Problem statement: If $F()$ is the Fourier transform of the funciton $f(x)$, then determine $F(f(x-a))$, where $a$ is a constant.
My attempt: $$F() = F(f(x)) = \int_{-\infty}^\infty f(x)e^{-i\omega x}dx$$ Hence, $$F(f(x-a)) = \int_{-\infty}^\infty f(x-a)e^{-i\omega x}dx$$ Substituting $x-a = t$ $$= \int_{-\infty}^\infty f(t)e^{-i\omega (t+a)}dt$$ $$= e^{-i\omega a}\int_{-\infty}^\infty f(t)e^{-i\omega t}dt$$ $$= e^{-i\omega a}\int_{-\infty}^\infty f(x)e^{-i\omega x}dx$$ $$= e^{-i\omega a}F() $$
My teacher's solution: $$F() = F(f(x)) = \int_{-\infty}^\infty f(x)e^{i\omega x}dx$$ Hence, $$F(f(x-a)) = \int_{-\infty}^\infty f(x-a)e^{i\omega x}dx$$ Substituting $x-a = t$ $$= \int_{-\infty}^\infty f(t)e^{i\omega (t+a)}dt$$ $$= e^{i\omega a}\int_{-\infty}^\infty f(t)e^{i\omega t}dt$$ $$= e^{i\omega a}\int_{-\infty}^\infty f(x)e^{i\omega x}dx$$ $$= e^{i\omega a}F() $$
The difference between the solutions is the minus sign in the exponent of $e.$ Which of these is correct? I use the minus sign because that's what the definition of fourier transform gives. Am I wrong? Should I use the minus sign or not? Which solution is correct? And why?