Fourier Transform Symmetry Misunderstanding

Question

I learned Fourier Transform about 2 years ago, but recently I found that I can't understand a simple property of that.

there is a lot of proof that shows Fourier Transform of even/odd signal is even/odd, but when I saw the proofs they missed a minus in the differential. here is my proof : (for even)

if $f(t) = f(-t)$

$F(-w) = \int_{-\infty}^{\infty} f(t)\:e^{jw2\pi t}\:dt = \int_{-\infty}^{\infty} f(-u)\:e^{-jw2\pi u}\:(-du) = \int_{-\infty}^{\infty} -f(u)\:e^{-jw2\pi u}\:du = -F(w)\:\:\:(!!)$

I use changing variable like all other proofs ( $u = -t$ ) but in other proofs they didn't do $du = -dt$ and they use $du = dt$ and get to the $F(-w) = F(w)$ easily.( I cant get why they did that)

please tell my what's my mistake.

Thanks!

Answer 1

The problem is that you have an incorrect change of variables.

Let's do it in the level of Riemann integrals.

Assuming absolute integrability and recall the definition $$ \int_{-\infty}^\infty f(x)dx=\lim_{N\to\infty}\int_{-N}^N f(x)\,dx $$

If you do the change of variables $u=-x$, then $$ \int_{-N}^N f(x)\,dx=\int_{N}^{-N}f(-u)\cdot(-1)du=\int_{-N}^N f(-u)\,du $$ which implies that $$ \int_{-\infty}^\infty f(x)dx =\int_{-\infty}^\infty f(-u)du =\int_{-\infty}^\infty f(-x)dx .\tag{1} $$

Note that we do not use any parity assumptions on $f$ here.

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So in your calculation, you should have $$ \begin{align} F(-w) &= \int_{-\infty}^{\infty} f(t)\:e^{jw2\pi t}\:dt\\ &= \int_{-\infty}^{\infty} \color{red}{f(-u)\:e^{-jw2\pi u}}\:du\\ &= \int_{-\infty}^{\infty} \color{red}{f(u)\:e^{-jw2\pi u}}\:du \\ &= F(w) \end{align} $$

Answer 2

When you change variables $u=-t$ you do, indeed, get $du = -dt$. But you also change the limits on the integral: $$ \int_{-\infty}^{+\infty} \phi(t)\;dt = \int_{+\infty}^{-\infty} \phi(-u)\;(-du) $$ then the convention says $$ -\int_{+\infty}^{-\infty} \phi(-u)\;du = +\int_{-\infty}^{+\infty} \phi(-u)\;du . $$