Integral in $n-$dimensional euclidean space I want to calculate this integral in $n$-dimensional euclidean space.
$$I(x)=\int_{\mathbb{R}^n}\frac{d^n k}{(2\pi)^n}\frac{e^{i(k\cdot x)}}{k^2+a^2},$$
where $k^2=(k\cdot k)$, $k=(k_1,\ldots,k_n)\in\mathbb{R}^n$, $x=(x_1,\ldots,x_n)\in\mathbb{R}^n$,$a\in \mathbb{R}$.
I've done this integral for $n=3$ by spherical coordinates and residue theorem. I have 
$$I(r)=\frac{1}{4\pi r}e^{-ar},$$
where $r=|x|$
But in $n$- dimensions I failed in using spherical coordinates, because I have never done it before. Also I see that this integral is Fourier transform of $\frac{1}{k^2+a^2}$, but I failed here too, because I can't find Fourier pair in my reference books.
If someone could guide me in this integration it would be great.
 A: One possibility is the following:


*

*Replace 
$$\frac{1}{k^2+a^2}=\int_0^{\infty}e^{-s(k^2+a^2)}ds.$$

*The integral over $n$ components of $k$ now factorizes into the product of $n$ Gaussian integrals
$$\prod_{\alpha=1}^n\int_{-\infty}^{\infty}e^{-sk_{\alpha}^2+ik_{\alpha}x_{\alpha}}dk_{\alpha}=\prod_{\alpha=1}^n\sqrt{\frac{\pi}{s}}\,e^{-x_{\alpha}^2/4s}=\left(\frac{\pi}{s}\right)^{\frac{n}{2}}e^{-r^2/4s}.$$

*Therefore the $n$-fold integral $I(x)$ can be rewritten as an ordinary integral
$$I(x)=\int_{0}^{\infty}\left(4\pi s\right)^{-n/2}e^{-a^2 s-r^2/4s}ds,$$
where $r^2=x\cdot x$. Making the change of variables $t=\frac{2a s}{r}$, this can be transformed into 
$$I(x)=\frac{1}{4\pi}\left(\frac{2\pi r}{a}\right)^{1-\frac{n}{2}}\int_0^{\infty}s^{-\frac{n}{2}}e^{-\frac{ar}{2}\left(s+s^{-1}\right)}ds.$$

*Using integral representation of the Macdonald function, the last integral reduces to
$$\boxed{I(x)=\frac{1}{2\pi}\left(\frac{2\pi r}{a}\right)^{1-\frac{n}{2}}K_{\frac{n}{2}-1}\left(ar\right)}$$
Remark: For odd dimension $n=2m+1$, Macdonald function $K_{m-\frac12}(ar)$ can be expressed in terms of elementary functions. In particular, $K_{\frac12}(z)=\sqrt{\frac{\pi}{2z}}\, e^{-z}$, which reproduces the  formula for $n=3$ mentioned above.
