Let $\gamma(\vartheta)=\mathrm{e}^{i\vartheta},\,\vartheta\in[0,2\pi]$, and consider the integral $$I(n)=\int_\gamma \frac{\cos{z}}{z^n},$$ where $n\in \{0,2,4,6,...\}$.

Is there any way to prove that $I(n)=0$ for all $n$, only by looking at the integral formula of Cauchy for $\cos{z}$ and the winding number of $\gamma$, i.e. only by looking at the expressions $$\int_\gamma \frac{\cos{z}}{z},\ \int_\gamma\frac{1}{z}$$

Remark: by expanding $\cos{z}$ in series, I know how to calculate $I(n)$. My question is whether it is possible to avoid this way.

  • 2
    $\begingroup$ Ignoring constant factors, it's the $n-1$-th derivative of $\cos$ in $0$. $\cos$ is even, so all odd order derivatives vanish in $0$. $\endgroup$ – Daniel Fischer Feb 13 '14 at 16:35

According the Cauchy's Integral Formula for $f(z)=\cos z$: $$ \cos^{(n)}(0)=\frac{n!}{2\pi i}\int_{|z|=1}\frac{\cos z\,dz}{z^{n+1}}=\frac{n!}{2\pi i} I(n+1) $$ But $$ \cos^{(2k)}(0)=(-1)^{k}\cos 0\quad\text{while}\quad \cos^{(2k-1)}=(-1)^{k}\sin (0). $$ Thus $$I(2n)=(-1)^n\sin(0)\frac{2\pi i}{(2n-1)!}=0,$$ while $$I(2n+1)=(-1)^n\cos (0)\frac{2\pi i}{n!}=(-1)^n\frac{2\pi i}{(2n)!}.$$

  • $\begingroup$ Not that it affects the answer to the question, but shouldn't the first equation be $$\cos^{(n)}(0)=\frac{n!}{2\pi i}\int_{|z|=1}\frac{\cos z\,dz}{z^{n+1}} =\frac{n!}{2\pi i}I(n+1)$$ ? $\endgroup$ – robjohn Feb 15 '14 at 13:56
  • $\begingroup$ @robjohn: I corrected it. $\endgroup$ – Yiorgos S. Smyrlis Feb 15 '14 at 14:45
  • $\begingroup$ I adjusted the subsequent formulae. $\endgroup$ – robjohn Feb 15 '14 at 14:50

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