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I am trying to find the value of $\int_{-\infty}^{\infty} \frac{\sin (x)}{x}$ using residue theorem and a contour with a kink around $0$. For this, I need to find $\int_{C_\epsilon} \frac {e^{iz}} {z}$ where $C_\epsilon$ is the semicircle centred at $0$ with radius $\epsilon$ from $-\epsilon$ to $\epsilon$. I guess it is equal to half the residue of $\frac {e^{iz}} {z}$ at $0$. Is this true? Any help is appreciated.

Shiva
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3 Answers3

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There is a theorem sometimes referred to as the fractional residue theorem.

If $z_{0}$ is a simple pole of $f(z)$, and $C_{\epsilon}$ is an arc of the circle $|z-z_{0}| = \epsilon$ of angle $\alpha$, then $\displaystyle\lim_{\epsilon \to 0} \int_{C_{\epsilon}} f(z) \ dz = \alpha i \ \text{Res} [f(z),z_{0}]$.

Random Variable
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  • Please give some "nice to read" reference to Fractional Residue Theorem. – MathArt Sep 09 '20 at 09:41
  • @MathArt Theodore Gamelin refers to it as the fractional residue theorem in his textbook on complex analysis. Most complex analysis textbooks mention it, but I don't think they give it a name. The proof is just a simple generalization of robjohn's answer. Also look at Shiva's comment under robjohn's answer. – Random Variable Sep 09 '20 at 13:39
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Note that $\dfrac{e^{iz}}{z}=\dfrac1z+O(1)$. Integrating this counter-clockwise around the semicircle of radius $\epsilon$ is $$ \begin{align} \int_\gamma\frac{e^{iz}}{z}\,\mathrm{d}z &=\int_0^\pi\left(\frac1\epsilon e^{-i\theta}+O(1)\right)\,\epsilon\,ie^{i\theta}\,\mathrm{d}\theta\\ &=\int_0^\pi\frac1\epsilon e^{-i\theta}\,\epsilon\,ie^{i\theta}\,\mathrm{d}\theta +\int_0^\pi O(1)\,\epsilon\,ie^{i\theta}\,\mathrm{d}\theta\\[9pt] &=\pi i+O(\epsilon)\\[12pt] \end{align} $$ The residue at $0$ is $1$, so integrating around the full circle would give $2\pi i$.

robjohn
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  • We're integrating clockwise about the origin. – Random Variable Apr 20 '13 at 20:45
  • @RandomVariable: If you're integrating clockwise, then negate the answer. Where does it say which direction? – robjohn Apr 20 '13 at 23:21
  • I assume he's integrating counterclockwise on the big half-circle and thus clockwise on the small half-circle. – Random Variable Apr 21 '13 at 00:17
  • @RandomVariable: there are 2 possibilities: including the singularity (counter-clockwise bump) or excluding the singularity (clockwise bump). – robjohn Apr 21 '13 at 00:25
  • Okay. So, once we know that $\frac {e^{iz}}{z}=\frac{1}{z}+g(z)$, where $g$ is holomorphic, we can separate the integral into two parts and since $g$ can be bounded, as $\epsilon$ goes to zero, $\int_{C_{\epsilon}} g(z)$ goes to zero and the other term can be found easily. Thank you for making this clear. – Shiva Apr 22 '13 at 06:00
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Another approach you can use is Jordan's lemma. They provided you the hint to use $\int_{C_\epsilon} \frac {e^{iz}} {z}$ because the integral $\int_{-\infty}^{\infty} \frac{\sin (x)}{x}$ is an improper integral and does not satisfy the conditions of p.v. We can use Jordan's lemma here. Thus we have $$\int_{C_\epsilon} \frac {e^{iz}} {z}$$ take the limit of the pole, which is at $z=0$. Apply Jordan lemma, and since we used the identity $e^{iz}$ to replace $\sin(z)$, you will take the imaginary parts and then you are through.

Lays
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