2.3 柯西积分公式【习题2.3 -5】柯西积分计算

5.求积分∮Cezzdz (C:∣z∣1)从而证明∫0πecos⁡θcos⁡(sin⁡θ)dθπ。解令f(z)ez则f(z)ez12πi∮Ceζζ−zdζf(0)112πi∮Ceζζdζ∴∮Cezzdz2πi∵z在圆周∣z∣1上可令zeiθ则∮Cezzdz∫02πeeiθeiθ⋅ieiθdθ∫02πieeiθdθ∫02πie(cos⁡θisin⁡θ)dθ∫02πiecos⁡θ[cos⁡(sin⁡θ)isin⁡(sin⁡θ)]dθ∫02π[iecos⁡θcos⁡(sin⁡θ)−ecos⁡θsin⁡(sin⁡θ)]dθ2πi旁注ecos⁡θsin⁡(sin⁡θ)是奇函数周期2π→0对比实部和虚部可得。∫02πecos⁡θcos⁡(sin⁡θ)dθ2π被积函数是个偶函数且周期为2π∴∫0πecos⁡θcos⁡(sin⁡θ)dθπ\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}dz2\pi i\\ \because z在圆周|z|1上可令ze^{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\thetai\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\theta2\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\theta2\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∮C​zez​dz(C:∣z∣1)从而证明∫0π​ecosθcos(sinθ)dθπ。解令f(z)ez则f(z)ez2πi1​∮C​ζ−zeζ​dζf(0)12πi1​∮C​ζeζ​dζ∴∮C​zez​dz2πi∵z在圆周∣z∣1上可令zeiθ则∮C​zez​dz∫02π​eiθeeiθ​⋅ieiθdθ∫02π​ieeiθdθ∫02π​ie(cosθisinθ)dθ∫02π​iecosθ[cos(sinθ)isin(sinθ)]dθ∫02π​[iecosθcos(sinθ)−ecosθsin(sinθ)]dθ2πi旁注ecosθsin(sinθ)是奇函数周期2π→0对比实部和虚部可得。∫02π​ecosθcos(sinθ)dθ2π被积函数是个偶函数且周期为2π∴∫0π​ecosθcos(sinθ)dθπ​这个积分是第一类零阶修正贝塞尔函数的特殊值。