when function f(x) of the generalized function is $f(x) = x+2$, the fourier transform of the generalized function will be like ($F$ is the Fourier transform)$$\int_{-∞}^{∞}(x+2)(F[k(x)])dx$$when $f(x) = \sin x$, it will be$$\int_{-∞}^{∞}\sin x(F[k(x)])dx$$Is there a way to simplify these? For example, I know when $f(x) = 1$, It will be$$\int_{-∞}^{∞}1(F[k(x)])dx = \int_{-∞}^{∞}e^{i0x}(F[k(x)])dx = 2πk(0)$$Can anyone help me?
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$\begingroup$With $g$ a Schwartz function and $F(g)$ its Fourier transform then the Fourier transform of $x+2$ in the sense of distributions is the unique distribution $T$ such that for all $g$ Schwartz $$\langle T,g\rangle = \langle x+2,F(g)\rangle$$where $\langle x+2,F(g)\rangle$ means the Riemann integral $ \int_{-\infty}^\infty (x+2) F(g)(x)dx$ because $x+2$ is a continuous function.
The Fourier inversion theorem for Schwartz functions yields$$ \int_{-\infty}^\infty 2 F(g)(x)dx= 4\pi g(0)$$ and$$ \int_{-\infty}^\infty x F(g)(x)dx= \int_{-\infty}^\infty (-i) F(g')(x)dx = -2i\pi g'(0)$$thus $$T = 4\pi\delta+2i\pi \delta'$$
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