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here, i know that

$$ 0=\frac{1}{2}+ \sum_{n=1}^{\infty}cos(nx) $$

so by simple integration term by term on the series

$$ \frac{A-x}{2}= \sum_{n=1}^{\infty}\frac{sin(nx)}{n} $$

so if i put x=0 the sum of sines is 0 and i get A=0

and so if i integrate another time i get

$$ \frac{x^2}{2}-\frac{Ax}{2}+B =\sum_{n=1}^{\infty}\frac{cos(nx)}{n^2}$$

then which are A and B ??

Jose Garcia
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    $A$ is still $0$, as it is the same as above. You will get $B$ by setting $x=0$. Where do you have a difficulty ? – LL 3.14 Nov 19 '22 at 10:53
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    $\cos(nx)$ does not converge to $0$ (except for e very few values of $x$) so the infinite series is divergent, so you cannot sum it, or worse integrate it. – zwim Nov 19 '22 at 11:03
  • Your initial series has a meaning in distribution theory. Further integrals are handled in this older answer. Hoping this will help, – Raymond Manzoni Nov 19 '22 at 11:14

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