应用留数定理计算实积分 $dps{I(x)=int_{-1}^1frac{ d t}{sqrt{1-t^2}(t-x)} (|x|>1,xinbR)}$ [华中师范大学2010年复变函数复试试题]
解答: $$eex ea I(x)&=int_{-1}^1 frac{ d t}{sqrt{1-t^2}(t-x)}\ &=int_{-frac{pi}{2}}^{frac{pi}{2}} frac{ d t}{sin t-x}quad(t=sin t)\ &=int_{frac{pi}{2}}^{frac{3pi}{2}} frac{ d au}{sin au-x}quad(pi- t= au)\ &=frac{1}{2}sez{int_{-frac{pi}{2}}^{frac{pi}{2}} +int_{frac{pi}{2}}^{frac{3pi}{2}}frac{ d t}{sin t-x}}\ &=frac{1}{2}int_{-frac{pi}{2}}^{frac{3pi}{2}} frac{ d t}{sin t-x}\ &=frac{1}{2}int_{|z|=1}frac{1}{frac{z-z^{-1}}{2i}-x}cdot frac{ d z}{iz}\ &=int_{|z|=1}frac{ d z}{z^2-2ixz-1}\ &=sedd{a{ll} 2pi icdot underset{z=i(x+sqrt{x^2-1})}{Res}cfrac{1}{z^2-2ixz-1},&x<-1\ 2pi icdot underset{z=i(x-sqrt{x^2-1})}{Res}cfrac{1}{z^2-2ixz-1},&x>1 ea}\ &=sedd{a{ll} cfrac{pi}{sqrt{x^2-1}},&x<-1\ -cfrac{pi}{sqrt{x^2-1}},&x>1 ea}\ &=-frac{pi}{xsqrt{1-frac{1}{x^2}}}. eea eeex$$