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How can I represent $\log^2\cos(x)$ as a (cosine) Fourier series over the interval $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$?

I know that the coefficients of the Taylor series of $\log^2(1+x)$ are related with the harmonic numbers $H_n$ and I know that $$\log\cos(x)=-\log(2)+\sum_{k\geq 1}\frac{(-1)^{k+1}}{k}\cos(2kx)$$ but I cannot figure out a way to exploit the last identity through a Cauchy product.

Jack D'Aurizio
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Mohammad Hussein
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  • Surely you mean $\log(\cos^2 x)$ and not $(\log\cos x)^2$, $\log$ isn't defined for negative arguments, unless you're using a branch cut for whatever reason. And how do you deal with $\log 0$? – Adam Hughes Apr 17 '17 at 16:38
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    Please improve your question by using MathJax and showing your attempts. Related: https://math.stackexchange.com/questions/292468/fourier-series-of-log-sine-and-log-cos – Jack D'Aurizio Apr 17 '17 at 16:39

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For technical reasons it is easier to find the Fourier cosine series of $\log^2\left(2\cos x\right)$ first.
By squaring and applying formal manipulations $$\begin{eqnarray*} \log^2\left(2\cos x\right) &=& \sum_{j,k\geq 1}\frac{(-1)^{j+k}}{jk}\cos(2jx)\cos(2kx)\\ &=& \frac{1}{2}\sum_{j,k\geq 1}\frac{(-1)^{j+k}}{jk}\left[\cos(2(j+k)x)+\cos(2|j-k|x)\right]\\&=&\frac{1}{2}\sum_{j\geq 1}\frac{\cos(2j x)+1}{j^2}+\sum_{j\geq 1}\sum_{d\geq 1}\frac{(-1)^d}{j(j+d)}\left[\cos(2(2j+d)x)+\cos(2dx)\right]\\&=&\frac{\pi^2}{12}+\sum_{m\geq 1}c_m \cos(2mx)\end{eqnarray*}$$ where the coefficient $c_m$ equals: $$ c_m = \frac{1}{2m^2}+\sum_{j\geq 1}\frac{(-1)^m}{j(j+m)}+\sum_{1\leq j<\frac{m}{2}}\frac{(-1)^m}{j(m-j)}$$ and by partial fraction decomposition this coefficient is related with the harmonic numbers.
For instance: $$\sum_{j\geq 1}\frac{1}{j(j+m)} = \frac{H_{m-1}}{m}.$$ I guess you may take it from here.

Jack D'Aurizio
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