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By application of calculus of residues, prove that $$ \int_{0}^{2\pi} \frac{\cos^{3}\left(3\theta\right)} {1 - 2p\cos\left(2\theta\right) + p^{2}} \,\mathrm{d}\theta = \frac{\pi\left(1 - p + p^{2}\right)}{1 - p} $$ I have attempted the above question using the substitution $z = \mathrm{e}^{\mathrm{i} \theta}$ and I have also used the substitution $\cos\left(x\right) = \mathrm{e}^{\mathrm{i}\theta}$ but I didn't get the solution correctly.

Solution \begin{equation*} \text{Let}\, z=e^{i\theta}\, \hspace{2mm} dz=ie^id\theta,\,\hspace{2mm} dz=izd\theta\\ \end{equation*} \begin{equation*} \cos\theta=\frac{z+z^{-1}}{2}\\ \end{equation*} \begin{equation*} \cos2\theta=\frac{z^2+z^{-2}}{2}\\ \end{equation*} \begin{equation*} \cos3\theta=\frac{z^3+z^{-3}}{2}\\ \end{equation*} \begin{equation*} \begin{split} \int_0^{2\pi} \frac{\cos ^3{3\theta}d\theta}{1-2p\cos 2\theta+p^2} &= \oint\frac{(\frac{z^3+z^{-3}}{2})^3}{1-2p(\frac{z^2+z^{-2}}{2})+p^2}\cdot\frac{dz}{iz}\\ &= \frac{1}{8}\oint\frac{-(z^3+z^{-3})^3i}{[z-p(z^3+z^{-1})+p^2z]}\cdot dz\\ &= \frac{1}{8}\oint\frac{(z^3+1)^3i}{z^8[pz^4+(p^2+1)z^2-p]}\cdot dz \end{split} \end{equation*} There is a pole at z=0 of order 8. Kindly help, to determine the remaining poles and the residues.

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    $\begingroup$ Integrate round the unit circle using $z=e^{i\theta}$. Use $\cos^3 3\theta=\frac{1}{8}(z^3+z^{-3})^3$ etc $\endgroup$ – ancientmathematician Oct 23 at 12:57
  • $\begingroup$ Thanks Christian. Using that substitution I observed the pole(s) will likely be of a high order. $\endgroup$ – Sodiq Mojeed Oct 23 at 13:03
  • $\begingroup$ I don't think that the pole at the origin of order 8 contributes anything, but I may be wrong. $\endgroup$ – ancientmathematician Oct 23 at 14:39
  • $\begingroup$ $\verb* Clear[p,t];Integrate[Cos[3t]^3/(1-2p Cos[2t]+p^2),{t,0,2Pi}] *$ ${\tt Mathematica}$ evaluates it as $\Large\color{red}{0}$. $\endgroup$ – Felix Marin Oct 24 at 2:18
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\bbox[5px,#ffd]{\int_{0}^{2\pi} {\cos^{3}\pars{3\theta} \over 1 - 2p\cos\pars{2\theta} + p^{2}} \,\dd\theta = \require{cancel} \cancelto{0}{{\pi\pars{1 - p + p^{2}} \over 1 - p}}}:\ {\Large ?}}$.


\begin{align} &\bbox[5px,#ffd]{\int_{0}^{2\pi} {\cos^{3}\pars{3\theta} \over 1 - 2p\cos\pars{2\theta} + p^{2}}\,\dd\theta} \\[5mm] = &\ -\int_{-\pi}^{\pi} {\cos^{3}\pars{3\theta} \over 1 - 2p\cos\pars{2\theta} + p^{2}}\,\dd\theta \\[5mm] = &\ -2\int_{0}^{\pi} {\cos^{3}\pars{3\theta} \over 1 - 2p\cos\pars{2\theta} + p^{2}}\,\dd\theta \\[5mm] = &\ 2\int_{-\pi/2}^{\pi/2} {\sin^{3}\pars{3\theta} \over 1 + 2p\cos\pars{2\theta} + p^{2}}\,\dd\theta = \bbx{\large 0} \\ & \end{align}
  • because the integrand is an $\ds{\underline{odd\ function}}$
  • and the integral is evaluated between symmetric limits $\ds{\pars{~\mbox{i.e.}\ \pm{\pi \over 2}~}}$.
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  • $\begingroup$ Thanks Felix. Is there any other way through which the prove can be done? To prove that the LHS is $$=\frac{\pi (1-p+p^2)}{1-p}$$ $\endgroup$ – Sodiq Mojeed Oct 24 at 11:41
  • $\begingroup$ Using calculus of residues. $\endgroup$ – Sodiq Mojeed Oct 24 at 12:09

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