# Thread: Proof of even fourier transform

1. ## Proof of even fourier transform

My problem is:

"Prove that if f is an even function, then f^ is also an even function"
f^ = fourier transform of f."

This is what I've done, after some hints from a teaching assistant:

$\hat{f}(w)=\int f(x)e^{(-iwx)}dx$
$\hat{f}(w)=\int f(-x)e^{(-iwx)}dx$, since f(x)=f(-x), even function

Choose u=-x ->dx = -du

$\hat{f}(w)=-\int f(-u)e^{(-iw(-u))}du$
$\hat{f}(-w)=-\int f(u)e^{(-i(-w)(-u))}du$

I can't see how this answers the problem, is there anything wrong in my calculations, or do you have any hints on how to continue? I'm thankful for any help...

2. Originally Posted by dreamsound
My problem is:

"Prove that if f is an even function, then f^ is also an even function"
f^ = fourier transform of f."

This is what I've done, after some hints from a teaching assistant:

$\hat{f}(w)=\int f(x)e^{(-iwx)}dx$
$\hat{f}(w)=\int f(-x)e^{(-iwx)}dx$, since f(x)=f(-x), even function

Choose u=-x ->dx = -du

$\hat{f}(w)=-\int f(-u)e^{(-iw(-u))}du$
$\hat{f}(-w)=-\int f(u)e^{(-i(-w)(-u))}du$

I can't see how this answers the problem, is there anything wrong in my calculations, or do you have any hints on how to continue? I'm thankful for any help...
You want to show that

$\hat{f}(-w)=\hat{f}(w)$ so you have
$\displaystyle \hat{f}(-w)=\int_{-\infty}^{\infty} f(-x)e^{(-i(-w)x)}dx=\int_{-\infty}^{\infty} f(x)e^{(iwx)}dx$

Now let $x=-t \implies dx=-dt$ this gives

$\displaystyle \int_{-\infty}^{\infty} f(x)e^{(iwx)}dx=\int_{\infty}^{-\infty} f(-t)e^{(-iw)}(-dt)=\int_{-\infty}^{\infty} f(x)e^{(-iwt)}dt$

Notice the switching of the limits of integration.

3. Thanks!