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I have noticed that the Fourier integral $\int_{-\infty}^\infty \mathscr{F}f(\omega) e^{i\omega t} d\omega $ of $f(t) = \mu(t) e^{-t}$, where $\mathscr{F}f(\omega) = \frac{1}{2\pi} \int_{-\infty}^\infty f(t) e^{-i\omega t} dt$ and $\mu(t)$ is the step function diverges at $t = 0$. Is there a beautiful notation using the Riemann-integral that fixes the problem? Here I mean the improper Riemann-integral of the first kind. I have considered the operator pair \begin{eqnarray} \mathscr{F}f(\omega) & = & \frac{1}{2\pi} \int_0^\infty (f(t) e^{-i\omega t} + f(-t) e^{i\omega t}) dt \\ \mathscr{F}'F(t) & = & \int_0^\infty (F(\omega) e^{i\omega t} + F(-\omega) e^{-i\omega t}) d\omega \ , \end{eqnarray} that is defined for functions $f$ that satisfy i) $\mathscr{F}f$ exists everywhere, ii) $\mathscr{F}' \mathscr{F}f$ exists everywhere, iii) $\mathscr{F}' \mathscr{F} f = f$. The Cauchy principal value would be ideal, but I don't quite like the notation P.V. in front of the integral and it might allow functions to satisfy all conditions without being locally bounded. The limit form $\lim_{M \rightarrow \infty} \int_{-M}^M f(x) dx$ isn't good either because it isn't notationally optimal for applications. I mean that for example the formula $De^x = e^x$ is simple but $D \lim_{n \rightarrow \infty} \big(1+\frac{x}{n}\big)^n = \lim_{n \rightarrow \infty} \big(1+\frac{x}{n}\big)^n$ is a little more complicated. I would want to find good expressions for Fourier and inverse Fourier transform that exist everywhere for some important functions familiar from signal processing including $\chi_{[-1,1]}(x)$ and $\textrm{sinc}(x)$.

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    $\begingroup$ Sorry, what is the question? $\endgroup$ – lcv Jun 7 '13 at 7:31
  • $\begingroup$ Is there a beautiful notation using the Riemann-integral that fixes the problem? $\endgroup$ – Juha-Matti Vihtanen Jun 7 '13 at 17:48
  • $\begingroup$ I guess I did not understand what the problem is. $\endgroup$ – lcv Jun 8 '13 at 19:49
  • $\begingroup$ The problem is that the imaginary part $\frac{1}{2\pi} \frac{-\omega}{1+\omega^2}$ of $\mathscr{F}f$, where $f(t)=\mu(t)e^{-t}$ causes the Fourier integral to diverge at $0$ with the conventional definitions of transform and inverse transform. $\endgroup$ – Juha-Matti Vihtanen Jun 20 '13 at 20:09
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I think you end up defining $f(t)=1/2$ for $t=0$. I will explain why below.

$$\hat{f}(\omega) = \int_{-\infty}^{\infty} dt \, f(t) \, e^{i \omega t} = \frac{1}{1-i \omega}$$

To recover $f(t)$, use the inverse transform:

$$\frac{1}{2 \pi} \int_{-\infty}^{\infty} d\omega \, \frac{e^{-i \omega t}}{1-i \omega}$$

Consider the following integral in the complex plane:

$$\oint_C dz \, \frac{e^{-i z t}}{1-i z}$$

When $t>0$, $C$ is a positively oriented semicircular contour in the lower half plane of radius $R$. The integral about the circular arc vanishes as $R \to \infty$. The integral over $C$ is $i 2 \pi$ times the residue at the pole at $z=-i$:

$$-\frac{1}{2 \pi} \int_{-\infty}^{\infty} d\omega \, \frac{e^{-i \omega t}}{1-i \omega} = \frac{i 2 \pi}{2 \pi i} e^{-i (-i) t} = e^{-t}$$

When $t<0$, $C$ is a semicircular contour in the upper half plane. Since there are no poles there, the integral there is zero.

When $t=0$, neither contour leads to convergence. In this case, we impose a sort of "analytic" continuation at $t=0$, even though we make no pretense at continuity at $t=0$. Rather, we impose the condition that

$$f(0) = \frac12 [f(0^-) + f(0^+)]$$

where $f(0^-)=0$ and $f(0^+)=1$. This then allows us to write $f(t)=\mu(t) e^{-t}$.

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  • $\begingroup$ Thanks for your answer. Yes, I have noticed also that the inverse transform should converge to $1/2$ at $0$. There is however a problem with the definition of integral because the integral of the imaginary part $\frac{\omega}{1+\omega^2}$ diverges. Now even the Lebesgue integral doesn't help. So I thougth to define the transform with a special expression. $\endgroup$ – Juha-Matti Vihtanen Jun 7 '13 at 18:08
  • $\begingroup$ So I thought to define the transform-inverse transform pair with special expressions. Note that by linearity we may analyze the imaginary part alone. Define now $f(\omega) = \frac{\omega}{1+\omega^2}$. Then $\mathscr{F}'f(0) = \int_0^\infty 0 d\omega = 0$. $\endgroup$ – Juha-Matti Vihtanen Jun 7 '13 at 18:19
  • $\begingroup$ For the real part $f(\omega) = \frac{1}{2\pi} \frac{1}{1+\omega^2}$ $\mathscr{F}'f(0) = \int_0^\infty \frac{1}{2\pi} \frac{2}{1+\omega^2} d\omega = \frac{1}{2}$. Yet corrections to the previous comments. The imaginary part equals to $\frac{1}{2\pi} \frac{\omega}{1+\omega^2}$. That's because of the definition of Fourier transform in our Complex analysis book "Complex Analysis for Mathematics and Engineering". $\endgroup$ – Juha-Matti Vihtanen Jun 7 '13 at 18:35
  • $\begingroup$ @Juha-MattiVihtanen: when confronted with a situation like this, a divergence at one point, I like to impose "continuity" in a Fourier sense. Think of it as some sort of analytic continuation. $\endgroup$ – Ron Gordon Jun 7 '13 at 18:45
  • $\begingroup$ It seems to me that we miss a minus sign. The transform $\mathscr{F}f(\omega) = \frac{1}{2\pi} \frac{1}{i+\omega}$ and its imaginary part is $\frac{1}{2\pi} \frac{-\omega}{1+\omega^2}$ using the definitions above that are equivalent to the definitions in my Comlpex nalysis book.. $\endgroup$ – Juha-Matti Vihtanen Jun 7 '13 at 18:54

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