Proving a symmetric property of the Fourier Transform

[sapsapz]

Hello

I'm trying to prove a symmetric property of the fourier transform, but I'm having some problems with that.
Here is the failed try of the proof: http://mathb.in/10634

Prove that if $f(x)$ is a function such that $f(x) = \overline{f}(-x)$ then it's Fourier Transform is real.

Try of proof:

\(\displaystyle F(w)\) will be the Fourier tranform of \(\displaystyle f(x).\)

We will try to prove that \(\displaystyle F(w) = \overline{F}(w)\), and in doing so we prove that \(\displaystyle F(w)\) is real.

So:

\(\displaystyle \overline{F}(w) = \overline{\int_{-\infty}^\infty \! f(x)e^{-iwx} \, \mathrm{d}x} = \int_{-\infty}^\infty \! \overline{f(x)e^{-iwx}} \, \mathrm{d}x =\)

\(\displaystyle \int_{-\infty}^\infty \! \overline{f(x)}*\overline{e^{-iwx}} \, \mathrm{d}x = \int_{-\infty}^\infty \! f(-x)e^{iwx} \, \mathrm{d}x.\)

Lets set a new variable \(\displaystyle x' = -x\), so \(\displaystyle {d}x' = -{d}x.\) Continuing the equalities:

\(\displaystyle \int_{-\infty}^\infty \! f(-x)e^{iwx} \, \mathrm{d}x = -\int_{-\infty}^\infty \! f(x')e^{-iwx'} \, \mathrm{d}x' = -F(w).\)

So we get \(\displaystyle \overline{F}(w) = -F(w)\), and not \(\displaystyle \overline{F}(w) = F(w).\)

What did I do wrong?

Any help would be much appreciated!

 

[Benjamin]

Re

Hello

I'm trying to prove a symmetric property of the fourier transform, but I'm having some problems with that.
Here is the failed try of the proof: http://mathb.in/10634

Prove that if $f(x)$ is a function such that $f(x) = \overline{f}(-x)$ then it's Fourier Transform is real.

Try of proof:

\(\displaystyle F(w)\) will be the Fourier tranform of \(\displaystyle f(x).\)

We will try to prove that \(\displaystyle F(w) = \overline{F}(w)\), and in doing so we prove that \(\displaystyle F(w)\) is real.

So:

\(\displaystyle \overline{F}(w) = \overline{\int_{-\infty}^\infty \! f(x)e^{-iwx} \, \mathrm{d}x} = \int_{-\infty}^\infty \! \overline{f(x)e^{-iwx}} \, \mathrm{d}x =\)

\(\displaystyle \int_{-\infty}^\infty \! \overline{f(x)}*\overline{e^{-iwx}} \, \mathrm{d}x = \int_{-\infty}^\infty \! f(-x)e^{iwx} \, \mathrm{d}x.\)

Lets set a new variable \(\displaystyle x' = -x\), so \(\displaystyle {d}x' = -{d}x.\) Continuing the equalities:

\(\displaystyle \int_{-\infty}^\infty \! f(-x)e^{iwx} \, \mathrm{d}x = -\int_{-\infty}^\infty \! f(x')e^{-iwx'} \, \mathrm{d}x' = -F(w).\)

So we get \(\displaystyle \overline{F}(w) = -F(w)\), and not \(\displaystyle \overline{F}(w) = F(w).\)

What did I do wrong?

Any help would be much appreciated!

There is a mistake in making the change of variable x = -x'.
When x goes to +oo, x' goes to - oo, and so on.

Hence, you should have

\(\displaystyle \int_{-\infty}^\infty \! f(-x)e^{iwx} \, \mathrm{d}x = -\int_{\infty}^{-\infty} \! f(x')e^{-iwx'} \, \mathrm{d}x' = F(w).\)

Benjamin