Fourier transform symmetry property proof

Question

I am trying to write my own proof that:

If $f(t)$ is odd then so is $F(\nu).$

Where $F$ denotes the Fourier transform of $f.$ What I want to show is that if $f(t)=-f(-t),$ then $F(\nu)=-F(-\nu).$ So assuming that $f(t)$ is odd the Fourier transform becomes:

$$F(\nu) = \int^\infty_{-\infty} f(t) e^{-j2 \pi \nu t} dt = -\int^\infty_{-\infty} f(-t) e^{-j2 \pi \nu t} dt \tag{1}$$

So, from here how do I get to $-F(-\nu)$?

Answer 1

I suggest you start from $F(-\nu)$ :

$$F(-\nu) = \int_{-\infty}^\infty f(t) e^{j2 \pi \nu t} \;dt \;=\; \int_{\infty}^{-\infty} f(-u) e^{-j2 \pi \nu u} \,\;(-du) \;=\; \int_{-\infty}^{\infty} -f(u) e^{-j2 \pi \nu u} \,\;du \;=\; -F(\nu)$$

where I used the change of variables $u=-t$.