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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.

user45453
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    You didn't change your integral limits when mapping $t \mapsto - \tau$. So you should have $X^{*} = X$. – Matthew Cassell Aug 06 '16 at 17:12
  • I don't quite understand, if $\tau = -t$, then $dt = -d\tau$, and $\int_{-\infty}^{+\infty} x^(t)e^{j\omega t},dt = \lim_{T\to+\infty} \int_{-T}^{+T} x^(t)e^{j\omega t},dt = -\lim_{T\to+\infty} \int_{-T}^{T} x^*(-\tau)e^{-j\omega\tau},d\tau = -\int_{-\infty}^{+\infty} x(\tau)e^{-j\omega\tau},d\tau$, right? – user45453 Aug 06 '16 at 17:19
  • OK so I worked it out with $\int_{-\pi/2}^{\pi/2} \cos x, dx$, and to my surprise, I can't simply do $\int_{-\pi/2}^{\pi/2} \cos x, dx = -\int_{-\pi/2}^{\pi/2} \cos (-y), dy$ (because one is 2 and the other -2), even though $-\pi/2 \leq x \leq \pi/2 \implies -\pi/2 \leq -x=y \leq \pi/2$. I have no idea how I'm only realizing this now....... Thank you anyway. – user45453 Aug 06 '16 at 17:33
  • Just to expand on my point, if $t = - \tau$ then as $t \to \infty, \tau \to - \infty$. Similarly, for $t \to \infty, \tau \to -(- \infty) = \infty$. So your integral will be of the form $$\int_{\infty}^{- \infty} - d \tau = \int_{-\infty}^{\infty} d \tau$$ not $$\int_{- \infty}^{ \infty} - d \tau$$ – Matthew Cassell Aug 07 '16 at 01:41

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