I try to understand the direct computation of the Fourier transform of the distribution `Cauchy principal value' $p.v. \frac{1}{x}$. I don't understand the following change of order of integration: $$ p.v.\int_\mathbb{R} \frac{1}{x}\Bigg(\int_\mathbb{R} e^{-kix}\varphi(k)dk\Bigg)dx=\int_\mathbb{R} \varphi(k)\Bigg(p.v.\int_\mathbb{R} \frac{e^{-kix}}{x}dx\Bigg)dk $$ where $\varphi$ is a Schwartz function and where p.v. denotes the principal value of the integral.
Why and how justify rigourously this change of order of integration?
We can't use Fubini directly as noted in the OP. Let $S_{\varepsilon,R}:=\{t\in\Bbb R, \varepsilon<|t|<R\}$. Using Fubini's theorem and a rewriting of the inner integral, we come up with the equality $$\int_{\Bbb R\times\Bbb R}\frac{e^{-isx}}x\varphi(s)\chi_{S_{\varepsilon,R}}(x)dxds=-\int_{\Bbb R}\varphi(s)\int_{s\varepsilon}^{sR}\frac{\sin u}ududs.$$ We can find $M$ such that for all $t$, $\left|\int_0^t\frac{\sin u}u du\right|<M$. This allows us to use the dominated convergence theorem in order to take the limit $R\to +\infty$ in the displayed equality. This gives $$\int_{\{|x|>\varepsilon\}}\int_{\Bbb R}\frac{e^{-isx}}x\varphi(s)dxds=-\int_{\Bbb R}\varphi(s)\int_{s\varepsilon}^{+\infty}\frac{\sin u}ududs.$$ Then use dominated convergence theorem again to take the limit with respect to $\varepsilon$.