How do I evaluate $$\int \frac{1 - \cos kx }{1 - \cos x}dx$$?
I realise that $\cos kx$ can be expressed as a Chebyshev polynomial but because it's part of a fraction I can't see that this helps very much.
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1I suppose that this would lead to hypergeometric functions. – Claude Leibovici Oct 12 '23 at 13:33
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Beside some nasty hypergeometric function it seems that $$\large\color{blue}{f_k=\int \frac{1-\cos(kx)}{1-\cos(x)}\,dx= kx+2 \sum_{n=1}^k \frac{k-n}n\, \sin(nx)}$$
You can easily check it using series expansion.
Claude Leibovici
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@EdGraham. Just looking at the very very first integrals. The pattern is clear. But notice that I wrote *"it seems"*. I did not have time to make a proof. – Claude Leibovici Oct 12 '23 at 14:38
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1I have now verified this formula by differentiating both sides and am satisfied that it is correct. Thank you! – Ed Graham Oct 12 '23 at 14:49
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The key identity to use is $\frac{1 - \cos kx}{1 - cos x} = k + 2\sum_{n=1}^{k-1}n\cos (k-n)x$. I was aware of this but didn't think to use it - oops! – Ed Graham Oct 12 '23 at 15:02