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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?

Rodrigo de Azevedo
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Skydiver
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  • Fourier Integral is the real representation of FT. https://math.stackexchange.com/questions/400679/difference-between-fourier-integral-and-fourier-transform – Kiryl Pesotski Nov 26 '17 at 15:26
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    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

2 Answers2

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

Vivek Thazhathattil
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Disintegrating By Parts
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The "only reason" to use complex numbers is that they are easier to handle, the drawback is they are more difficult to interpret. Below I will show how you can make the transition from "a real Fourier transform" to the complex notation.
(forgive me if the expressions below are not correctly normalized)

The "real Fourier transform" looks like this: $$a{\left(\omega\right)}=2\int_{0}^{\infty}x{\left(t\right)}\cos{\left(\omega t\right)}$$ $$b{\left(\omega\right)}=2\int_{0}^{\infty}x{\left(t\right)}\sin{\left(\omega t\right)}$$ In fact these are the coefficients of the Fourier series when the period becomes infinite.

The "inverse real Fourier transform" looks like this: $$x{\left(t\right)}=\frac{1}{2\pi}\int_{0}^{\infty}\left(a{\left(\omega\right)}\cos{\left(\omega t\right)}+b{\left(\omega\right)}\sin{\left(\omega t\right)}\right)\ d\omega$$ This the Fourier series when the period becomes infinite.

The next expression nicely shows that you can interpret $\left(a{\left(\omega\right)}\cos{\left(\omega t\right)}+b{\left(\omega\right)}\sin{\left(\omega t\right)}\right)$ as a phase-shifted cosine:
$$x{\left(t\right)}=\frac{1}{2\pi}\int_{0}^{\infty}{c{\left(\omega\right)}\cos{\left(\omega t+\varphi\left(\omega\right)\right)}}\ d\omega$$ $$c{\left(\omega\right)}=\sqrt{{a{\left(\omega\right)}}^2+{b{\left(\omega\right)}}^2},\ \varphi\left(\omega\right)=atan2{\left(b{\left(\omega\right)},a{\left(\omega\right)}\right)}$$

Now we jump to complex numbers by replacing $cos()$ by its complex expression: $$x{\left(t\right)}=\frac{1}{2\pi}\int_{0}^{\infty}{c{\left(\omega\right)}\frac{\left(e^{j\left(\omega t+\varphi\left(\omega\right)\right)}+e^{-j\left(\omega t+\varphi\left(\omega\right)\right)}\right)}{2}}\ d\omega$$

$$x{\left(t\right)}=\frac{1}{2\pi}\int_{-\infty}^{\infty}{c{\left(\omega\right)}\frac{\left(e^{j\left(\omega t+\varphi\left(\omega\right)\right)}\right)}{2}}\ d\omega$$

$$x{\left(t\right)}=\frac{1}{2\pi}\int_{-\infty}^{\infty}{c{\left(\omega\right)e^{j\varphi\left(\omega\right)}}\ \frac{e^{j\omega t}}{2}}\ d\omega = x{\left(t\right)}=\frac{1}{2\pi}\int_{-\infty}^{\infty}{\left(a{\left(\omega\right)}+j\ b{\left(\omega\right)}\ \right)\ \frac{e^{j\omega t}}{2}}\ d\omega$$

So the (inverse) Fourier transform expressed in complex numbers is just a formulation that is a lot easier to handle (to derive, integrate, sum, multiply) than an expression in sines and cosines. It also allows squeezing two real numbers into one complex number: $a(\omega)$ and $b(\omega)$ are squeezed into one complex number $a(\omega) + b(\omega) i = c{\left(\omega\right)e^{j\varphi\left(\omega\right)}}$

Bart Vanderbeke
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