# Is there a real representation of the fourier transform?

The fourier transform is commonly given by $$\hat{f}(\xi)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(x)e^{-i\xi x}\mathrm{d}x$$ with its inverse being $$f(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\hat{f}(\xi)e^{i\xi x}\mathrm{d}\xi$$ We also know that a fourier series of a function has a real as well as a complex representation: $$f(x)=\frac{a_0}{2} + \sum_1^{\infty}a_n\cos(n\omega x)+\sum_1^{\infty}b_n\sin(n\omega x) = \sum_{-\infty}^{\infty}c_ne^{-in\omega x}$$ The complex form of the fourier series can be directly derived from its real form using Euler's formula.

It can also be shown that the real coefficients of the fourier series can be calculated by the following integrals: $$a_n=\frac{2}{T}\int_{T}f(x)\cos(n\omega x)\mathrm{d}x$$ $$b_n=\frac{2}{T}\int_{T}f(x)\sin(n\omega x)\mathrm{d}x$$ While deriving the complex fourier series, the relation $$c_n=\frac{a_n-ib_n}{2}=\frac{1}{T}\int_{-\infty}^{\infty}f(x)e^{-in\omega x}\mathrm{d}x=\frac{1}{T}\int_{-\infty}^{\infty}f(x)e^{-i\xi x}\mathrm{d}x=\hat{f}(\xi)$$ becomes apparent. This would imply that the discrete inverse fourier transform can be represented using real functions, as that is where the derivation for the complex form of the fourier series came from. While this connection may not be completly correct, I do hope that it is understandable where my question originated:

Is it possible to express the continous inverse fourier transfrom as well as the continous fourier transform using real functions?

• Fourier Integral is the real representation of FT. math.stackexchange.com/questions/400679/… – Kiryl Pesotski Nov 26 '17 at 15:26
• Use $e^{i \omega x} = \cos(\omega x)+i \sin(\omega x)$, assume $f$ is real (so that $\hat{f}(\omega) = \overline{\hat{f}(-\omega)}$) and look at $\Re(\hat{f}(\omega))$ and $\Im(\hat{f}(\omega))$ separately... – reuns Nov 26 '17 at 15:36

## 1 Answer

If $f$ is an even function, then \begin{align} f(x) &= \frac{1}{2\pi}\int_{-\infty}^{\infty}\left(\int_{0}^{\infty}f(t)e^{-ist}ds\right)e^{isx}ds \\ &= \frac{1}{2\pi}\int_{-\infty}^{\infty}\left(2\int_{0}^{\infty}f(t)\cos(st)dt\right)e^{isx}ds \\ &= \frac{2}{\pi}\int_{0}^{\infty}\left(\int_{0}^{\infty}f(t)\cos(st)dt\right)\cos(sx)ds. \end{align} If $f$ is an odd function, then $$f(x) = \frac{2}{\pi}\int_{0}^{\infty}\left(\int_{0}^{\infty}f(t)\sin(st)dt\right)\sin(sx)ds.$$