Principal value not appearing in Fourier transform of the Heaviside function I am trying to determine the Fourier transform of the Heaviside function $H$ as a tempered distribution. Consequently, for $\phi\in\mathcal{S}(\mathbb{R})$, I get
$$\begin{aligned}
(\mathcal{F}H)(\phi)&=H(\mathcal{F}\phi)
\\
&=\int_0^\infty(\mathcal{F}\phi)(\xi)\,d\xi
\\
&=\int_0^\infty\left(\int_{-\infty}^\infty\phi(x)e^{-i\xi x}\,dx\right)\,d\xi
\\
&=\int_{-\infty}^{\infty}\left(\int_0^\infty e^{-i\xi x}\,d\xi\right)\phi(x)\,dx\qquad\text{by Fubini}
\\
&=\lim_{N\to\infty}\int_{-N}^N\left(\frac{i}{x} e^{-i\xi x}\bigg|_{\xi=0}^{\infty}\right)\phi(x)\,dx
\\
&=\lim_{N\to\infty}\int_{-N}^N\left(\frac{i}{x}\lim_{R\to\infty}e^{-i\xi x}\bigg|_{\xi=0}^R\right)\phi(x)\,dx
\\
&=i\lim_{N\to\infty}\int_{-N}^N\frac{\phi(x)}{x}\lim_{R\to\infty}e^{-iRx}\,dx-i\lim_{N\to\infty}\int_{-N}^N\frac{\phi(x)}{x}\,dx,
\end{aligned}$$
so where am I going wrong? If I had $\epsilon\downarrow 0$ instead of $N\to\infty$, then this would certainly look closer to what I want, but I cannot see the mistake.
 A: $$\int_0^\infty\left(\int_{-\infty}^\infty\phi(x)e^{-i\xi x}\,dx\right)\,d\xi
=\int_{-\infty}^{\infty}\left(\int_0^\infty e^{-i\xi x}\,d\xi\right)\phi(x)\,dx$$ 
is not correct since $\int_0^\infty e^{-i\xi x}\,d\xi$ diverges for $\xi \in \mathbb{R}$.
You can regularize it with $\xi \in \mathbb{C},Im(\xi) < 0$ which is the same as saying that as $a \to 0$ : $$e^{-a^2 x} 1_{x  > 0} \to 1_{x > 0}\qquad \text{in the sense of distributions}$$
So that $$( \mathcal{F} 1_{x > 0})(\phi) = \lim_{a \to 0} ( \mathcal{F} e^{-a^2 x} 1_{x  > 0})(\phi)=\lim_{a \to 0}\int_{-\infty}^\infty \phi(x) \int_0^\infty e^{-a^2 t} e^{-itx}dtdx $$
$$= \lim_{a \to 0}\int_{-\infty}^\infty  \frac{\phi(x)}{a^2+ix}dx
=\lim_{a \to 0}\int_{-\infty}^\infty  \frac{\phi(x)-\phi(0)1_{|x| < 1}}{a^2+ix}dx+\int_{-1}^1 \frac{\phi(0)}{a^2+ix}dx$$
$$= \int_{-\infty}^\infty\frac{\phi(x)-\phi(0)1_{|x| < 1}}{ix}dx+\lim_{a \to 0}\frac{\phi(0)\log(a^2+ix)}{i}|_{-1}^1$$
$$ = \lim_{\epsilon \to 0} \int_{|x| > |\epsilon|} \frac{\phi(x)}{ix}dx+ \pi \phi(0)$$
and hence $$\mathcal{F} 1_{x > 0} =  p.v.(\frac{1}{i x})+\pi \delta(x)$$
