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Let $f(t)$ be an even function, so that $f(t) = f(-t)$. Let $F(\omega)$ be its Fourier Transform. Prove that $F(\omega)$ is real and even too (in $\omega$).

An attempt could be the following:

$$F(\omega) = \int_{-\infty}^{+\infty} f(t) e^{-j \omega t} dt = \int_{-\infty}^0 f(t) e^{-j \omega t} dt + \int_0^{+\infty} f(t) e^{-j \omega t} dt$$

The variable $-t$ can be used in the first integral instead of $t$:

$$F(\omega) = \int_{+\infty}^0 - f(-t) e^{j \omega t} dt + \int_0^{+\infty} f(t) e^{-j \omega t} dt = \int_0^{+\infty} f(-t) e^{j \omega t} dt + \int_0^{+\infty} f(t) e^{-j \omega t} dt$$

$$F(\omega) = \int_0^{+\infty} f(t) \left( e^{j \omega t} + e^{-j \omega t} \right) dt = 2\int_0^{+\infty} f(t) \cos (\omega t) dt$$

$f(t) \cos (\omega t)$ is even in $t$, being the product of two even functions. $\cos (\omega t)$ is even in $\omega$. But now? What considerations can be added to prove that the resulting $F(\omega)$ is even in $\omega$?

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    Note that since $f$ is real-valued, you have $\overline{F(\omega)} = F(-\omega)$. Use the substitution $u = -t$ in the Fourier integral. – 2017-01-26
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    @DanielFischer Isn't it enough to simply note that $\cos( \omega t) = \cos (- \omega t)$, being the cosine an even function of its argument? – 2017-01-26
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    Yes, in the form you have reached, it suffices to note that $\cos$ is even. – 2017-01-26

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Use the substitution $z=-t$. Here a very detailed solution: $F(w)=\int_{-\infty}^{\infty}{f(t)e^{-iwt}dt} = \int_{-\infty}^{\infty}{f(-t)e^{-iwt}dt} = -\int_{-\infty}^{\infty}{f(-t)e^{iw(-t)}(-1)dt} = -\int_{\infty}^{-\infty}{f(z)e^{iwz}dz}=\int_{-\infty}^{\infty}{f(z)e^{iwz}dz}=\int_{-\infty}^{\infty}{f(z)e^{-i(-w)z}dz}=F(-w)$