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So the task is to integrate $${\frac{1}T}\int_0^{2\pi}tf(t)\,dt$$ where $T=2\pi$ and $$f(t) = \left\{ \begin{array}{ll} -t^2 & \quad -\pi < t < 0, \\ t^2 & \quad 0 < t < \pi \end{array} \right.$$ So I got $$-{\frac{1}{2\pi}}\int_0^{2\pi}tf(t)\,dt= -{\frac{1}{2\pi}}\left({\int_0^{\pi}t^3\,dt}-{\int_{-\pi}^0 t^3\,dt}\right)=-{\frac{1}4}\pi^3.$$ Since $f(t)$ is $2\pi$ periodic, I thought that $${\int_\pi^{2\pi}f(t)\,dt}={\int_{-\pi}^0 f(t)\,dt}.$$ When I looked up the solution, however, the integral was handled differently $$-{\frac{1}{2\pi}}{\int_0^{2\pi}tf(t)\,dt}=-{\frac{1}{2\pi}}\left({\int_0^\pi t^3\,dt} - {\int_\pi^{2\pi}t(t-2\pi)^2\,dt}\right)={\frac{\pi^3}{12}}.$$ Obviously, the result differs. I do understand, why it says $(t-2\pi)^2$ (shifting from $[\pi;2\pi]$ to $[-\pi;0]$).

As I stated above, I did believe that it does not matter whether I integrate $f(t)$ from $\pi$ to $2\pi$ or from $-\pi$ to $0$ (because of the periodicity, also considering $f(t)$ being uneven).

So where is my mistake causing the first result to be wrong?

Przemysław Scherwentke
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VGD
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  • One question: if the function is defined on $;[-\pi,\pi];$ , why do you integrate on $;[0,2\pi];$ ? – Timbuc Oct 31 '14 at 16:17
  • @Timbuc In this context, which is likely fourier analysis, we're always considering periodic extensions to $\mathbb R$ of functions defined on an interval of finite length, so here $f$ is actually the periodic extension of that function defined on $]-\pi, \pi[$. – Git Gud Oct 31 '14 at 16:18
  • I know that, @GitGud. Thanks. Yet this looks like an exercise in Fourier series and thus the more usual, or even natural, thing to do is to integrate on $;[-\pi,\pi];$ , which is the interval on which the function's definition's given. – Timbuc Oct 31 '14 at 16:21
  • $f(t)$ is periodic, but $tf(t)$ is not because of the factor $t$ ! –  Oct 31 '14 at 16:22
  • @Timbuc This is an exercise, I actually think it is a good idea to ask to integrate outside of $]-\pi, \pi[$ to force the reader to think. It may be mathematically unnatural, but pedagogically it's OK, in my opinion. – Git Gud Oct 31 '14 at 16:32
  • I agree with that, @GitGud. Thank you. – Timbuc Oct 31 '14 at 16:33
  • another question for the OP: @VGD, why the minus sign in fron of the integral on the left in your 6th line? – Timbuc Oct 31 '14 at 16:34
  • @VGD I'll believe you when you say that the integrals are equal, but that's a coincidence and not at all a consequence of $f$ be $2\pi$-periodic. Consider what? That equality is what I wrote before, with $T=2\pi$. – Git Gud Oct 31 '14 at 16:47
  • @GitGud I do not think that is my mistake. Both integrals ${\int_\pi^{2\pi}f(t)dt}={\int_{-\pi}^0 f(t)dt}$ have equal values (which of course is no proof) and by looking at the graph of f(t) on can see that both areas (the one from $-\pi$ to $0$ and the other from $\pi$ to $2\pi$ are equally shaped because of f being $2\pi$-periodically continued and thus have equal values. – VGD Oct 31 '14 at 16:52
  • sorry for double posting, could not edit anymore. – VGD Oct 31 '14 at 16:52
  • @VGD, immediately after you write the definition of $;f;$, you write "So I got" and then there's a minus sign in front of the very first integral. Why? BTW, there's also a minus sign in what you call "the solution" – Timbuc Oct 31 '14 at 16:53
  • @Timbuc that is (more or less) a formula given in our lecture. It gives me the fourier coefficient c_0 of the antiderviative of $f$. – VGD Oct 31 '14 at 16:55
  • @GitGud Granted, the fact that both integrals are equal is not directly caused by the periodicity but by the symmetry of $f$. Still, whether caused by periodicity or by symmetry I disagree on saying that it's a coinincidence that both have equal values. Moreover, since I think that this equality is given (because of the facts I mentioned above) that discussion is not helping too much finding my mistake either. – VGD Oct 31 '14 at 17:04

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