $\,\mathcal{S}'\,$ as a linear functional. Say, a sesquilinear form
$$
\langle f,\varphi\rangle=\int\limits_{\mathbb{R}^3}f(x)\overline{\varphi(x)}\,dx\quad
\forall\,\varphi\in\mathcal{S},\tag{5}
$$
with a complex conjugate $\,\overline{\varphi}$, is absolutely unacceptable as an alternative to the bilinear form $(4)$, since $(5)$ defines a functional antilinear in $\varphi$, and hence living outside $\mathcal{S}'$ traditionally defined as a space of linear functionals on $\mathcal{S}$. 
So, the only definition we can have is
$$
\bigl\langle F[\delta_{S_R}],\varphi\bigr\rangle
=\bigl\langle \delta_{S_R},\hat{\varphi}\bigr\rangle                           
=\int\limits_{S_R}\hat{\varphi}(x)\,ds_x 
=\int\limits_{\mathbb{R}^3}\Bigl(\int\limits_{S_R}  
e^{-ix\cdot \xi}ds_x\Bigr) \varphi (\xi)\,d\xi\quad
\forall\,\varphi\in \mathcal{S}
$$
where the order of integration has been interchanged. Hence we find
$$
 F[\delta_{S_R}]=\int\limits_{S_R}  
e^{-ix\cdot\xi}ds_x=\int\limits_{S_R} e^{-ix_3|\xi|}ds_x
$$
with the RHS integral invariant with respect to rotations of the sphere $S_R$.
To establish the rather obvious identity
$$
\int\limits_{S_R}  
e^{-ix\cdot\xi}ds_x=\int\limits_{S_R} e^{-ix_3|\xi|}ds_x
\quad\forall\, x\in\mathbb{R}^3,\tag{6}
$$
take a rotation $\,T\,$ of the sphere $\,S_R\,$ in $\,\mathbb{R}^3$.  
We have $\,T^{-1}=T^{\ast}\,$ and $\,TT^{\ast}=T^{\ast}T=I\,$ 
while $\,TS_R=S_R\,$ and $\,|{\rm det\,}T|=1$. Hence, for the inner 
product in $\mathbb{R}^3$, we have 
$$
(Tz)\cdot\xi=z\cdot T^{\ast}\xi\;\;
\forall\,z\in S_R\,.
$$
 Now fix arbitarary $\,\xi\in\mathbb{R}^3\,$ and choose 
the rotation $\,T=T_{\xi}\,$ such that $\,T_{\xi}^{\ast}\xi=(0,0,|\xi|)$. 
Change of variables $\,x=T_{\xi}z\,$ yields 
$$
\int\limits_{S_R}e^{-ix\cdot\xi}ds_x=
\int\limits_{S_R}e^{-i(T_{\xi}z)\cdot\xi}ds_z=
\int\limits_{S_R}e^{-iz\cdot (T_{\xi}^{\ast}\xi)}ds_z=
\int\limits_{S_R}e^{-iz_3|\xi|}ds_z\,,
$$
which implies $(6)$.  Finally, changing to spherical coordinates, we get
$$
\begin{align*}
\int\limits_{S_R} e^{-ix_3|\xi|}ds_x=
\int\limits_{0}^{2\pi}\int\limits_0^\pi e^{-i |\xi| R \cos \theta} R^2
\sin\theta\, d\theta d\varphi =2\pi  R^2  \int\limits_0^\pi  e^{-i|\xi|R  \cos  \theta}\sin\theta\, d\theta\\
=4\pi  R^2  \int\limits_0^{\pi/2}  \cos  (|\xi|R\cos  \theta)\sin
\theta\, d\theta = 4\pi R\frac{\sin (|\xi|R)}{|\xi|}.
\end{align*}
$$
Thus we have found the Fourier transform of $\delta_{S_R}$ in $\mathcal{S}'$, namely,
$$
F[\delta_{S_R}]=4\pi R\frac{\sin (|\xi|R)}{|\xi|}\quad \forall\,R>0,
$$ 
whence follows
$$
\Phi_t\overset{\rm def}{=}F^{-1}\Bigl[\frac{\sin (|\xi|t)}{|\xi|}\Bigr]=
\frac{1}{4\pi t}\delta_{S_t}\quad \forall\, t>0
$$
which implies that
$$
\langle \Phi_t,h\rangle=\frac{1}{4\pi t}\langle\delta_{S_t},h\rangle=
\frac{1}{4\pi t}\!\int\limits_{S_t}\,h(x)\,ds_x=
\frac{t}{4\pi}\!\int\limits_{S_1}\,h(ty)\,ds_y=
\frac{t}{4\pi} \int\limits_{S^2} h(t\omega)\, d\omega\quad
\forall\,h\in \mathcal{S}
$$
where $S^2\overset{\rm def}{=}S_1\,$.   Q.E.D.
