I'm stuck with the following problem
Let $\varphi_m \in \mathcal{S}'(\mathbb{R}^{n})$, $n \in \mathbb{N}$, $m\in \mathbb{C}$, $0 >\text{Re}(m)>-n$ the distribution defined by $$\varphi_m(u)=\int_{\mathbb{R}^m}|x|^mu(x)dx$$ Find the fourier transform $\mathcal F(\varphi_m) \in \mathcal{S}'(\mathbb{R}^{n})$.
I would like to use Fubbini's theorem to get $\mathcal{F}({|x|^m})\int u(x)dx$ (not completely sure), but I need to check if that is possible in that case. so I tried to use polar coordinates to get
$$\mathcal{F}\varphi_m(u)=\int_{\mathbb{R}^m} |x|^m\mathcal{F}u(x)dx=\int_{\mathbb{S}^{n-1}}\int_{0}^\infty r^{m+n-1}\mathcal{F}u(r\hat{x})drdS$$
then if we separate the integral
$$\mathcal{F}\varphi_m(u)=\int_{\mathbb{S}^{n-1}}\int_{1}^\infty r^{m+n-1}\mathcal{F}u(r\hat{x})drdS+\int_{\mathbb{S}^{n-1}}\int_{0}^{1} r^{m+n-1}\mathcal{F}u(r\hat{x})drd$$
Where $\hat{x}:=x/|x|\in \mathbb{R}^n$, note that $n>\text{Re}(m)+n>0 $ and if $r^{-1}<1$ the first member of RHS equation is bounded, because if $u \in \mathcal{S}(\mathbb{R}^n)$ then $|r\hat{x}|^n|\mathcal{F}u(r\hat{x})|$ is bounded, but I don't know how control the other member of the RHS of equation.
Let me change your notation a little bit. In what follows I'm going to use $-m$ as $m$ in your statement. First of all I have to say that the solution of your question is quite long, so I'm just going to give you the hints for the case $m\in\mathbb{R}$ satisfying your hypothesis. First of all prove that in $\mathcal{S}'(\mathbb{R}^n)$ the following holds: $$ \dfrac{\pi^{m/2}}{\Gamma(\tfrac{m}{2})}\int_\varepsilon^R e^{-\pi t\vert x\vert^2}t^{m/2}\tfrac{dt}{t} \ \xrightarrow{\varepsilon\to 0, \ R\to\infty} \ \vert x\vert^{-m}. $$ Of course this is not direct. However, to prove the previous limit it is enought to notice that, on the one hand you have: $$ \left\vert\int_0^\varepsilon e^{-\pi t \vert x\vert^2}t^{m/2}\tfrac{dt}{t}\right\vert\leq C_m \varepsilon^{m/2}, $$ while on the other hand, $$ \left\vert\int_R^\infty e^{-\pi t \vert x\vert^2}t^{m/2}\tfrac{dt}{t}\right\vert \leq C_{m,b}\int_R^\infty(t\vert x\vert^2)^{-b/2}t^{m/2}\tfrac{dt}{t}\leq R^{(m-b)/2}\vert x\vert^{-b} $$ for any $b\in(m,n)$. With the previous computations in mind (and assuming the FT of $e^{-\pi t\vert x\vert^2}$ to be known) you should be able to obtain: $$ \langle \mathcal{F}\varphi,u\rangle=\langle \varphi,\mathcal{F}u\rangle=\int_{\mathbb{R}^n}\vert \xi\vert^{-m}\hat{u}(\xi)d\xi=\dfrac{(2\pi)^{m}}{c_{n,m}}\int_{\mathbb{R}^n}\vert x\vert^{-(n-m)}u(x)dx. $$ where $$ c_{n,m}=\pi^{n/2}2^{m}\dfrac{\Gamma(\tfrac{m}{2})}{\Gamma\big(\tfrac{n-m}{2}\big)}. $$ Thus, $$ \mathcal{F}\varphi(u)=\dfrac{(2\pi)^{m}}{c_{n,m}}\int_{\mathbb{R}^n}\vert x\vert^{-(n-m)}u(x)dx. $$ Note that the assumptions in $m$ are exactly the ones for everything to be well-defined as tempered-distributions.
Ps: Be careful with the constants. I may be forgetting some of them on my computations (depending on your definition of Fourier transform).