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I've tried writing this integral as sum of two integrals, the first integral on the interval $[-\pi/2, 0]$ and the second one on $[0, \pi/2]$ and then tried showing that the sum of these integrals is equal to 0. I've tried with variable substitution, but neither of my attempts gave results and I don't know how to continue.

H-a-y-K
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  • Are there any other constraints on $f$? If not, it trivially fails in general, because e.g. we can force the integrand to be positive for $|x|\in(0,,1]$. – J.G. Apr 26 '21 at 16:13
  • @J.G. I actually just realized that I had misunderstood a notation in the book. But I'll post a new question for that, because initially I was trying to solve the problem of this post. – H-a-y-K Apr 26 '21 at 16:21
  • What about $f(x)=\frac{1+x}2$ ? – zwim Apr 26 '21 at 16:23
  • It turns out there is already an answer that I couldn't find on google for the real problem. https://math.stackexchange.com/questions/159381/evaluate-int-0-pi-xf-sin-xdx – H-a-y-K Apr 26 '21 at 16:28

1 Answers1

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Let $f(x)=0$ for $-\pi/2\leq x\leq 0$ and $f(x)=1$ for $0<x\leq\pi/2$, the integral becomes $\int_{0}^{\pi/2}xdx>0$.

user284331
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