Fourier transform and inverse Fourier transform is given as -
$f(x)=\int_{-\infty}^{\infty}g(\alpha)e^{i\alpha x}d\alpha$
$g(\alpha)=\frac{1}{2\pi}\int_{-\infty}^{\infty}f(x)e^{-i\alpha x}dx$
I am having a trouble in proving the following theorem:
If $f(x)$ is even then so is $g(\alpha)$. Likewise if $f(x)$ is odd, then so is $g(\alpha)$
My attempt.
$g(-\alpha)=\frac{1}{2\pi}\int_{-\infty}^{\infty}f(x)e^{i\alpha x}dx$
Changing the variable $x\to -x$, we have
$g(-\alpha)=\frac{1}{2\pi}\int_{-\infty}^{\infty}f(-x)e^{-i\alpha x}(-dx)$
$\implies g(-\alpha)=-\frac{1}{2\pi}\int_{-\infty}^{\infty}f(-x)e^{-i\alpha x}dx$
So we can see that if $f(x)$ is even $(f(-x)=f(x))$ then $g(-\alpha)=-g(\alpha)$
But this is contrary to what we have to prove.
Can someone please help me where I have gone wrong?
When you made the change of variables $x\rightarrow-x$ you neglected application of this to the limits of the integral, i.e., you should have written the integral limits as $\int_\infty^{-\infty}$.