# Fourier transform of even/odd function

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?

• Just a change of variables. Commented Aug 19, 2012 at 17:40

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.
• How are you getting that $\hat{f}$ is real-valued or purely imaginary from that? Commented Dec 5, 2022 at 19:05
• Try solving the equation $\bar{z}=-z$ for #z=x+iy \in \mathbb{C}$, with$x,y \in\mathbb{R}$. Commented Dec 5, 2022 at 22:20 Define$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. Consequently,$F(p)$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. Consequently,$F(p)$is imaginary. 1. The FT of even functions are also even; The FT of odd functions are also odd. 2. The real part of the FT of a real function is even; The imaginary part of the FT of a real function is odd. So the Fourier Transform$F(\omega)$of a real and even function$f(x)$must satisfy both: 1. Because$f(x)$is even:$F(\omega)$is even (for both real and imaginary parts) 2. Because$f(x)$is real: the real part of$F(\omega)$is even, and the imaginary part is odd Now for the imaginary part of$F(\omega)$to be both even and odd, it must be zero, thus$F(\omega)\$ is real-only.