TITLE: Fourier transform of Cauchy principal value
QUESTION [6 upvotes]: I try to understand the direct computation of the Fourier transform of the distribution `Cauchy principal value' $v.p \frac{1}{x}$. I don't understand the following change of order of integration: 
$$
v.p.\int_\mathbb{R} \frac{1}{x}\Bigg(\int_\mathbb{R} e^{-kix}\varphi(k)dk\Bigg)dx=\int_\mathbb{R} \varphi(k)\Bigg(v.p.\int_\mathbb{R} \frac{e^{-kix}}{x}dx\Bigg)dk
$$ 
where $\varphi$ is a Schwartz function and where v.p. denotes the principal value of the integral.
Why and how justify rigourously this change of order of integration?

REPLY [7 votes]: 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 $\varepsilon$.