Fourier transform of Cauchy principal value

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?

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1 Answer

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 $\varepsilon$.

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I was readin your solution, I dont understand the minus signal in the first equality. can you explain ? please –  math student Nov 18 '13 at 23:55
We integrate $e^{-isx}/x$ with respect to $x$ on $(-R,-\varepsilon)\cup (\varepsilon,R)$, and we reduce this to an integral over $(\varepsilon,R)$, making a substitution on the first integral. –  Davide Giraudo Nov 19 '13 at 9:46
please correct me if i am wrong: –  math student Nov 19 '13 at 12:42