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Evaluate:

$\int\limits_{0}^{\pi}\dfrac{\cos 2017x}{5-4\cos x}~dx$

I thought of using some series but could not get it

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    Try to prove, for integers $n$, that $$\int_{0}^{\pi} \frac{\cos nx}{5-4\cos x}=\frac{\pi}{3} \times \frac{1}{2^n}$$ – 2017-02-24
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    @S.C.B. How is it done – 2017-02-24
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    @S.C.B. So are you saying that your formula also works if $n$ is negative? Because $cos(nx)=cos(-nx)$ but the answer will be different... – 2017-02-25
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    @imranfat Sorry, positive integers. – 2017-02-25

2 Answers 2

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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}^{\pi}{\cos\pars{2017x} \over 5 - 4\cos\pars{x}}\,\dd x = \left.\Re\int_{0}^{\pi}{z^{2017} \over 5 - 4\pars{z+1/z}/2} \,{\dd z \over \ic z}\right\vert_{\ z\ =\ \exp\pars{\ic x}} \\[5mm] = &\ \left.-\,{1 \over 2}\,\Im\int_{0}^{\pi}{z^{2017} \over z^{2} - \pars{5/2}z + 1} \,\dd z\,\right\vert_{\ z\ =\ \exp\pars{\ic x}} = \left.-\,{1 \over 2}\,\Im\int_{0}^{\pi}{z^{2017} \over \pars{z - 1/2}\pars{z - 2}}\,\dd z\,\right\vert_{\ z\ =\ \exp\pars{\ic x}} \\[5mm] = &\ {1 \over 2}\,\Im\lim_{\epsilon \to 0^{+}} \int_{\pi}^{0}{\pars{1/2 + \epsilon\expo{\ic\theta}}^{2017} \over \epsilon\expo{\ic\theta}\pars{1/2 + \epsilon\expo{\ic\theta} - 2}} \epsilon\expo{\ic\theta}\ic\,\dd\theta = {1 \over 2}\pars{-\pi}{\pars{1/2}^{2017} \over 1/2 - 2} \\[5mm] = &\ \bbx{\ds{{2^{-2017} \over 3}\,\pi}} \approx 6.9587 \times 10^{-608} \end{align}

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    Hi Felix. How did the half circular contour with radius $1$ collapse around the pole at $1/2$? If we want a closed contour to begin, we can exploit the evenness of the integrand. But then, we need a multiplicative factor of $1/2$. – 2017-02-25
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    @Dr.MV There are two more contributions: 1) From $-1$ to $1/2 - \epsilon$ and 2) from $1/2 + \epsilon$ to $1$. Those contributions are integrals $\color{#f00}{\in \mathbb{R}}$. So, its imaginary part vanishes out. Thanks for your remark. – 2017-02-25
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    Felix, yes of course. (+1) for this nice solution. – 2017-02-25
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    @Dr.MV Thanks. It came, indeed, as a surprise !!!. – 2017-02-25
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You may go "the other way round".

Lemma (S). Since $\cos(x)=\frac{e^{ix}+e^{-ix}}{2}$, $$ \sum_{n\geq 0}\frac{2\cos(nx)}{2^n}=1+\frac{3}{5-4\cos x}.$$

Since $\int_{0}^{\pi}\cos(nx)\cos(mx)\,dx = \frac{\pi}{2}\,\delta(m,n)$, the Fourier cosine series shown by (S) directly gives the value of the wanted integral, $\color{red}{\frac{\pi}{3}\cdot 2^{-2017}}$.

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    Jack, just curious. It is easy to show the validity of the equation in the Lemma. So, was it purely experience that led you to this way forward, or was there a way one can "engineer" this? That is, if the coefficients in the denominator of the integrand were arbitrary, what would you propose then? – 2017-02-24
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    @Dr.MV: I did it by pure experience, but the same can be spotted by recognizing a Poisson kernel in action. – 2017-02-25
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    Can you please proof your second lemma(the summation one) and please add more explanation on how that leads to the answer.Thanks. – 2017-04-05
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    @Navin A bit late, but the infinite sum can probably be proven by using $\Re e^{ix}=\cos x$ and using the infinite geometric sequence. Or use the identity$$\cos x=\frac {e^{ix}+e^{-ix}}2$$Which Jack provided – 2018-06-24