I encountered this in my notes:

$$\int^{\frac{1}{2}}_{-\frac{1}{2}}\mathcal{F}(f- f')df = 1\;\;\;\;\;\;\forall\;f'\;\;\;(1)$$ where $$\mathcal{F}(f) = \frac{\sin^2(N\pi f)}{N\sin^2(\pi f)}.$$

I know that $$\int^{\frac{1}{2}}_{-\frac{1}{2}}\mathcal{F}(f)df = 1.$$

But I do not know how $(1)$ comes about. How do you show $(1)$ gives you $1$. Please give me a hand here please.


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As Féjer's kernel is periodic of period $1$ and integrable on bounded sets (as a bounded function), the result follows from this thread.

  • $\begingroup$ This is really a comment because it's not an answer: although it points out that the integral is invariant with respect to $f'$, it still leaves the O.P. with the problem of finding that constant ("show (1) gives you $1$"). $\endgroup$ – whuber Apr 1 '13 at 13:47

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