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Problem of the Week #26 - November 26th, 2012

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Chris L T521

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Jan 26, 2012
995
Here's this week's problem.

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Problem: For $a>1$, show that\[\int_0^{2\pi}\frac{\,d\theta}{(a+\cos\theta)^2} = \frac{2\pi a}{(a^2-1)^{3/2}}.\]

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Hint:
Use the substitution $z=\exp(i\theta)$ to rewrite the definite integral as a contour integral over the unit circle $|z|=1$.

Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Chris L T521

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Jan 26, 2012
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This week's question was correctly answered by Sudharaka.

Here's my solution:
Let $z=e^{i\theta}$. Then $\cos\theta=\tfrac{1}{2}(z+z^{-1})$ and $\,dz=ie^{i\theta}\,d\theta$. Therefore, the integral becomes

\[\int_{|z|=1}\frac{\,dz}{iz\left(a+\frac{1}{2}(z+1/z)\right)^2} = \int_{|z|=1} \frac{-4iz\,dz}{(z^2+2az+1)^2}.\]
Thus, $f(z)$ has two poles of order $2$ at $z=-a-\sqrt{a^2-1}$ and $z=-a+\sqrt{a^2-1}$. Clearly, $-a-\sqrt{a^2-1}$ is not inside the unit circle, so we can disregard it. However, note that $a\leq 1+\sqrt{a^2-1}$ (due to the triangle inequality), which implies that
\[a-\sqrt{a^2-1}\leq 1 \implies -a+\sqrt{a^2-1}\geq-1.\]
So $-a+\sqrt{a^2-1}$ is contained in the unit circle and now the residue at that point is
\[\begin{aligned}\text{res}_{-a+\sqrt{a^2-1}}f(z) &= \lim_{z\to-a+\sqrt{a^2-1}}\frac{\,d}{\,dz}\left[\frac{-4iz}{(z+a+\sqrt{a^2-1})^2}\right]\\ &=\lim_{z\to-a+\sqrt{a^2-1}}\frac{-4i(z+a+\sqrt{a^2-1})^2+8iz(z+a+\sqrt{a^2-1})}{(z+a+\sqrt{a^2-1})^4}\\ &=\lim_{z\to-a+\sqrt{a^2-1}}\frac{-4i(z+a+\sqrt{a^2-1})+8iz}{(z+a+\sqrt{a^2-1})^3}\\ &=\frac{-4i(2\sqrt{a^2-1})+8i(-a+\sqrt{a^2-1})}{(2\sqrt{a^2-1})^3}\\&=\frac{-ia}{(a^2-1)^{3/2}}\end{aligned}\]
Therefore,
\[\int_0^{2\pi}\frac{\,d\theta}{(a+\cos\theta)^2}= \int_{|z|=1}\frac{-4iz}{(z^2+2az+1)^2} = 2\pi i\left(\frac{-ia}{(a^2-1)^{3/2}}\right) = \frac{2\pi a}{(a^2-1)^{3/2}}\]


Here's Sudharaka's solution:

Substitute \(z=\exp(i\theta)\) and we get,

\begin{eqnarray}

\int_0^{2\pi}\frac{\,d\theta}{(a+\cos\theta)^2} &=& \frac{4}{i}\int_{C}\frac{z}{\left[z-(\sqrt{a^2-1}-a)\right]^{2}\left[z+\sqrt{a^2-1}+a\right]^{2}}\mbox{ where }C\mbox{ is the contour }|z|=1\\

&=& \frac{4}{i}\int_{C}\frac{f(z)}{\left[z-(\sqrt{a^2-1}-a)\right]^{2}}\mbox{ where }f(z)=\frac{z}{\left[z+\sqrt{a^2-1}+a\right]^{2}}\\


\end{eqnarray}

It could be shown that, \(|\sqrt{a^2-1}-a|<1\) and therefore the point \(\sqrt{a^2-1}-a\) lie within the contour \(|z|=1\). Using Cauchy's integral formula we get,

\begin{eqnarray}

\int_0^{2\pi}\frac{\,d\theta}{(a+\cos\theta)^2}&=&\frac{4}{i}\left[2\pi i f'\left(\sqrt{a^2-1}-a\right)\right]\\

&=&8\pi \left[\frac{(z+a+\sqrt{a^2-1})^2-2z(z+a+\sqrt{a^2-1})}{(z+a+\sqrt{a^2-1})^4}\right]_{z=\sqrt{a^2-1}-a}\\

\end{eqnarray}

Simplifying this we get,

\[\int_0^{2\pi}\frac{\,d\theta}{(a+\cos\theta)^2} = \frac{2\pi a}{(a^2-1)^{3/2}}\]
 
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