Consider the operator $K(\varphi)(t)=\int_{s=-\pi}^{\pi}\sin(t-s)\varphi(s)ds$. If we will look at $K(e^{int})$ we will get zeros $ \forall n\in N$, thus it means that we got the zero operator? (because this is a basis to $L^2[-\pi,\pi]$)
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Who says you get $0$ for all $n$? – Daniel Fischer May 28 '17 at 18:30
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Maybe I made a terrible mistake, but isn't the integral zero? http://www.wolframalpha.com/input/?i=integral+from+-pi+to+pi+of+sin(t-s)e%5E(in*t) – user May 28 '17 at 18:35
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The integration variable is $s$, you need to input "sin(t-s)*e^(ins)". And you should probably tell Alpha that $n$ is an integer. And that $s$ is the variable of integration. – Daniel Fischer May 28 '17 at 18:39
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For $n = \pm 1$ you don't get zero: instead, $$ K(e^{is}) = i\pi e^{it} $$ and $$ K(e^{-is}) = -i\pi e^{-it}, $$ so these are eigenfunctions with eigenvalues $\pm i\pi$. These do span the image of $K$, however, as is slighly clearer if you use the angle addition formula and use sine and cosine, or use the exponential formula for the sine.
Chappers
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