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How can I show that the Fourier transform of an even integrable function $f\colon \mathbb{R}\to\mathbb{R}$ is even real-valued function? And the Fourier transform of an odd integrable function $f\colon \mathbb{R}\to\mathbb{R}$ is odd and purely imaginary function?

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Just a change of variables. – Ahriman Aug 19 '12 at 17:40
Welcome to the site. Not a bad question for a first-timer. An upvote to get you started. – Tom Au Aug 19 '12 at 20:46
up vote 5 down vote accepted

Let $f: \mathbb{R} \to \mathbb{R}$ be an integrable function and let $\hat{f}$ denote its Fourier transform, i.e. $$ \hat{f}(\xi)=\int_\mathbb{R}e^{ix\xi}f(x)dx. $$ We have $$ \overline{\hat{f}(\xi)}=\hat{f}(-\xi)=\int_\mathbb{R}e^{-ix\xi}f(x)dx=\int_\mathbb{R}e^{iy\xi}f(-y)dy. $$ If $f$ is even then $$ \overline{\hat{f}(\xi)}=\hat{f}(-\xi)=\int_\mathbb{R}e^{iy\xi}f(y)dy=\hat{f}(\xi), $$ i.e. $\hat{f}$ is an even real-valued function.

If $f$ is odd then $$ \overline{\hat{f}(\xi)}=\hat{f}(-\xi)=\int_\mathbb{R}-e^{iy\xi}f(y)dy=-\hat{f}(\xi), $$ i.e. $\hat{f}$ is an odd purely imaginary function.

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Define $F(p) = \int_{-L}^{L} f(x)e^{ipx} dx$. Note that: (let $u=-x$ so $x=-u$ and $dx=-du$ etc..)

$$ F(-p) = \int_{-L}^{L} f(x)e^{-ipx} dx = \int_{L}^{-L} f(-u)e^{ipu}(-du)=\int_{-L}^{L} f(-u)e^{ipu}du$$

Clearly $f(-x) = \pm f(x)$ implies $F(-p) = \pm F(p)$.

Now we turn to the reality part of the claim. Recall $e^{ipx} = \cos(px)+i\sin(px)$. Also, remark sine is an odd function whereas cosine is an even function. We know from elementary calculus that the integral of an odd function on $[-L,L]$ vanishes.

  1. When $f$ is even then $f(x)\cos(px)$ is even and $f(x)\sin(px)$ is odd. It follows that the imaginary part of Fourier transform vanishes. Consquently, $F(p)$ is real.

  2. When $f$ is odd then $f(x)\cos(px)$ is odd and $f(x)\sin(px)$ is even. It follows that the real part of Fourier transform vanishes. Consquently, $F(p)$ is imaginary.

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