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The following integral comes up in the theory of Brownian motion in 1D: $$\int_{-\pi}^{\pi} \frac{dk}{2 \pi} \frac{\cos(nk)-1}{\cos(k)-1} = n$$ where $n \in \mathbf{N}$. Does anybody know a short proof of this formula?

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Let $$I_n = \int_{-\pi}^{\pi} \frac{ \cos(nk) - 1}{\cos k - 1} \,dk.$$

Then $$ \begin{align*} \frac{I_{n+1} + I_{n-1}}{2} &= \int_{-\pi}^{\pi} \frac{-2 + \cos(n+1)k + \cos(n-1)k}{2( \cos k -1)} \,dk \\ &= \int_{-\pi}^{\pi} \frac{-1+ \cos nk \cos k}{\cos k - 1} \,dk \\ &= \int_{-\pi}^{\pi} \frac{(\cos nk -1) + \cos nk(\cos k-1)}{ \cos k -1} \,dk \\ &= I_n + \int_{-\pi}^{\pi} \cos nk \,dk \\ &= I_n \end{align*} $$ Thus, $$I_n = \frac{1}{2} (I_{n+1} + I_{n-1})$$ and this is an arithmetic sequence.

One checks that $I_0 = 0, I_1 = 2 \pi, I_2 = 4 \pi$ so dividing by $2\pi$ yields the result you wanted.

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  • $\begingroup$ Nice solution! (+1) $\endgroup$ – Niklas Feb 26 at 14:42
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Hints: It is a nice exercise to show that it holds $$\frac{1}{n} \left(\frac{1 - \cos(nx)}{1 - \cos x}\right) = \frac{1}{n} \left(\frac{\sin \frac{n x}{2}}{\sin \frac{x}{2}}\right)^2 = \sum_{|j|\le n-1}\left(1-\frac{|j|}{n}\right)e^{ijx}$$ for every $n\in \mathbf N$. Now use the third expression to calculate your integral. By the way, the function $x\mapsto \frac{1}{n} \left(\frac{1 - \cos(nx)}{1 - \cos x}\right)$ is called the Fejer kernel.

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Let $z=e^{ik}$ and then $$ \cos k=\frac12(z+\frac1z), \cos nk=\frac12(z^n+\frac1{z^n}). $$ So \begin{eqnarray*} \int_{-\pi}^{\pi} \frac{dk}{2 \pi} \frac{\cos(nk)-1}{\cos(k)-1}&=&\frac{1}{2\pi}\int_{|z|=1}\frac{\frac12(z^n+\frac1{z^n})-1}{\frac12(z+\frac1z)-1}\frac{dz}{iz}\\ &=&\frac{1}{2\pi i}\int_{|z|=1}\frac{(1-z^n)^2}{z^n(1-z)^2}dz\\ &=&\frac{1}{2\pi i}\int_{|z|=1}\frac{1}{z^n}(\sum_{k=0}^{n-1}z^{k})^2\\ &=&\frac{1}{2\pi i}\int_{|z|=1}\frac{1}{z^n}(\cdots+nz^{n-1}+\cdots)dz\\ &=&\frac{1}{2\pi i}2\pi i n=n. \end{eqnarray*}

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  • $\begingroup$ (+1) for exploiting contour integration. $\endgroup$ – Mark Viola Feb 26 at 15:46

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