What is the $d$-dimensional Fourier transform of $\cos|x|$? More specifically, what does $$ \int_{\mathbb{R}^d}\text{d}^dx\cos|x|e^{-i\vec{k}\cdot\vec{x}}\ , $$ where $|x|=\sqrt{x_1^2+\dots+x_d^2}$, evaluate to? I know that the Fourier transform of $e^{-t|x|}$ with positive $t>0$ gives me the Poisson kernel (c.f. this answer on MO) and I would like to know if there exists a generalization to imaginary $t$.
Addendum
For $d=1$ the above integral reduces to the Fourier transform of the ordinary cosine. It can be expressed via the delta distribution $$ \int_{-\infty}^{\infty}\text{d}x\cos x\,e^{-ikx}=\pi\big(\delta(k-1)+\delta(k+1)\big)\ , $$ as can be seen by applying the inverse transformation to the r.h.s. $$ \frac{1}{2\pi}\int_{-\infty}^{\infty}\text{d}k\,\pi\big(\delta(k-1)+\delta(k+1)\big)e^{ikx}=\frac12\big(e^{ix}+e^{-ix}\big)=\cos x\ . $$
For completeness I include here a dodgy calculation for the $d=3$ case. Help from someone who actually knows about distributions is needed.
We can use spherical coordinates to obtain \begin{align} \int_{\mathbb{R}^3}\text{d}^3x\cos|x|e^{-i\vec{k}\cdot\vec{x}}=&\,2\pi\int_{0}^{\infty}\text{d}r\int_{-1}^{1}\text{d}(\cos\theta)\,r^2\cos r\,e^{-ikr\cos\theta}\nonumber\\ =&\,2\pi\int_0^\infty\text{d}r\,r^2\cos r\,\frac{1}{ikr}\big(e^{ikr}-e^{-ikr}\big)\nonumber\\ =&\,\frac{4\pi}k\int_{0}^\infty\text{d}r\,r\cos r\sin kr\nonumber\\ =&\,\frac{4\pi}k\int_0^\infty\text{d}r\,-\frac{\partial}{\partial k}\frac12\Big[\cos\big(r(1+k)\big)+\cos\big(r(1-k)\big)\Big]\nonumber\\ =&\,-\frac\pi k\frac{\partial}{\partial k}\int_{-\infty}^\infty\text{d}r\,\frac12\Big[e^{ir(1+k)}+e^{-ir(1+k)}+e^{ir(1-k)}+e^{-ir(1-k)}\Big]\nonumber\\ =&\,-\frac{2\pi^2} k\frac{\partial}{\partial k}\big[\delta(1+k)+\delta(1-k)\big]\ . \end{align} Thus we finally obtain $$ \int_{\mathbb{R}^3}\text{d}^3x\cos|x|e^{-i\vec{k}\cdot\vec{x}}=\frac{2\pi^2}{k}\big(\delta'(1-k)-\delta'(1+k)\big)\ . $$
I found an answer to my question which I would like to share with those interested. For odd dimension $d$ we have $$ \int_{\mathbb{R}^d}\text{d}^dxe^{ifx}e^{-i\vec{k}\cdot\vec{x}}=(-2\pi)^{\frac{d-1}{2}}\left(\frac{1}{k}\frac{\partial}{\partial k}\right)^{\frac{d-1}{2}}\left[\pi\big(\delta(f+k)+\delta(f-k)\big)-\frac{2if}{f^2-k^2}\right]\ , $$ where $x=|\vec{x}|$ and $k=|\vec{k}|$. It might even be applicable to all dimensions if one defines the fractional derivative in some way.
