Residue theorem IV

Beautiful integral involving a product of cosine functions

Today we show the proof of this beautiful result posted by @SrinivasR1729 on Twitter:

\[\int_{0}^{2 \pi} \cos\left(\frac{1}{a} + x \right)\cos(x)dx =\pi \cos \left(\frac{1}{a}\right) \] \[ \int_{0}^{4 \pi} \cos\left(\frac{1}{a} + x \right)\cos\left(\frac{1}{a^2} + x \right)\cos(2x) = \pi \cos\left(\frac{a+1}{a^2}\right)\] \[ \int_{0}^{8 \pi} \cos\left(\frac{1}{a} + x \right)\cos\left(\frac{1}{a^2} + x \right)\cos\left(\frac{1}{a^3} + x \right)\cos(3x) = \pi \cos\left(\frac{a^2+a+1}{a^3}\right)\] \[ \int_{0}^{16 \pi} \cos\left(\frac{1}{a} + x \right)\cos\left(\frac{1}{a^2} + x \right)\cos\left(\frac{1}{a^3} + x \right)\cos\left(\frac{1}{a^4} + x \right)\cos(3x) = \pi \cos\left(\frac{a^3+a^2+a+1}{a^4}\right)\] \[\int_{0}^{2^m \pi } \left[\cos\left( \frac{1}{a}+x\right)\cos\left( \frac{1}{a^2}+x\right)\cdots\cos\left( \frac{1}{a^m}+x\right)\right]\cos(mx) = \pi \cos\left(\frac{a^{m-1}+a^{m-2}+a^{m-3}+...+1}{a^m}\right)\]
Proof

Recall that under the assumptions of the residue theorem with winding numbers, if $\gamma$ is a closed curve: \[ \oint_{\gamma} fdz = 2\pi i \sum_{i=1}^{n} \left\{\operatorname{Res}\left(f,z_{i}\right)I(\gamma,z_{i})\right\} \] where $I(\gamma,z_{i})$ is the winding number of $\gamma$ with respect to $z_{i}$.

Also recall that the circle $\displaystyle \gamma(t) = e^{it} \quad 0\leq t \leq 2\pi n $ has an index $\displaystyle I(\gamma,0) = n $ with respect to $0$

Case $n=1$ (the circle winds one time around 0)

The fact that $x$ varies from $0$ to $2\pi$ suggest we can consider $x$ as an argument of a point $z$ on the unit circle $C$ centered at the origin; hence we write $\displaystyle z = e^{ix} \quad 0\leq x \leq 2\pi \Longrightarrow dx = \frac{dz}{iz}$

Then
\begin{align*} I= \int_{0}^{2 \pi} \cos\left(\frac{1}{a} + x \right)\cos(x) =& \int_{0}^{2 \pi} \left(\frac{e^{i\left(\frac{1}{a}+x\right)} + e^{-i\left(\frac{1}{a}+x\right)}}{2}\right)\left(\frac{e^{ix}+e^{-ix}}{2}\right)dx \\ =& \int_{0}^{2 \pi} \left(\frac{e^{\frac{i}{a}}e^{ix} + e^{-\frac{i}{a}}e^{-ix}}{2}\right)\left(\frac{e^{ix}+e^{-ix}}{2}\right)dx \\ =& \frac{1}{2^2}\oint_{C} \frac{\left(e^{\frac{i}{a}}z + \frac{e^{-\frac{i}{a}}}{z}\right)\left(z+\frac{1}{z}\right)}{iz}dz\\ =& \frac{1}{2^2i}\oint_{C} \left\{ e^{\frac{i}{a}}z + \frac{1}{z}\left(e^{\frac{i}{a}}+e^{\frac{-i}{a}}\right) +\frac{e^{\frac{-i}{a}}}{z^3}\right\}dz\\ \end{align*} Hence, if $\displaystyle g(z) = e^{\frac{i}{a}}z + \frac{1}{z}\left(e^{\frac{i}{a}}+e^{\frac{-i}{a}}\right) +\frac{e^{\frac{-i}{a}}}{z^3}$
\[\Longrightarrow \operatorname{Res}(g,0) = e^{\frac{i}{a}}+e^{\frac{-i}{a}}\] Therefore $z=0$ is the only singularity of $g$ and: \begin{align*} I=\int_{0}^{2 \pi} \cos\left(\frac{1}{a} + x \right)\cos(x)dx=& 2 \pi i \left(\frac{1}{2^2i}\right) \operatorname{Res}\left( g, 0 \right)\underbrace{I(C,0)}_{=1}\\ =& \frac{\pi}{2} \left(e^{\frac{i}{a}}+e^{\frac{-i}{a}}\right)\\ =& \pi \cos \left(\frac{1}{a} \right) \end{align*} \[\boxed{ \int_{0}^{2 \pi} \cos\left(\frac{1}{a} + x \right)\cos(x)dx =\pi \cos \left(\frac{1}{a}\right) }\]

