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I need to prove that if absolutely integrable function on a segment $$[-\pi;\pi]$$ has next condition: $$f(x+\pi)=f(x)$$ then $$a_{2n-1}=b_{2n-1}=0, n\in N$$ On the fingers, it seems like that this is some shift and the coefficients in front of the $\sin$ go to zero. But it's't formal and most likely wrong. How to say correct proof?

DumbSimon
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  • Try to prove that the coefficients for all even terms collapse to zero. Possibly, a choice of integration by parts will prove advantageous. Substitute in the periodicity requirement to form an integral identity that will yield the answer. I believe such a problem is an exercise in the Strauss PDEs Textbook. – Talmsmen Jun 17 '21 at 19:11

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$\int_{-\pi}^0f(x)trig((2n-1)x)dx$$=-\int_0^{\pi}f(x)trig((2n-1)x)dx$

Here $trig$ is either $sin$ or $cos$.
The trig functions at each point $(x)$ in the lower interval have the same magnitude at the corresponding point $(x+\pi)$ of the upper interval, but of opposite sign, while $f(x)$ has the same values in the corresponding points with the same sign.

herb steinberg
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