So I was trying to prove/disprove the following claim: If $x^*(-t)=x(t)$ (where $^*$ denotes the complex conjugate) then the Fourier transform $$X(\omega) = \int_{-\infty}^{+\infty} x(t) e^{-j \omega t}\, dt$$ is a real-valued function, assuming $\omega \in \mathbb{R}$.
I managed to "disprove" it, but I know there must be something wrong with my argument. I'm also pretty sure the claim is true because I can't find a counter-example.
Anyways, this is my "proof": \begin{align*} X^*(\omega) &= \left[\int_{-\infty}^{+\infty} x(t)e^{-j\omega t}\,dt \right]^*\\ &= \int_{-\infty}^{+\infty} \left[x(t)e^{-j\omega t}\right]^*dt\\ &= \int_{-\infty}^{+\infty} x^*(t)e^{j\omega t}\, dt\\ &= -\int_{-\infty}^{+\infty} x^*(-\tau)^{-j\omega\tau}\, d\tau\\ &= -\int_{-\infty}^{+\infty} x(\tau)e^{-j\omega\tau}\, d\tau\\ &= - X(\omega), \end{align*} and therefore $X(\omega)$ must be purely imaginary.
I know this can't be true, because given any real-valued even function $x(t)$, such as the normalized sinc function $x(t)=\frac{\sin(\pi t)}{\pi t}$, clearly satisfies $x^*(-t)=x(t)$, but $X(\omega)$ must be real-valued too.
This isn't homework, any help understanding what's wrong with my "proof" would be appreciated.