What is the integral of this expression in the sense of distribution? Consider this integral:
$$
\int_{\mathbb{R}^n} e^{-i2\pi(\xi\cdot x + |\xi|)} d\xi 
$$
What is this integral in the sense of distribution ? I guess it is some form of Dirac but I failed several times to solve it.
 A: A good way to write the object you are looking at is as the Fourier transform of a tempered distribution, i.e.
$$
f_i(x) = {\cal F}_\xi(e^{-2i\pi|\xi|})(x)
$$
Defining more generally $f_z(x) = \cal F_\xi(e^{-2\pi\,z\,|\xi|})(x)$, as proved for example here one has for any $z > 0$
$$\label{eq}\tag{1}
f_z(x) = \frac{2 \, z}{\omega_{d+1}\,( z^2 + |x|^2)^{(d+1)/2}} =: g_z(x)
$$
Now for every Schwartz function $\varphi$, $\langle f_z,\varphi\rangle = ∫ e^{-2π\,z\,|\xi|}\,\widehat{\varphi}(\xi)\,\mathrm d \xi$ and $\langle g_z,\varphi\rangle = ∫ g_z(x)\,\varphi(x)\,\mathrm d x$ are analytic functions of $z$ on the domain $\{z\in\Bbb C, \mathrm{Re}(z)> 0\}$. Therefore by the identity theorem for analytic function, the formula also holds with $z=i+\varepsilon$ and so
$$
{\cal F}_\xi(e^{-2(i+\varepsilon)\pi|\xi|})(x) = \frac{2 \, (i+\varepsilon)}{\omega_{d+1}\,(|x|^2 + \varepsilon^2 - 1 + 2\,i\,\varepsilon)^{(d+1)/2}}
$$
In particular, when $|x|≠ 1$ (or more precisely, restricting the space of tests functions to the functions $\varphi$ such that $\varphi(x)=0$ when $|x|=1$) it holds
$$
{\cal F}_\xi(e^{-2\,i\,\pi|\xi|})(x) = \frac{2 \, i}{\omega_{d+1}\,(|x|^2 - 1)^{(d+1)/2}}.
$$
