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I need to evaluate the integral $$ \int_0^1 \frac{6\pi}{\lvert{2 - 3e^{2\pi i t}}\rvert^2} \,\mathrm{d}t.$$ The problem I am having is that I can't find a nice way to re-write the denominator to invoke a substitution. Perhaps I can view this as a contour integral in the complex plane? Any ideas appreciated.

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  • $\begingroup$ Contour integral won't likely work nice as the integrand is a real function... $\endgroup$ – DonAntonio Apr 4 '17 at 18:07
  • $\begingroup$ @DonAntonio by contour I mean techniques like integrating $\int_\infty^\infty \sin x/x dx$ using $e^z/z$ $\endgroup$ – user369210 Apr 5 '17 at 1:42
  • $\begingroup$ Well, you can see in my answer that you can now substitute $\;z=e^{2\pi it}\;$ and integrate around the unit circle in the complex plane... $\endgroup$ – DonAntonio Apr 5 '17 at 8:37
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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \int_{0}^{1}{6\pi \over \verts{2 - 3\expo{2\pi\ic t}}^{2}}\,\dd t & = {2\pi \over 3}\int_{0}^{1} {\dd t\over \pars{1- 2\expo{-2\pi\ic t}/3}\pars{1- 2\expo{2\pi\ic t}/3}} \\[5mm] & = {2\pi \over 3}\sum_{k = 0}^{\infty}\sum_{j = 0}^{\infty} \pars{2 \over 3}^{k + j}\ \overbrace{\int_{0}^{1}\expo{-2\pars{k - j}\pi\ic t}\,\dd t}^{\ds{\delta_{kj}}} = {2\pi \over 3}\sum_{k = 0}^{\infty}\pars{4 \over 9}^{k} \\[5mm] & = {2\pi \over 3}\,{1 \over 1 - 4/9} = \bbx{\ds{6\pi \over 5}} \end{align}

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  • $\begingroup$ Too clever; is this a technique that you find widely applicable? $\endgroup$ – user369210 Apr 9 '17 at 4:17
  • $\begingroup$ @user321210 Thanks. When Residue Theorem isn't 'available' it's better to try another approach. This one yields a straightforward answer. $\endgroup$ – Felix Marin Apr 9 '17 at 22:20
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An idea:

$$2-3e^{2\pi it}=2-3\cos2\pi t-3i\sin2\pi t\implies |2-3e^{2\pi it}|^2=13-12\cos2\pi t\implies$$$${}$$

$$\int_0^1\frac{6\pi}{|2-3e^{2\pi it}|^2}dt=\int_0^1\frac{6\pi}{13-12\cos2\pi t}dt$$

You can now try some trigonometric substitution.

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  • $\begingroup$ And now contour integration is viable. ;-)) (+1) $\endgroup$ – Mark Viola Apr 4 '17 at 18:10
  • $\begingroup$ @Dr.MV Shhh.... :) . Indeed so, and it will probably be the easiest way. $\endgroup$ – DonAntonio Apr 4 '17 at 18:17

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