So over at the Signal Processing Stack Exchange, I posted this question and the background behind it, of interest to electrical engineers and signal processing geeks. If you go there, remember that EEs like to call the imaginary unit "$j$" instead of "$i$". Also we like this "unitary" definition of the Fourier Transform (with "$f$" instead of "$\omega$") because of the symmetry between forward and inverse Fourier Transform and the Duality theorem.

But here is the mathematical bottom line... (keep in mind that we gotta use this Cauchy principal value for the Hilbert transform integral.) Without using complex variable analysis (i.e. contour integration or residue theory or anything else in Levison and Redheffer), can someone prove this fact? :

$$\begin{align} \arctan(\omega) &= -\mathscr{H}\big\{ \tfrac12 \log(1 + \omega^2) \big\} \\ \\ &= \frac{-1}{2\pi} \lim_{\epsilon \to 0^+} \int\limits_{-\infty}^{-\epsilon} \frac{\log(1 + (\omega-x)^2)}{x} \, dx + \int\limits_{+\epsilon}^{\infty} \frac{\log(1 + (\omega-x)^2)}{x} \, dx \\ \\ &= \frac{-1}{2\pi} \lim_{\epsilon \to 0^+} \int\limits_{-\infty}^{\omega-\epsilon} \frac{\log(1 + x^2)}{\omega-x} \, dx + \int\limits_{\omega+\epsilon}^{\infty} \frac{\log(1 + x^2)}{\omega-x} \, dx \\ \end{align}$$

($\log(\cdot)$ is natural, base-$e$, of course.)

I have an approach where maybe I can show that the derivatives of $\arctan(\omega)$ and $\tfrac12 \log(1 + \omega^2)$ are a Hilbert pair. Is there another or better or more direct way, using just calculus (without more advanced mathematics), to do it? The cleanest, simpliest proof with the least advanced post-calculus mathematics (like something I can explain to bonehead electrical engineers) is preferred.

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    $\begingroup$ What about non bonehead electrical engineers? $\endgroup$ – copper.hat Mar 23 '17 at 6:13
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    $\begingroup$ then there is all sortsa handwaving regarding the Hilbert Transform. $\endgroup$ – robert bristow-johnson Mar 23 '17 at 6:17

The following is not really rigorous but may be along the lines of what you want, and the same trick can be used quite often in practice. The correct way to justify these steps is via complex analysis (in particular analytic continuation) but this will do for motivation.

The famous "Feynman trick" seems to work here. If we throw in a dummy variable $y$, for example $$f(w;y) := -\frac{1}{2\pi} \int_{-\infty}^{\infty} \frac{\log(y + x^2)}{w - x} \, \mathrm{d}x$$ (this denoting the principal value) and differentiate with respect to $y$, swapping $\frac{\partial}{\partial y}$ with $\int \, \mathrm{d}x$, we get $$\partial_y f = -\frac{1}{2\pi} \int_{-\infty}^{\infty} \frac{1}{(w-x)(y+x^2)} \, \mathrm{d}x.$$ This can be solved by partial fractions decomposition, namely $$\frac{1}{(w-x)(y+x^2)} = \frac{1}{w^2 + y} \Big( \frac{1}{w-x} + \frac{w+x}{x^2 + y} \Big).$$ The Cauchy principal value of the first term is $0$; and the second can be integrated directly to give $$\int_{-\infty}^{\infty} \frac{1}{(w-x)(y+x^2)} \, \mathrm{d}x = \frac{1}{w^2 + y} \Big( \frac{w}{\sqrt{y}} \mathrm{arctan}(x / \sqrt{y}) + \frac{1}{2} \log(x^2 + y) \Big)\Big|_{-\infty}^{\infty} = \frac{\pi w}{(w^2 + y)\sqrt{|y|}} \mathrm{sgn}(y),$$ again as a Cauchy principal value. The integral of this with respect to $y$ is again elementary and $$f(w;y) = -\mathrm{sgn}(y)\mathrm{arctan}(\sqrt{|y|}/w) + C.$$ We find $C$ by letting $y \rightarrow -\infty$, which sends the integral to $0$ (again in a rather weird "principal value" sense, essentially since the integral of $\frac{1}{w-x}$ itself is $0$ and $\log(y+x^2)$ becomes independent of $x$); so $$0 = \lim_{y \rightarrow -\infty} f(w;y) = -\pi/2 + C.$$ It follows that $$-\frac{1}{2\pi} \int_{-\infty}^{\infty} \frac{\log(1 + x^2)}{w - x} \, \mathrm{d}x = f(w;1) = \frac{\pi}{2} - \mathrm{arctan}(1/w) = \mathrm{arctan}(w).$$

  • $\begingroup$ not bad for someone with a rep=1. did you just stumble by here? $\endgroup$ – robert bristow-johnson Mar 23 '17 at 7:12
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    $\begingroup$ @robertbristow-johnson I answer questions in Chrome's "incognito mode" and tend to lose my accounts again and again. I don't collect rep, just here for fun $\endgroup$ – user428512 Mar 23 '17 at 7:13
  • $\begingroup$ hey, now you got 11. $\endgroup$ – robert bristow-johnson Mar 23 '17 at 7:14
  • $\begingroup$ Worth noting this trick is called Differentiation Under the Integral Sign, and predates it's populariser by some centuries. $\endgroup$ – Chappers Mar 23 '17 at 13:56

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