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I understand that the fourier transform of f(x) can be broken up into odd and even parts such that the transform of $f(x)$ can be represented by $$F(s) = 2\int_{0}^{\infty}E(x)cos(2\pi x s)dx - 2i\int_{0}^{\infty}O(x)sin(2\pi xs)dx$$

Where $E(x)$ is the even part of $f(x)$ and $O(x)$ is the odd part. Likewise, the inverse transform is $$f(x) = 2\int_{0}^{\infty}E(s)cos(2\pi x s)ds - 2i\int_{0}^{\infty}O(s)sin(2\pi xs)ds$$

I'm assuming that, since $f(x)$ is real, this reduces to $$f(x) = 2\int_{0}^{\infty}E(s)cos(2\pi x s)ds$$

Does this mean that $f(x)$ is even, and can I use this information to say anything about $F(s)$? Am I heading in the right direction with this proof?

JayJay
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  • Assuming that $f(x)$ is real doesn't imply that $F(s)$ is real. And it is no clear how you'd decompose a complex function into even/odd components (second equation). It is true however that the spectrum of a real, even function is real. – Weaam Oct 26 '15 at 16:06

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Since $F(s) = 2\int_0^\infty E(x)\cos(2\pi xs) dx-2i\int_0^\infty O(x)\sin(2\pi xs) dx$, define $A(s)=2\int_0^\infty E(x)\cos(2\pi xs) dx$, $B(s)=2\int_0^\infty O(x)\sin(2\pi xs) dx$, we know $A, B$ are all real functions because $f(x)$ is real. Then we can calculate $|F(s)|^2= A(s)^2+B(s)^2$ and found that $|F(-s)|^2 = A(-s)^2+B(-s)^2 = |F(s)|^2$. That is $|F(s)|^2$ is even function.

Mia
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