In his derivation of the Fourier Transform of $\dfrac{1}{x}$, Bracewell starts with

$$\mathscr{F}\left\{\dfrac{1}{x}\right\} = P.V. \int_{-\infty}^{\infty}{\dfrac{e^{-i2\pi sx}}{x}}dx$$

And goes on to consider the contour integral with the expected semicircular contour (but Bracewell omits specifying which half-plane, upper or lower) with a small semicircular arc excursion around the pole at the origin:

$$\int_{C}{\dfrac{e^{-i2\pi sz}}{z}}dz$$

When evaluating the integral for the semi-circular arc around the origin, with $\epsilon$ a vanishingly small positive number, Bracewell states the integral "equals $\pm i\pi$ according to the sign of s", so we have

$$\int_{\pi}^{0}{ie^{-i2\pi s \epsilon e^{i\theta}}d\theta} = i\pi \space\mathrm{sgn}(s)$$

Could someone please show me in more detail how the $\mathrm{sgn}(s)$ arises in the evaluation of this integral? You can assume $\epsilon \rightarrow 0$ in the limit.

Is it that the sign of $s$ dictates which half plane, upper or lower, the contour should be in, and hence affects this integral around this small semi-circular arc around $0$?

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    $\begingroup$ As you have noticed, the claim is wrong. The integrand converges to $i$ uniformly in $\theta$, therefore the integral tends to $\int_\pi^0 i \,d\theta$. $\endgroup$ – Maxim Jan 25 '19 at 18:27

Edited: Your guess is right, we should take upper semi-circle contour when $\text{sgn}(s)<0$ and lower one when $\text{sgn}(s)>0$. It is related to the behavior of the integral $$ \int_{C_R^+} \frac{e^{isz}}{z}dz,\quad s=\pm 1 $$ as $R\to\infty$ where $C_R^+: Re^{it}, 0\le t\le \pi$ is an upper semi-circle contour. When we choose this contour, we expect that the limit of the above integral over $C_R^+$ tends to $0$ so that $\int_{-\infty}^\infty \frac{e^{\pm iz}}{z}dz$ can be computed using the singularity at $z=0$. When $s=1$, we can do it since $$ \int_{C_R^+} \frac{e^{iz}}{z}dz=i\int_0^\pi e^{iR\cos t-R\sin t}dt\stackrel{R\to\infty}\longrightarrow 0. $$ This happens essentially because $|e^{iz}|=e^{-\text{Im}(z)}=e^{-R\sin t}\to 0$ as $R\to\infty$. However, when $s=-1$, it is not the case anymore because $|e^{-iz}|=e^{\text{Im}(z)}=e^{R\sin t}\to \infty$ for $t\ne 0,\pi$, making us unable to control $\int_{C_R^+} \frac{e^{-iz}}{z}dz$ as $R\to \infty$. In this case, it is natural to take the lower semi-circle contour $C_R^-$to have $\int_{C_R^-} \frac{e^{isz}}{z}dz \to 0$. I hope this will help you.

We have $$ \text{p.v.}\int \frac{e^{-2\pi isz}}{z}dz=\text{p.v.}\int \frac{\cos (2\pi sz)-i\sin(2\pi sz)}{z}dz=-i\text{p.v.}\int \frac{\sin(2\pi sz)}{z}dz $$ since $z\mapsto \frac{\cos(2\pi sz)}{z}$ is an odd function. Note that if $a>0$, then by making change of variables $t=ax$, $$ \int_{-\infty}^\infty \frac{\sin ax}{x}dx=\int_{-\infty}^\infty \frac{\sin t}{t}dt=\pi $$ and if $a=-|a|<0$, then $$ \int_{-\infty}^\infty \frac{\sin ax}{x}dx=-\int_{-\infty}^\infty \frac{\sin |a|x}{x}dx=-\pi. $$ Therefore it follows $$ \text{p.v.}\int \frac{e^{-2\pi isz}}{z}dz=-i\pi \cdot\text{sgn}(s) $$

  • $\begingroup$ Have an upvote for a clear and concise derivation of the particular Fourier Transform given in the background information. However, the question is how sgn() could emerge from the integral in the title. I now believe that it can't. It is likely Bracewell was hand-waving over the detail of having to use two different closed contours depending on the sign of $s$, which is required to get the contour integral along the infinitely large semi-circular arc to vanish to $0$. $\endgroup$ – Andy Walls Jan 25 '19 at 13:53
  • $\begingroup$ @AndyWalls I've edited my answer explaining what you have asked. I hope this will help ... $\endgroup$ – Myeonghyeon Song Jan 25 '19 at 14:37

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