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My attempt was : $f(t) = \sum_{n=1}^{\infty}\frac{\sin(nx)}{2n+1}t^{2n+1}$, where $|t|\le 1$. So we can find $f'(t) = \displaystyle \Im\sum_{n=0}^{\infty}(e^{ix}t^2)^{n} = \Im\frac{1}{1-e^{ix}t^2} = \frac{t^2\sin(x)}{t^4-2\cos(x)t^2+1}$. So $f(t) = \displaystyle \int\frac{t^2\sin(x)}{t^4-2\cos(x)t^2+1}dt = \int\frac{t^2 \Im e^{ix}}{(t^2-e^{ix})(t^{2}-e^{-ix})}dt$.

Now if we decompose it and integrate, we will get $\displaystyle \frac{\pi-x}{8}t = \frac{\pi - x}{8}$(when $t = 1$), am I right?

Jack D'Aurizio
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openspace
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1 Answers1

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Let $z=\rho e^{ix}$ with $\rho\in(0,1)$. Then $$\sum_{n\geq 0}\frac{\rho^n\sin(nx)}{2n+1}=\text{Im}\sum_{n\geq 0}\frac{z^n}{2n+1} =\text{Im}\left(\frac{\text{arctanh}\sqrt{z}}{\sqrt{z}}\right)$$ and by considering the limit as $\rho\to 1^-$ we get $$ \sum_{n\geq 0}\frac{\sin(nx)}{2n+1}=\frac{\pi}{4}\cos\left(\frac{x}{2}\right)-\frac{1}{2}\sin\left(\frac{x}{2}\right)\log\cot\left(\frac{x}{4}\right) $$ for any $x\in(0,2\pi)$.

Jack D'Aurizio
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