The integral $$\int_0^\pi x(\pi -x) \sin 2nx$$ evaluates to zero, but the function can't be said to be even or odd. What argument, other than pure calculation (as I did) would give the value $0$ immediately? The thing multiplying the sine in the integrand is symmetric (w.r.t to reflection against $x=\pi/2$) on the interval, but that's about it I can really say. Can something more general be extrapolated from this?
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Out of curiosity...are you computing the Fourier series of that function? :-) – Avitus Jun 24 '13 at 15:51
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Yeah, it popped up in my earlier question: http://math.stackexchange.com/questions/428341/uniform-convergence-of-fourier-series-how-do-i-check-it – Spine Feast Jun 24 '13 at 15:58
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You're on the right track. You need to think about the symmetries and antisymmetries about $x= \pi/2$. The quadratic part you've already noted, now think about the sine function. You may want to draw a few of the sine functions out if you're stuck. After this it should be clear why the integral is zero.
john
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Put $x=y+\frac{\pi}{2}$; the integral is equal to
$$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(-y^2+\frac{\pi^2}{4})\sin(2\pi y+n\pi)dy;$$
as $\sin(2\pi y+n\pi)=\sin(2\pi y)\cos(n\pi)=(-1)^{n}\sin(2\pi y)$ and $(-y^2+\frac{\pi^2}{4})(-1)^{n}\sin(2\pi y)$ is odd, then the result follows.
Avitus
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