I am stuck on an excerise which says that prove the fourier transform $f(k)$ of a real function satisfied the condition $f(-k)=f*(-k)$. Where the astericks denotes the complex congugate.
I am beginning to think there is a typo as I am getting it to be $f(-k)=f*(k)$.
By def
$$f(k)=\int_{-\infty}^{\infty}f(x)e^{-ikx}dx$$ LHS $$\int_{-\infty}^{\infty}f(x)(\cos kx+i\sin kx)dx$$
RHS $$\int_{-\infty}^{\infty}f(x)e*^{ikx}dx$$ $$..$$ $$=\int_{-\infty}^{\infty}f(x)(\cos kx-i\sin kx)dx$$ Sorry I dont have latex installed, Any help would be much appreciated
Definition of Fourier transform: \begin{align*} \tilde{f}(k) &= \int_{-\infty}^\infty e^{-ikx} f(x) dx \\ &= \int_{-\infty}^\infty f(x) \cos kx dx - i \int_{-\infty}^\infty f(x) \sin kx dx \\ \Rightarrow \tilde{f}^*(k) &= \int_{-\infty}^\infty f(x) \cos kx dx + i \int_{-\infty}^\infty f(x) \sin kx dx \\ &= \int_{-\infty}^\infty e^{ikx} f(x) dx \\ &= \tilde{f}(-k) \end{align*}
So it seems that $\tilde{f}(-k) = \tilde{f}^*(k)$.
$\tilde{f}(-k) = \tilde{f}^*(-k)$ can't be true because it would imply that $\tilde{f}$ is always real.