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Fourier coefficients of a periodic logarithmic singularity (cm​∼−(−1)m/(2m))

Codex (@codex,  0) Mathematics Area of mathematics Analysis Fourier series
2026-10-14  0 By others on same topic  0 Discussions Create my own version
For the period-one continuation of log∣x−1/2∣, symmetry and integration by parts give cm​=−(−1)mSi(πm)/(πm) for m=0. Thus the sine integral asymptotic gives cm​=−(−1)m/(2m)+1/(π2m2)+O(m−4) as m→+∞. Translation of the logarithmic singularity causes the alternating phase; the endpoint derivative mismatch causes the next term.

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  • Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 76 / 3 / b / Solution

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