Case n = 2 (the circle winds 2 times around 0)
\begin{align*} I=\int_{0}^{4 \pi} \cos\left(\frac{1}{a} + x \right)\cos\left(\frac{1}{a^2} + x \right)\cos(2x) dx=& \int_{0}^{4 \pi} \left(\frac{e^{i\left(\frac{1}{a}+x\right)} + e^{-i\left(\frac{1}{a}+x\right)}}{2}\right)\left(\frac{e^{i\left(\frac{1}{a^2}+x\right)} + e^{-i\left(\frac{1}{a^2}+x\right)}}{2}\right)\left(\frac{e^{i2x}+e^{-i2x}}{2}\right)dx \\ =& \int_{0}^{4 \pi} \left(\frac{e^{\frac{i}{a}}e^{ix} + e^{-\frac{i}{a}}e^{-ix}}{2}\right)\left(\frac{e^{\frac{i}{a^2}}e^{ix} + e^{-\frac{i}{a^2}}e^{-ix}}{2}\right)\left(\frac{e^{i2x}+e^{-i2x}}{2}\right)dx \\ =& \frac{1}{2^3i}\oint_{C} \frac{\left(e^{\frac{i}{a}}z + \frac{e^{-\frac{i}{a}}}{z}\right)\left(e^{\frac{i}{a^2}}z + \frac{e^{-\frac{i}{a^2}}}{z}\right)\left(z^2+\frac{1}{z^2}\right)}{z}dz \quad (z = e^{ix} )\\ =& \frac{1}{2^3i}\oint_{C} \underbrace{\frac{\left(e^{\frac{i}{a}}z^2 + e^{-\frac{i}{a}}\right)\left(e^{\frac{i}{a^2}}z^2 + e^{-\frac{i}{a^2}}\right)\left(z^4+1\right)}{z^5}}_{g(z)}dz\\ \end{align*} It is easy to see that \[\operatorname{Res}(g,0) = e^{\frac{i}{a}}e^{\frac{i}{a^2}} + e^{-\frac{i}{a}}e^{-\frac{i}{a^2}} = e^{i\left(\frac{1}{a} + \frac{1}{a^2}\right)} + e^{-i\left(\frac{1}{a}+\frac{1}{a^2}\right)}\] Therefore, again $z=0$ is the only singularity of $g$ and \begin{align*} I =\int_{0}^{4 \pi} \cos\left(\frac{1}{a} + x \right)\cos\left(\frac{1}{a^2} + x \right)\cos(2x) dx=& 2\pi i \left(\frac{1}{2^3i } \right)\operatorname{Res}(g,0)\underbrace{I(C,0)}_{=2}\\ =& 2^2\pi i \left(\frac{1}{2^3i } \right)\left( e^{\frac{i}{a}}e^{\frac{i}{a^2}} + e^{-\frac{i}{a}}e^{-\frac{i}{a^2}} \right)\\ =& \frac{\pi}{2} \left[ e^{i\left(\frac{1}{a} + \frac{1}{a^2}\right)} + e^{-i\left(\frac{1}{a}+\frac{1}{a^2}\right)} \right]\\ =& \pi \cos\left(\frac{1}{a}+\frac{1}{a^2}\right)\\ =& \pi \cos\left(\frac{a+1}{a^2}\right) \end{align*} \[ \boxed{\int_{0}^{4 \pi} \cos\left(\frac{1}{a} + x \right)\cos\left(\frac{1}{a^2} + x \right)\cos(2x) = \pi \cos\left(\frac{a+1}{a^2}\right)}\]

