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I have been trying to follow steps of a book (Mathematical Physics by Butkov) to understand how Fourier integral is derived.

$$f(x)=\dfrac{1}{\sqrt{2\pi}}\int_{-\infty}^0 F(k)e^{-ikx}dk+\dfrac{1}{\sqrt{2\pi}}\int_{0}^{\infty} F(k)e^{-ikx}dk$$ after considering that $F(-k)=F^*(k)$

$$f(x)=\dfrac{1}{\sqrt{2\pi}}\int_0^{\infty}[F(k)e^{-ikx}+F^*(k)e^{-ikx}]dk$$

And book immediately follows with

$$F(k)e^{-ikx}=\dfrac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(\xi)e^{ik(\xi-x)}d\xi$$ but I would expect the following relation $$F(k)=\mathfrak{F}\{f(x-1)\}=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(\xi)e^{ik(\xi-1)}d\xi$$

I don't understand where the book gets the $x$ term. Then it builds on it.

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    $\begingroup$ From the FT inversion theorem we have $F(k)=\frac1{\sqrt {2\pi}}\int_{-\infty}^\infty f(\xi)e^{ik\xi}\,d\xi$. Now just multiply both sides by $e^{-ikx}$ $\endgroup$ – Mark Viola Oct 26 '18 at 0:57
  • $\begingroup$ if you put your comment as an answer I mark it as solved. Also I was wondering if what I was doing correct as well? Maybe not for deriving Fourier integral theorem but I was using the shifting property of Fourier transforms $\endgroup$ – GGphys Oct 26 '18 at 19:29
  • $\begingroup$ Sure. I've added a bit more to supplement the comment. $\endgroup$ – Mark Viola Oct 26 '18 at 21:55
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From the FT inversion theorem we have

$$F(k)=\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty f(\xi)e^{ik\xi}\,d\xi \tag1$$

Multiplying both sides of $(1)$ by $e^{-ikx}$ and absorbing this factor inside the integral yields

$$F(k)e^{-ikx}=\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty f(\xi)e^{ik(\xi-x)}\,d\xi$$

as was to be shown.


In addition, we can write

$$\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty f(\xi+x)e^{ik\xi}\,d\xi=\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty f(\xi)e^{ik(\xi-x)}\,d\xi=F(k)e^{-ikx}$$

Hence, we can write the following sifting property:

$$\mathscr{F}\{f(\xi+x)\}(k)=e^{-ikx}\mathscr{F}\{f(\xi)\}(k)$$

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