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Question Number 179435 by Tawa11 last updated on 29/Oct/22

Commented by a.lgnaoui last updated on 29/Oct/22

$$\mathrm{V}?$$

Answered by Peace last updated on 31/Oct/22

$$\sum _{n⩾1}\frac{sin\left(2\pi nt\right)cos\left({tan}^{-}\left(4n\pi \right)\right)-cos\left(2\pi nt\right)sin\left({\tan }^{-1}\left(4n\pi \right)\right)}{n\sqrt{1+16n^2{\pi }^2}}=S\left(t\right)$$$$S\left(t\right)=A-B;A=\sum _{n⩾1}\frac{sin\left(2\pi nt\right).\sqrt{\frac{1}{1+{tg}^2\left({\tan }^{-1}\left(4n\pi \right)\right)}}}{n\sqrt{1+16n^2{\pi }^2}}$$$$B=\sum _{n⩾1}\frac{cos\left(2\pi nt\right).\sqrt{\frac{{tg}^2\left({tan}^{-}\left(4n\pi \right)\right)}{1+{tg}^2\left({tan}^{-}\left(4\mathit{n}\mathit{\pi }\right)\right)}}}{n\sqrt{1+16n^2{\pi }^2}}$$$$=\sum _{n⩾1}\frac{cos\left(2\pi nt\right)4\pi }{(1+16n^2{\pi }^2)}=Re\mathrm{\Sigma }\frac{e^{2in\pi t}.4\pi }{(4n\pi +i)(4n\pi -i)}$$$$=-\frac{1}{2\pi }\mathrm{\Sigma }(\frac{e^{2in\pi t}}{n-\frac{i}{4\pi }}-\frac{e^{2in\pi t}}{n+\frac{i}{4\pi }})$$$$A=\sum _{n⩾1}\frac{sin\left(2\pi nt\right)}{n(1+16n^2{\pi }^2)}=Im\sum _{n⩾1}\frac{e^{i2\pi nt}}{n(1+16n^2{\pi }^2)}$$$$\frac{1}{n(1+16n^2{\pi }^2)}=\frac{1}{n(4n\pi -\mathit{i})(4\mathit{n}\mathit{\pi }+i)}=\frac{1}{n}+\frac{-2\pi }{(4n\pi -i)}.-\frac{2\pi }{4n\pi +i}$$$$A=\sum _{n⩾1}\frac{e^{i2\pi nt}}{n}-\frac{1}{2\pi }\sum _{n⩾1}\frac{e^{2in\pi t}}{n-\frac{i}{4\pi }}-\frac{1}{2\pi }\sum _{n⩾1}\frac{e^{i2n\pi t}}{n+\frac{i}{4\pi }}$$$$-ln(1-x)=\sum _{n⩾1}\frac{x^n}{n}$$$$\mathrm{\Phi }(z,s,a)=\sum _{n⩾0}\frac{z^n}{{(n+a)}^s}LerchTranscendentFunction$$$$=-ln(1-e^{2i\pi nt})-\frac{1}{2\pi }(\mathrm{\Phi }(e^{2i\pi t},1,-\frac{i}{4\pi })+\mathrm{\Phi }(e^{2i\pi t},1,\frac{i}{4\pi }))=A$$$$B=-\frac{1}{2\mathit{\pi }}.(\mathrm{\Phi }(e^{2i\pi t},1,-\frac{i}{4\pi })-8\pi -\mathrm{\Phi }(e^{2i\pi t},1,\frac{i}{4\pi }))$$$$S=-Re\left(B\right)+Im\left(A\right)$$$$$$$$$$$$$$$$$$$$$$$$$$

Commented by Tawa11 last updated on 01/Nov/22

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

Commented by mindispower last updated on 02/Nov/22

$$withepleasurthanlyouallhappinesseforyou$$

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