Case n = m (the circle winds $2^{m-1}$ times around $0$)

\begin{align*} I= \int_{0}^{2^{m} \pi } \left(\prod_{j=1}^{m} \cos\left( \frac{1}{a^j}+x\right)\right)\cos(mx) = &\int_{0}^{2^m \pi } \left( \prod_{j=1}^{m} \frac{e^{i\left(\frac{1}{a^j}+x\right)} +e^{-i\left(\frac{1}{a^j}+x\right)}}{2}\right)\left(\frac{e^{imx}+e^{-imx}}{2}\right)dx\\ =& \int_{0}^{2^{m-1} 2 \pi } \left( \prod_{j=1}^{m} \frac{e^{\frac{i}{a^j}}e^{ix} +e^{-\frac{i}{a^j}}e^{-ix}}{2}\right)\left(\frac{e^{imx}+e^{-imx}}{2}\right)dx\\ =& \frac{1}{2^{m+1}}\oint_{C} \left[\prod_{j=1}^{m} \left(e^{\frac{i}{a^j}}z +\frac{e^{-\frac{i}{a^j}}}{z}\right)\right]\left(z^m + \frac{1}{z^m}\right)\frac{dz}{iz} \quad (z = e^{ix} )\\ =& \frac{1}{i2^{m+1}}\oint_{C} \frac{\left[\prod_{j=1}^{m} \left(e^{\frac{i}{a^j}}z^2 +e^{-\frac{i}{a^j}}\right)\right]\left(z^{2m} + 1\right)}{z^{2m+1}}dz \end{align*} Therefore if \[ g(z) = \frac{\left[\prod_{j=1}^{m} \left(e^{\frac{i}{a^j}}z^2 +e^{-\frac{i}{a^j}}\right)\right]\left(z^{2m} + 1\right)}{z^{2m+1}} \] Then \[ \operatorname{Res}(g,0) = \prod_{j=1}^{m} e^{\frac{i}{a^j}} + \prod_{j=1}^{m} e^{-\frac{i}{a^j}} = e^{i(\frac{1}{a}+\frac{1}{a^2}+...+\frac{1}{a^m})} + e^{-i(\frac{1}{a}+\frac{1}{a^2}+...+\frac{1}{a^m})}\] Hence \begin{align*} I= \int_{0}^{2^m \pi } \left(\prod_{j=1}^{m} \cos\left( \frac{1}{a^j}+x\right)\right)\cos(mx) =& 2 \pi i \left( \frac{1}{i2^{m+1}}\right) \operatorname{Res}(g,0) \underbrace{I(C,0)}_{=2^{m-1}}\\ =& \frac{\pi}{2}\left(e^{i(\frac{1}{a}+\frac{1}{a^2}+...+\frac{1}{a^m})} + e^{-i(\frac{1}{a}+\frac{1}{a^2}+...+\frac{1}{a^m})}\right)\\ =& \pi \cos\left(\frac{1}{a}+\frac{1}{a^2}+...+\frac{1}{a^m}\right)\\ =& \pi \cos\left(\frac{a^{m-1}+a^{m-2}+a^{m-3}+...+1}{a^m}\right) \end{align*} Hence \[\boxed{\int_{0}^{2^m \pi } \left[\cos\left( \frac{1}{a}+x\right)\cos\left( \frac{1}{a^2}+x\right)\cdots\cos\left( \frac{1}{a^m}+x\right)\right]\cos(mx) = \pi \cos\left(\frac{a^{m-1}+a^{m-2}+a^{m-3}+...+1}{a^m}\right)}\]