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Question Number 69238 by ~ À ® @ 237 ~ last updated on 21/Sep/19

$$UseResidusTheoremtoexplicit$$$$f\left(a\right)=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{n}sin\left(na\right)}{n^3}$$

Commented by mathmax by abdo last updated on 22/Sep/19

$$letfirstfinds\left(x\right)=\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^{n}sin\left(nx\right)}{n}$$$$s\left(x\right)=Im\left(\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n{e}^{inx}}{n}\right)=Im\left(\sum _{n=1}^{\mathrm{\infty }}\frac{{(-e^{ix})}^n}{n}\right)$$$$letw\left(z\right)=\sum _{n=1}^{\mathrm{\infty }}\frac{z^n}{n}with\mid z\mid ⩽1\Rightarrow w^{{}^{\prime }}\left(z\right)=\sum _{n=1}^{\mathrm{\infty }}{z}^{n-1}=\frac{1}{1-z}\Rightarrow$$$$w\left(z\right)=\int \frac{dz}{1-z}+c=-ln(1-z)+c(c=w\left(0\right)=0)\Rightarrow w\left(z\right)=-ln(1-z)$$$$\Rightarrow \sum _{n=1}^{\mathrm{\infty }}\frac{{(-e^{ix})}^n}{n}=-ln(1+e^{ix})=-ln(1+cosx+isinx)$$$$=-ln(2{cos}^2\left(\frac{x}{2}\right)+2isin\left(\frac{x}{2}\right)cos\left(\frac{x}{2}\right))$$$$=-ln\left(2cos\left(\frac{x}{2}\right)e^{\frac{ix}{2}}\right)=-ln\left(2cos\left(\frac{x}{2}\right)\right)-\frac{ix}{2}\Rightarrow s\left(x\right)=-\frac{x}{2}\Rightarrow$$$$\int s\left(x\right)dx=-\frac{x^2}{4}+c=-\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{n^2}cos\left(nx\right)$$$$x=0\Rightarrow c=-\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{n^2}=-(2^{1-2}-1)\xi \left(2\right)=\frac{1}{2}\frac{{\pi }^2}{6}=\frac{{\pi }^2}{12}\Rightarrow$$$$\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{n^2}cos\left(nx\right)=-\frac{x^2}{4}+\frac{{\pi }^2}{12}\Rightarrow$$$$\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{n^3}sin\left(nx\right)=-\frac{x^3}{12}+\frac{{\pi }^2}{12}x+c$$$$x=0\Rightarrow c=0\Rightarrow \sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{n^3}sin\left(nx\right)=-\frac{x^3}{12}+\frac{{\pi }^{2}x}{12}\Rightarrow$$$$\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{n^3}sin\left(nx\right)=\frac{x}{12}({\pi }^2-x^2)$$$$$$

Answered by mind is power last updated on 22/Sep/19

$$Wehave$$$$limx\to n\frac{\pi (x-n)}{sin\left(\pi x\right)}=\frac{\pi }{(sin{\left(\pi x\right)}_{x=n}^{{}^{\prime }}}={(-1)}^n$$$$letf\left(z\right)=\frac{\pi sin\left(za\right)}{z^{3}sin\left(n\pi \right)}$$$$fhaspolesatz=nwithen\in {\mathbb{Z}}^{\ast }oforder1andinz=0oforder3$$$$let{C}_{r}squarofsiderandcenterin0$$$$\Rightarrow {\int }_{C_r}f\left(z\right)dz=\sum _{Z_k}2i\pi (Res(f\mid z_k){}^{″}{Z}_k\in C_r^{″}$$$$andwehavelim\mid zf\left(z\right)\mid \Rightarrow 0$$$$ByjordanlemmawegetLimr\to +\mathrm{\infty }{\int }_{C_r}f\left(z\right)dz=0$$$$Residusth\Rightarrow Limr\to +\mathrm{\infty }{\int }_{cr}f\left(z\right)dz=\sum _{n\hat{\in }IZ}2i\pi Res(f\left(z\right)\mid \mid n)$$$$\Rightarrow 0=Res(f\mid 0)+\sum _{n\in {IZ}^{\ast }}Res(f\mid n)$$$$Res(f\mid n)=limz\to n\frac{(z-n)\pi sin\left(za\right)}{z^{3}sin\left(\pi z\right)}=\frac{\pi sin\left(na\right)}{n^3\pi cos\left(n\pi \right)}=\frac{{(-1)}^{n}sin\left(na\right)}{n^3}$$$$Res(f\mid 0)weuselaurentserie$$$$\frac{sin\left(za\right)}{z^3}.\frac{\pi }{sin\left(\pi z\right)}=\frac{(za-\frac{{\left(za\right)}^3}{6}...)}{z^3}.\frac{1}{\pi z-\frac{{\pi }^3{z}^3}{6}..}$$$$=\frac{1}{z^3}(a-\frac{z^2{a}^3}{6}...)(1+\frac{{\pi }^2{z}^2}{6}...)=\frac{a}{z^3}+\frac{1}{z}(\frac{a{\pi }^2}{6}-\frac{a^3}{6})....$$$$Res(f\mid 0)=\frac{a}{6}(-a^2+{\pi }^2)$$$$\Rightarrow \sum _{\mid n\mid ⩾1}\frac{{(-1)}^{n}sin\left(na\right)}{n^3}-\frac{a}{6}(a^2-{\pi }^2)=0$$$$\Rightarrow 2\sum _{n=1}^{+\mathrm{\infty }}\frac{{(-1)}^{n}sin\left(na\right)}{n^3}=\frac{a}{6}(a^2-{\pi }^2)$$$$\Rightarrow \sum _{n=1}^{+\mathrm{\infty }}\frac{{(-1)}^{n}sin\left(na\right)}{n^3}=\frac{a}{12}(a^2-{\pi }^2)$$$$$$$$$$$$$$

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