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In the above figure, consider the path $\gamma\equiv \gamma_1+\gamma_{12}+\gamma_2+\gamma_{21}$. Now write $z=Re^{i\theta}$. On an arc, $dz=iRe^{i\theta }\,d\theta $ and on the x-Axis, $dz=e^{i\theta}\,dR$. Write
\[ \int_{-\infty}^\infty {\sin x\over x}\,dx = \Im \int_\gamma {e^{iz}\over z}\,dx, \]
where $\Im$ denotes the Imaginary Part. Now define
\begin{align*}I&\equiv\int_\gamma {e^{iz}\over z}\,dz\\
&=\lim_{R_1\to 0} \int_\pi^0 {\mathop{\rm exp}\nolimits (iR_1e^{i\theta })\over R_1e^{i\theta }}\, i\theta R_1 e^{i\theta }\,d\theta\\
&\phantom=\mathop{+} \lim_{R_1\to 0}\lim_{R_2\to\infty} \int_{R_1}^{R_2} {e^{iR}\over R}\,dR\\
&\phantom= \mathop{+} \lim_{R_2\to\infty} \int_0^\pi {\mathop{\rm exp}\nolimits (iz)\over z}\,dx + \lim_{R_1\to 0} \int_{R_2}^{R_1} {e^{-iR}\over -R} \,(-dR),\end{align*}
where the second and fourth terms use the identities $e^{i0}=1$ and $e^{i\pi}=-1$. Simplifying,
\begin{align*}I&=\lim_{R_1\to 0} \int_\pi^0 \mathop{\rm exp}\nolimits (iR_1e^{i\theta})i\theta \,d\theta+\int_{0^+}^\infty {e^{iR}\over R}\,dR\\
& \phantom{=}+ \lim_{R_2\to \infty} \int_0^\pi {\mathop{\rm exp}\nolimits (iz)\over z}\,dz+ \int_\infty^{0^+} {e^{-iR}\over -R}\,(-dR)\\&= -\int_0^\pi i\theta \,d\theta +\int_{0^+}^\infty {e^{iR}\over R}\, dR + 0+\int^{0^-}_{-\infty} {e^{iR}\over R}\,dR,\end{align*}
where the third term vanishes by Jordan's Lemma. Performing the integration of the first term and combining the others yield
\[ I=-i\pi+\int_{-\infty}^\infty {e^{iz}\over z}\, dz = 0. \]
Rearranging gives
\[ \int_{-\infty}^\infty {e^{iz}\over z}\,dz = i\pi, \]
so
\[ \int_{-\infty}^\infty {\sin z\over z}\,dz = \pi. \]
The same result is arrived at using the method of Residues by noting
\begin{align*}I&=0+{\textstyle{1\over 2}}2\pi i\,{\rm Res} [f(z)]_{z=0}\\
&=i\pi\left[{(z-0) {e^{iz}\over z}}\right]_{z=0}\\&= i\pi [e^{iz}]_{z=0}\\
&=i\pi,\end{align*}
so
\[ \Im(I)=\pi. \]
Since the integrand is symmetric, we therefore have
\[ \int_0^\infty {\sin x\over x}\,dx={\textstyle{1\over 2}}\pi, \]
giving the Sine Integral evaluated at 0 as
\[ \mathop{\rm si}(0)=-\int_0^\infty {\sin x\over x}\,dx=-{\textstyle{1\over 2}}\pi. \] |
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