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 Post subject: Line integral 01Posted: Mon May 15, 2017 5:27 pm
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Joined: Mon Nov 09, 2015 1:36 am
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Location: Ioannina, Greece
Evaluate $$\displaystyle\oint_{\gamma}{{\rm{Log}}\,\big(z-\tfrac{1}{2}\big)\,dz}\,,$$ where $\gamma$ has parametric representation $$\gamma(t)=\displaystyle(1-t)\cos({t\pi})+\frac{1}{2}\,\big(-1+(2t-4)\sin(t\pi)\big)\,i\,,\quad t\in\big[-\tfrac{1}{2},\tfrac{1}{2}\big]\,.$$

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Grigorios Kostakos

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 Post subject: Re: Line integral 01Posted: Tue Jun 27, 2017 8:29 am
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Joined: Mon Nov 09, 2015 1:36 am
Posts: 443
Location: Ioannina, Greece
We give and a solution to that -not so difficult- exercise:

The complex function ${\rm{Log}}\,\big(z-\tfrac{1}{2}\big)$ is holomorphic on ${\mathbb{C}}\setminus\big\{z\in{\mathbb{C}}\;|\;\Re(z)\leqslant\tfrac{1}{2}\,,\; \Im({z})=0\big\}$.
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Because the curve $\gamma$ with parametric representation $\gamma(t)=\displaystyle(1-t)\cos({t\pi})+\frac{1}{2}\,\big(-1+(2t-4)\sin(t\pi)\big)\,i\,,\quad t\in\big[-\tfrac{1}{2},\tfrac{1}{2}\big]\,.$ lies entirely on a simply connected, open set, which does not contains the half line $\big\{z\in{\mathbb{C}}\;|\;\Re(z)\leqslant\tfrac{1}{2}\,,\; \Im({z})=0\big\}$, and also is homotopic to the semicircle $c(t)=2\,{\rm{e}}^{-t\,\pi\,i}\,,\quad t\in\big[-\tfrac{1}{2},\tfrac{1}{2}\big]$, we have that
\begin{align*}
\displaystyle\oint_{\gamma}{{\rm{Log}}\,\big(z-\tfrac{1}{2}\big)\,dz}&=\oint_{c}{{\rm{Log}}\,\big(z-\tfrac{1}{2}\big)\,dz}\\
&=\int_{-\frac{1}{2}}^{\frac{1}{2}}{{\rm{Log}}\,\big(2\,{\rm{e}}^{-t\pi\,i}-\tfrac{1}{2}\big)\,\big(2\,{\rm{e}}^{-t\pi\,i}\big)'\,dt}\\
&\mathop{=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=}\limits^{\begin{subarray}{c}
{u\,=\,2\,{\rm{e}}^{-t\pi\,i}} \\
{du\,=\,(2\,{\rm{e}}^{-t\pi\,i})'\,dt} \\
\end{subarray}}\,\int_{2i}^{-2i}{{\rm{Log}}\,\big(u-\tfrac{1}{2}\big)\,du}\\
&=\int_{2i}^{-2i}{u'\,{\rm{Log}}\,\big(u-\tfrac{1}{2}\big)\,du}\\
&=\Big[u\,{\rm{Log}}\,\big(u-\tfrac{1}{2}\big)\Big]_{2i}^{-2i}-\int_{2i}^{-2i}{\frac{u}{u-\frac{1}{2}}\,du}\\
&=-2i\log\tfrac{17}{4}-\int_{2i}^{-2i}{du}-\frac{1}{2}\int_{2i}^{-2i}{\frac{1}{u-\frac{1}{2}}\,du}\\
&=-2i\log\tfrac{17}{4}+4i-\frac{1}{2}\Big[{\rm{Log}}\,\big(u-\tfrac{1}{2}\big)\Big]_{2i}^{-2i}\\
&\stackrel{(*)}{=}-2i\log\tfrac{17}{4}+4i-i\,({\rm{Arctan}}\,4-\pi)\\
&=\big(4+\pi-{\rm{Arctan}}\,4-2\log\tfrac{17}{4}\big)\,i\,.
\end{align*}

$(*)\quad {\rm{Arctan}}\,{z}=\dfrac{i}{2}\big({\rm{Log}}\,(1-i\,z)-{\rm{Log}}\,(1+i\,z)\big)={\rm{Arccot}}\,{\tfrac{1}{z}}\,.$

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Grigorios Kostakos

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