应用留数定理计算实积分 \dps{I(x)=\int_{-1}^1\frac{\rd t}{\sqrt{1-t^2}(t-x)}\ (|x|>1,x\in\bbR)} [华中师范大学2010年复变函数复试试题]
解答: \beex \bea
I(x)&=\int_{-1}^1 \frac{\rd t}{\sqrt{1-t^2}(t-x)}\\
&=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\rd
\tt}{\sin\tt-x}\quad(t=\sin\tt)\\ &=\int_{\frac{\pi}{2}}^{\frac{3\pi}{2}}
\frac{\rd \tau}{\sin\tau-x}\quad(\pi-\tt=\tau)\\
&=\frac{1}{2}\sez{\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}
+\int_{\frac{\pi}{2}}^{\frac{3\pi}{2}}\frac{\rd \tt}{\sin\tt-x}}\\
&=\frac{1}{2}\int_{-\frac{\pi}{2}}^{\frac{3\pi}{2}} \frac{\rd
\tt}{\sin\tt-x}\\
&=\frac{1}{2}\int_{|z|=1}\frac{1}{\frac{z-z^{-1}}{2i}-x}\cdot \frac{\rd
z}{iz}\\ &=\int_{|z|=1}\frac{\rd z}{z^2-2ixz-1}\\ &=\sedd{\ba{ll} 2\pi
i\cdot \underset{z=i(x+\sqrt{x^2-1})}{\Res}\cfrac{1}{z^2-2ixz-1},&x<-1\\
2\pi i\cdot
\underset{z=i(x-\sqrt{x^2-1})}{\Res}\cfrac{1}{z^2-2ixz-1},&x>1 \ea}\\
&=\sedd{\ba{ll} \cfrac{\pi}{\sqrt{x^2-1}},&x<-1\\
-\cfrac{\pi}{\sqrt{x^2-1}},&x>1 \ea}\\
&=-\frac{\pi}{x\sqrt{1-\frac{1}{x^2}}}. \eea \eeex
原文：http://www.cnblogs.com/zhangzujin/p/3593997.html