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

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 $$0$$?

• As you have noticed, the claim is wrong. The integrand converges to $i$ uniformly in $\theta$, therefore the integral tends to $\int_\pi^0 i \,d\theta$. – 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 $$\text{p.v.}\int \frac{e^{-2\pi isz}}{z}dz=-i\pi \cdot\text{sgn}(s)$$

• 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 $0$. – Andy Walls Jan 25 '19 at 13:53
• @AndyWalls I've edited my answer explaining what you have asked. I hope this will help ... – Myeonghyeon Song Jan 25 '19 at 14:37