Continuous Fourier Transform of a real and odd signal - proof

On my book there is the following proof relating the continuous Fourier Transform X(f) of a real and odd signal x(t)=x(-t) (goal X(f)=X(-f)):

$$X(f)=\int_{-\infty}^{+\infty}{x(t) \exp(-j2\pi ft)dt}$$

$$X(-f)=\int_{-\infty}^{+\infty}{x(t) \exp(j2\pi ft)dt}$$

Now $\alpha=-t$, thus:

$$X(-f)=-\int_{+\infty}^{-\infty}{x(-\alpha) \exp(-j2\pi f\alpha)d\alpha}=\int_{-\infty}^{+\infty}{x(-\alpha) \exp(-j2\pi f\alpha)d\alpha}=\int_{-\infty}^{+\infty}{x(\alpha) \exp(-j2\pi f\alpha)d\alpha}=X(f)$$

In the last operation I use $x(\alpha)=x(-\alpha)$ because $x(\alpha)$ is an odd signal.

Why $X(f)=\int_{-\infty}^{+\infty}{x(t) \exp(-j2\pi ft)dt}=\int_{-\infty}^{+\infty}{x(\alpha) \exp(-j2\pi f\alpha)d\alpha}$? $\alpha\neq t$

Thank you very much.

Firs of all, $x$ is an even signal, not an odd signal.
As for your question, $t$ and $\alpha$ are mute variables. You could use any symbol and the integral would be the same, like in the following example: $$\int_0^1x\,dx=\int_0^1y\,dy=\int_0^1t\,dt=\int_0^1\square\,d\square=\frac12.$$