Symmetry of Fourier Transform Spectrum

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Given the continuous Fourier Transform $X(f)=\int_{-\infty}^{\infty}x(t)e^{-j2\pi ft}dt$

A sufficient condition for it to exist is $E_{x}=\int_{-\infty}^{\infty}|x(t)|^{2}dt < +\infty$.

We can write $X(f) = R(f) + jI(f)$ where R(f) is the real part and I(f) is the imaginary part.

Assuming x(t) is a real function, we can then write:

$R(f)=\int_{-\infty}^{\infty}x(t)cos(2\pi ft)dt$

$I(f)=-\int_{-\infty}^{\infty}x(t)sin(2\pi ft)dt$

The book states that from these two expressions above it is clearly visible that:

$R(f)=R(-f)$

$I(f)=-I(-f)$

I don't see it. Shouldn't it depends on whether the x(t) is an even or odd function?

E.g. in case of R(f), $cos(2\pi ft)$ is even, so:

  • odd times even = odd hence R(f)=-R(-f)
  • even times even = even hence R(f)=R(-f)

What's wrong?