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I want to calculate the integral of the complex function $\dfrac 1 {\sin\frac{1}{z}} $ along the curve $|z|=\frac{1}{5}$. I wanted to use cauchy's residues theorem but the function has an essential singularity at $z=0$ which lies inside the curve so I don't know what to do.

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  • $\begingroup$ What is the function $sen$? $\endgroup$ – Amitai Yuval Jun 25 '17 at 3:53
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    $\begingroup$ $sen$ is $\sin$ in $\texttt{spanish}$. $\endgroup$ – Felix Marin Jun 25 '17 at 4:03
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\oint_{\verts{z} = 1/5}{\dd z \over \sin\pars{1/z}} \,\,\,\stackrel{z\ \mapsto\ 1/z}{=}\,\,\, \oint_{\verts{z} = 5}{\dd z \over z^{2}\sin\pars{z}} \\[5mm] & = 2\pi\ic\braces{% \lim_{z \to 0}{1 \over 2!}\totald[2]{}{z}\bracks{z \over \sin\pars{z}} + \lim_{z \to -\pi} \bracks{z + \pi \over z^{2}\sin\pars{z}} + \lim_{z \to \pi}\bracks{z - \pi \over z^{2}\sin\pars{z}}} \\[5mm] & = 2\pi\ic\pars{{1 \over 6} + {-1\phantom{^{2}} \over \pi^{2}} + {-1\phantom{^{2}} \over \pi^{2}}} = \bbx{\pars{{1 \over 3}\,\pi - {4 \over \pi}}\ic} \end{align}

There are three poles inside the contour $\ds{\braces{z\ \mid\ \verts{z} = 5}}$: single ones at $\ds{\pm\pi}$ and a third order one at $\ds{z = 0}$.

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  • $\begingroup$ yes but the integral I want is of the function 1/sin(1/z) $\endgroup$ – allizdog Jun 25 '17 at 4:09
  • $\begingroup$ @allizdog Sorry. I'll try to fix it. $\endgroup$ – Felix Marin Jun 25 '17 at 4:10
  • $\begingroup$ no problem thanks any way, hadn't tried anything like that $\endgroup$ – allizdog Jun 25 '17 at 4:11
  • $\begingroup$ @allizdog Fixed. $\endgroup$ – Felix Marin Jun 25 '17 at 4:22
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    $\begingroup$ @AmitaiYuval $\color{#f00}{\mathrm{N0}}$. $z \mapsto 1/z$ makes the integral a 'clockwise' one. So, we have to multiply by $-1$ to rewrite the integral as a 'counterclockwise' one which is required by Cauchy$\ldots$ $\endgroup$ – Felix Marin Jun 25 '17 at 4:48

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