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