how can I get the fourier transform of this generalized function?

Question

when function f(x) of the generalized function is f(x)=x+2$f(x)=x+2$, the fourier transform of the generalized function will be like (F$F$ is the Fourier transform)

∫∞−∞(x+2)(F[k(x)])dx$${\int }_{-\infty }^{\infty }(x+2)(F[k(x)])dx$$

when f(x)=sinx$f(x)=\sin x$, it will be

∫∞−∞sinx(F[k(x)])dx$${\int }_{-\infty }^{\infty }\sin x(F[k(x)])dx$$

Is there a way to simplify these? For example, I know when f(x)=1$f(x)=1$, It will be

∫∞−∞1(F[k(x)])dx=∫∞−∞ei0x(F[k(x)])dx=2πk(0)$${\int }_{-\infty }^{\infty }1(F[k(x)])dx={\int }_{-\infty }^{\infty }{e}^{i0x}(F[k(x)])dx=2\pi k(0)$$

Can anyone help me?

Answer 1

With g$g$ a Schwartz function and F(g)$F(g)$ its Fourier transform then the Fourier transform of x+2$x+2$ in the sense of distributions is the unique distribution T$T$ such that for all g$g$ Schwartz

⟨T,g⟩=⟨x+2,F(g)⟩$$⟨T,g⟩=⟨x+2,F(g)⟩$$

where ⟨x+2,F(g)⟩$⟨x+2,F(g)⟩$ means the Riemann integral ∫∞−∞(x+2)F(g)(x)dx${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}(x+2)F(g)(x)dx$ because x+2$x+2$ is a continuous function.

The Fourier inversion theorem for Schwartz functions yields

∫∞−∞2F(g)(x)dx=4πg(0)$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}2F(g)(x)dx=4\pi g(0)$$

and

∫∞−∞xF(g)(x)dx=∫∞−∞(−i)F(g′)(x)dx=−2iπg′(0)$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}xF(g)(x)dx={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}(-i)F(g^{\prime })(x)dx=-2i\pi g^{\prime }(0)$$

thus

T=4πδ+2iπδ′$$T=4\pi \delta +2i\pi {\delta }^{\prime }$$