Derivation
Starting with $$ I(\vec{k})=\int_{\mathbb{R}^d}\text{d}^dxe^{ifx}e^{-i\vec{k}\cdot\vec{x}}=(2\pi)^{\frac{d}{2}}\int_{0}^{\infty}\text{d}r\,\left(\frac{r}{k}\right)^{\frac{d-2}{2}}re^{ifr}J_{\frac{d-2}{2}}(kr)\ , $$ where $J_\nu$ denotes the $\nu$-th Bessel function of the first kind, we can use $$ J_{\nu+m}(z)=(-1)^{m}z^{\nu+m}\left(\frac1z\frac{\partial}{\partial z}\right)^{m}\big(z^{-\nu}J_\nu(z)\big) $$ with $\nu=-1/2$ and $m=(d-1)/2$ to write $$ I(\vec{k})=(-1)^{\frac{d-1}{2}}(2\pi)^{\frac{d}{2}}\sqrt{\frac{2}{\pi}}\frac{1}{k^{d-1}}\int_{0}^{\infty}\text{d}r\,r^{d-1}e^{ifr}\left(\frac1r\frac{\partial}{\partial r}\right)^{\frac{d-1}{2}}\cos(kr)\ . $$ Expanding $$ \left(\frac1r\frac{\partial}{\partial r}\right)^{n}=\sum_{j=1}^{n}a_j^{(n)}r^{-(2n-j)}\frac{\partial^j}{\partial r^j}\ , $$ where the coefficients $a_j^{(n)}$ satisfy the relation $$ a_j^{(n)}=a_{j-1}^{(n-1)}-\big[2(n-1)-j\big]a_j^{(n-1)}\ ,\quad a_1^{(1)}=1\ ,\ a_0^{(n)}=a_{n+1}^{(n)}=0\ , $$ we can calculate (using the shorthand $n=(d-1)/2$) \begin{align} \left(\frac{r}{k}\right)^{2n}\left(\frac1r\frac{\partial}{\partial r}\right)^{n}\cos(kr)=&\,\left(\frac{r}{k}\right)^{2n}\sum_{j=1}^{n}a_j^{(n)}r^{-(2n-j)}(ik)^{j}\frac12\big(e^{ikr}+(-1)^{j}e^{-ikr}\big)\nonumber\\ =&\,\sum_{j=1}^{n}a_j^{(n)}k^{-(2n-j)}\frac{\partial^{j}}{\partial k^{j}}\frac12\big(e^{ikr}+e^{-ikr}\big)\nonumber\\ =&\,\left(\frac1k\frac{\partial}{\partial k}\right)^{n}\cos(kr)\ . \end{align} (Is there a more obvious way to see this?) With that we obtain $$ I(\vec{k})=(-1)^{\frac{d-1}{2}}(2\pi)^{\frac{d-1}{2}}\left(\frac1k\frac{\partial}{\partial k}\right)^{\frac{d-1}{2}}\int_0^\infty\text{d}r\,e^{ifr}\big(e^{ikr}+e^{-ikr}\big)\ . $$ For the integral over $r$ we now use $$ \int_0^\infty\text{d}r\,e^{iar}=\pi\delta(a)+\frac{1}{ia} $$ to finally arive at $$ I(\vec{k})=(-2\pi)^{\frac{d-1}{2}}\left(\frac1k\frac{\partial}{\partial k}\right)^{\frac{d-1}{2}}\left[\pi\big(\delta(f+k)+\delta(f-k)\big)+\frac{1}{i(f+k)}+\frac{1}{i(f-k)}\right]\ , $$ which gives the desired result.
Connection to the Poisson kernel
One nice thing about the above result is that it can be connected to the Poisson kernel mentioned in my original question. To see this we let $f=i\beta$ for some positive $\beta>0$ such that $$ \int_{\mathbb{R}^{d}}\text{d}^dxe^{-\beta x}e^{-i\vec{k}\cdot\vec{x}}=(-2\pi)^{\frac{d-1}{2}}\left(\frac1k\frac{\partial}{\partial k}\right)^{\frac{d-1}{2}}\left[\pi\big(\delta(i\beta+k)+\delta(i\beta-k)-\frac{2\beta}{\beta^2+k^2}\right]\ . $$ Focusing on what used to be the imaginary part of the answer and using $$ \frac1k\frac{\partial}{\partial k}=2\frac{\partial}{\partial(k^2)} $$ we obtain $$ \int_{\mathbb{R}^{d}}\text{d}^dx\,e^{-\beta x}e^{-i\vec{k}\cdot\vec{x}}=-\frac{(2\pi)^d}{\pi^{\frac{d+1}{2}}}\Gamma\left(\frac{d+1}{2}\right)\frac{\beta}{(\beta^2+k^2)^{\frac{d+1}{2}}}\ , $$ which is precisely the Poisson kernel (up to a negative sign which I would be very happy if someone can point to where it comes from).