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Given $H(x)$ is the Heaviside Theta, the tables give the following Fourier transforms for it:

$$ H(x+a)\to -PV\frac{i e^{i a w} }{w}+\pi \delta (w)$$

while from Sokhotski–Plemelj theorem it follows (seemingly) that

$$ H(x+a)\to~PV\frac{-i}{w} +\pi\delta(w)+ |a|e^{iax/2}\operatorname {sinc} \left({\frac {w a }{2\pi}}\right)$$

The first result seems questionable to me because it does not depend on $a$ at $w=0$. But this should not be the case because $H(x+a)$ can be represented as a sum of $H(x)$ and the $rect$ function, and the Fourier transform of the later depends on $a$.

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application of the Sokhotski–Plemelj formula does in fact give the first of the two expressions:

$$\int_{-\infty}^\infty H(x+a)e^{-ix w}dx=\int_{-a}^\infty e^{-i xw}dx=e^{iwa}\int_0^\infty e^{-ix w}dx$$ $$=e^{iwa}\left(\text{PV}\frac{-i}{w}+\pi\delta(w)\right)=-\text{PV}\frac{ie^{i aw}}{w}+\pi\delta(w)\qquad(\ast)$$

where PV indicates the principal value; so this is the first of the two formulas in the OP; it may equivalently be written as $$(\ast)=-\text{PV}\,\frac{i\cos aw}{w}+\pi\delta(w)+a\,\text{sinc}\,aw$$ which is different from the second formula


the OP asks whether I can reach the same result by splitting $H(x+a)$ into $H(x)$ plus a rect function:

Fourier transform of $H(x)$: $\int_{0}^\infty e^{-ixw}dx=-\text{PV}\,\frac{i}{w}+\pi\delta(w)$

Fourier transform of rect: $\int_{-a}^0 e^{-ixw}dx=\frac{i}{w}\left(1-e^{iwa}\right)=\text{PV}\,\frac{i}{w}(1-\cos aw)+a\,\text{sinc}\,aw$
since $w^{-1}(1-\cos aw)=\text{PV}\,w^{-1}(1-\cos aw)$

summing up gives the same result as above, $(\ast)=-\text{PV}\,\frac{i\cos aw}{w}+\pi\delta(w)+a\,\text{sinc}\,aw$

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  • $\begingroup$ But with the addition "PV", yes? Is it the same as the second result? $\endgroup$ – Anixx May 8 '18 at 11:53
  • $\begingroup$ If we take $a=0$, we will obtain the second result (without the sinc part), and then if we represent $H(x+a)=H(x)+rect(x/a)$ we should obtain the second result? $\endgroup$ – Anixx May 8 '18 at 11:56
  • $\begingroup$ I corrected the second result (which seems to have had an error). The basic idea, it is obtained as a sum of Fourier transforms of H(x) and the rect function. $\endgroup$ – Anixx May 8 '18 at 12:02
  • $\begingroup$ Yor result seems questionable to me because it does not depend on $a$ at $w=0$. But this should not be the case because $H(x+a)$ can be represented as a sum of $H(x)$ and the rect function, and the Fourier transfor of the later depends on $a$. $\endgroup$ – Anixx May 8 '18 at 12:06
  • $\begingroup$ Do you obtain the same result if you represent H(a+x) as H(x) plus a rect function? $\endgroup$ – Anixx May 8 '18 at 14:02

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