# 利用留数定理计算定积分（四）：积分路径上有奇点的情形

$\int_0^{\infty}\frac{\sin x}{x}dx=\frac{1}{2}\int_{-\infty}^{\infty}$

\begin{array}{ll}\Gamma_R: |z|=R, \text{Im}z>0, &\Gamma_{\epsilon}:|z|=\epsilon, \text{Im}z>0\\ \Gamma_1: -R\le x\le -\epsilon,&\Gamma_2:\epsilon \le x\le R\end{array}

$\oint_{\Gamma}\frac{e^{iz}}{z}=0$也就是

$\left(\int_{\Gamma_R}+\int_{\Gamma_1}+\int_{\Gamma_{\epsilon}}+\int_{\Gamma_2}\right)\frac{e^{iz}}{z}=0$

\begin{align*}\left(\int_{\Gamma_1}+\int_{\Gamma_2}\right)\frac{e^{iz}}{z}&=\int_{-R}^{-\epsilon}\frac{e^{ix}}{x}dx+\int_{\epsilon}^{R}\frac{e^{ix}}{x}dx\\ &=\int_{\epsilon}^{R}\frac{e^{ix}-e^{-ix}}{x}dx\\ &=2i\int_{\epsilon}^{R}\frac{\sin x}{x}dx\end{align*}

$\frac{e^{iz}}{z}=\frac{1}{z}(1+iz+\frac{(iz)}{2}+\cdots)=\frac{1}{z}+h(z)$

\begin{align*}\int_{\Gamma_{\epsilon}}\frac{e^{iz}}{z}dz&=\int_{\Gamma_{\epsilon}}\frac{1}{z}dz+\int_{\Gamma_{\epsilon}}h(z)dz\\ &=\int_{\pi}^0e^{-i\theta}ie^{i\theta}d\theta+\int_{\Gamma_{\epsilon}}h(z)dz\\ &=-\pi i+\int_{\Gamma_{\epsilon}}h(z)dz\end{align*}

$\left|\int_{\Gamma_{\epsilon}}h(z)dz\right|\le \int_0^{\pi}|h(z)| |i\epsilon e^{i\theta}|d\theta\le M\pi \epsilon\to 0$

\begin{align*}\left|\int_{\Gamma_R}\frac{e^{iz}}{z}dz\right|&\le \frac{1}{R}\left|\int_{\Gamma_R}e^{iz}dz\right|=\frac{1}{R}\left|\int_{\Gamma_R}e^{iR\cos \theta-R\sin\theta}iRe^{i\theta}d\theta\right|\\ &\le \int_0^{\pi}e^{-R\sin\theta}Rd\theta=\frac{2}{R}\int_0^{\frac{\pi}{2}}e^{-R\sin\theta}Rd\theta\\ &\le 2\int_0^{\frac{\pi}{2}}e^{-R\cdot\frac{2}{\pi}\theta}=-\frac{2\pi}{2R}e^{-R\cdot\frac{2}{\pi}\theta}\Big|_0^{\frac{\pi}{2}}\\ &=\frac{1}{R}(1-e^{-R})\to 0, R\to\infty\end{align*}

$\left(\int_{\Gamma_R}+\int_{\Gamma_1}+\int_{\Gamma_{\epsilon}}+\int_{\Gamma_2}\right)\frac{e^{iz}}{z}=0$

$\left(\int_{\Gamma_1}+\int_{\Gamma_2}\right)\frac{e^{iz}}{z}=-\left(\int_{\Gamma_R}+\int_{\Gamma_{\epsilon}}\right)\frac{e^{iz}}{z}$

$2i\int_{\epsilon}^{R}\frac{\sin x}{x}dx=\pi i$

$\int_{\epsilon}^{R}\frac{\sin x}{x}dx=\frac{\pi}{2}$