5.求积分∮Cezzdz (C:∣z∣=1)，从而证明∫0πecosθcos(sinθ)dθ=π。解：令f(z)=ez，则f(z)=ez=12πi∮Ceζζ−zdζf(0)=1=12πi∮Ceζζdζ∴∮Cezzdz=2πi∵z在圆周∣z∣=1上，可令z=eiθ，则∮Cezzdz=∫02πeeiθeiθ⋅ieiθdθ=∫02πieeiθdθ=∫02πie(cosθ+isinθ)dθ=∫02πiecosθ[cos(sinθ)+isin(sinθ)]dθ=∫02π[iecosθcos(sinθ)−ecosθsin(sinθ)]dθ=2πi（旁注：ecosθsin(sinθ)是奇函数，周期2π→0）对比实部和虚部可得。∫02πecosθcos(sinθ)dθ=2π被积函数是个偶函数，且周期为2π∴∫0πecosθcos(sinθ)dθ=π\begin{aligned} &5.求积分\oint_{C}\frac{e^{z}}{z}dz\ (C:|z|=1)，从而证明\\ &\int_{0}^{\pi}e^{\cos\theta}\cos(\sin\theta)d\theta=\pi。\\ &解：令f(z)=e^{z}，则\\ &f(z)=e^{z}=\frac{1}{2\pi i}\oint_{C}\frac{e^{\zeta}}{\zeta-z}d\zeta\\ &f(0)=1=\frac{1}{2\pi i}\oint_{C}\frac{e^{\zeta}}{\zeta}d\zeta\\ &\therefore\oint_{C}\frac{e^{z}}{z}dz=2\pi i\\ &\because z在圆周|z|=1上，可令z=e^{i\theta}，则\\ &\oint_{C}\frac{e^{z}}{z}dz=\int_{0}^{2\pi}\frac{e^{e^{i\theta}}}{e^{i\theta}}\cdot ie^{i\theta}d\theta\\ &=\int_{0}^{2\pi}ie^{e^{i\theta}}d\theta\\ &=\int_{0}^{2\pi}ie^{(\cos\theta+i\sin\theta)}d\theta\\ &=\int_{0}^{2\pi}ie^{\cos\theta}[\cos(\sin\theta)+i\sin(\sin\theta)]d\theta\\ &=\int_{0}^{2\pi}[ie^{\cos\theta}\cos(\sin\theta)-e^{\cos\theta}\sin(\sin\theta)]d\theta=2\pi i\\ &（旁注：e^{\cos\theta}\sin(\sin\theta)是奇函数，周期2\pi\to0）\\ &对比实部和虚部可得。\\ &\int_{0}^{2\pi}e^{\cos\theta}\cos(\sin\theta)d\theta=2\pi\\ &被积函数是个偶函数，且周期为2\pi\\ &\therefore\int_{0}^{\pi}e^{\cos\theta}\cos(\sin\theta)d\theta=\pi\\ \end{aligned}5.求积分∮Czezdz (C:∣z∣=1)，从而证明∫0πecosθcos(sinθ)dθ=π。解：令f(z)=ez，则f(z)=ez=2πi1∮Cζ−zeζdζf(0)=1=2πi1∮Cζeζdζ∴∮Czezdz=2πi∵z在圆周∣z∣=1上，可令z=eiθ，则∮Czezdz=∫02πeiθeeiθ⋅ieiθdθ=∫02πieeiθdθ=∫02πie(cosθ+isinθ)dθ=∫02πiecosθ[cos(sinθ)+isin(sinθ)]dθ=∫02π[iecosθcos(sinθ)−ecosθsin(sinθ)]dθ=2πi（旁注：ecosθsin(sinθ)是奇函数，周期2π→0）对比实部和虚部可得。∫02πecosθcos(sinθ)dθ=2π被积函数是个偶函数，且周期为2π∴∫0πecosθcos(sinθ)dθ=π
这个积分是第一类零阶修正贝塞尔函数的特殊值。