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Question Number 142191 by iloveisrael last updated on 27/May/21

Commented by Mathspace last updated on 27/May/21

$$whytan\left(\frac{x}{2}\right)=i+\frac{2i}{1+e^{ix}}?$$$$andwhatisvalueof{\int }_0^{2\pi }\frac{sin\left(nx\right)}{1+e^{ix}}dx$$$$allthisneedproof...$$

Commented by MJS_new last updated on 28/May/21

$$\mathrm{typo}$$$$\tan \frac{x}{2}=-\mathrm{i}+\frac{2\mathrm{i}}{1+{\mathrm{e}}^{\mathrm{i}x}}$$$$\mathrm{proof}:$$$$\tan y=\frac{\sin y}{\cos y}=\frac{\frac{{\mathrm{e}}^{\mathrm{i}y}-{\mathrm{e}}^{-\mathrm{i}y}}{2\mathrm{i}}}{\frac{{\mathrm{e}}^{\mathrm{i}y}+{\mathrm{e}}^{-\mathrm{i}y}}{2}}=-\mathrm{i}\frac{{\mathrm{e}}^{2\mathrm{i}y}-1}{{\mathrm{e}}^{2\mathrm{i}y}+1}=-\mathrm{i}(1-\frac{2}{{\mathrm{e}}^{2\mathrm{i}y}+1})=$$$$[y=\frac{x}{2}]$$$$=-\mathrm{i}+\frac{2}{1+{\mathrm{e}}^{\mathrm{i}x}}$$

Commented by iloveisrael last updated on 27/May/21

Answered by mathmax by abdo last updated on 28/May/21

$${\mathrm{A}}_{\mathrm{n}}={\int }_0^{2\pi }\log (1+\mathrm{cosx})\cos \left(\mathrm{nx}\right)\mathrm{dx}\mathrm{by}\mathrm{parts}\mathrm{we}\mathrm{get}$$$${\mathrm{A}}_{\mathrm{n}}={\left[\frac{1}{\mathrm{n}}\sin \left(\mathrm{nx}\right)\log (1+\mathrm{cosx})\right]}_0^{2\pi }-{\int }_0^{2\pi }\frac{1}{\mathrm{n}}\sin \left(\mathrm{nx}\right)\times \frac{-\mathrm{sinx}}{1+\mathrm{cosx}}\mathrm{dx}$$$$=\frac{1}{\mathrm{n}}{\int }_0^{2\pi }\frac{\mathrm{sinx}.\sin \left(\mathrm{nx}\right)}{1+\mathrm{cosx}}\mathrm{dx}\Rightarrow {\mathrm{nA}}_{\mathrm{n}}={\int }_0^{2\pi }\frac{\mathrm{sinx}.\sin \left(\mathrm{nx}\right)}{1+\mathrm{cosx}}\mathrm{dx}$$$$={\int }_0^{2\pi }\frac{{\mathrm{e}}^{\mathrm{ix}}-{\mathrm{e}}^{-\mathrm{ix}}}{2\mathrm{i}}.\frac{{\mathrm{e}}^{\mathrm{inx}}-{\mathrm{e}}^{-\mathrm{inx}}}{2\mathrm{i}}.\frac{1}{1+\frac{{\mathrm{e}}^{\mathrm{ix}}+{\mathrm{e}}^{-\mathrm{ix}}}{2}}\mathrm{dx}$$$$=-\frac{1}{2}{\int }_0^{2\pi }\frac{({\mathrm{e}}^{\mathrm{ix}}-{\mathrm{e}}^{-\mathrm{ix}})({\mathrm{e}}^{\mathrm{inx}}-{\mathrm{e}}^{-\mathrm{inx}})}{2+{\mathrm{e}}^{\mathrm{ix}}+{\mathrm{e}}^{-\mathrm{ix}}}\mathrm{dx}$$$${=}_{{\mathrm{e}}^{\mathrm{ix}}=\mathrm{z}}-\frac{1}{2}{\int }_{\mid \mathrm{z}\mid =1}\frac{(\mathrm{z}-{\mathrm{z}}^{-1})({\mathrm{z}}^{\mathrm{n}}-{\mathrm{z}}^{-\mathrm{n}})}{(2+\mathrm{z}+{\mathrm{z}}^{-1})}\frac{\mathrm{dz}}{\mathrm{iz}}$$$$=\frac{\mathrm{i}}{2}{\int }_{\mid \mathrm{z}\mid =1}\frac{{\mathrm{z}}^{\mathrm{n}+1}-{\mathrm{z}}^{-\mathrm{n}+1}-{\mathrm{z}}^{\mathrm{n}-1}-{\mathrm{z}}^{-\mathrm{n}-1}}{{(\mathrm{z}+1)}^2}\mathrm{dz}$$$$=\frac{\mathrm{i}}{2}\left(2\mathrm{i}\pi \right)\mathrm{Res}(\phi ,-1)\mathrm{with}\phi \left(\mathrm{z}\right)=\frac{{\mathrm{z}}^{\mathrm{n}+1}-{\mathrm{z}}^{-\mathrm{n}+1}-{\mathrm{z}}^{\mathrm{n}-1}-{\mathrm{z}}^{-\mathrm{n}-1}}{{(\mathrm{z}+1)}^2}$$$${\mathrm{Res}(\phi ,-1)={\lim }_{\mathrm{z}\to -1}\frac{1}{(2-1)!}\left\{{(\mathrm{z}+1)}^2\right\}\phi \left(\mathrm{z}\right)\}}^{\left(1\right)}$$$$={\lim }_{\mathrm{z}\to -1}{\{{\mathrm{z}}^{\mathrm{n}+1}-{\mathrm{z}}^{-\mathrm{n}+1}-{\mathrm{z}}^{\mathrm{n}-1}-{\mathrm{z}}^{-\mathrm{n}-1}\}}^{\left(1\right)}$$$$={\lim }_{\mathrm{z}\to -1}(\mathrm{n}+1){\mathrm{z}}^{\mathrm{n}}-(-\mathrm{n}+1){\mathrm{z}}^{-\mathrm{n}}-(\mathrm{n}-1){\mathrm{z}}^{\mathrm{n}-2}+(\mathrm{n}+1){\mathrm{z}}^{-\mathrm{n}-2}$$$$=(\mathrm{n}+1){(-1)}^{\mathrm{n}}+(\mathrm{n}-1){(-1)}^{\mathrm{n}}-(\mathrm{n}-1){(-1)}^{\mathrm{n}}+(\mathrm{n}+1){(-1)}^{\mathrm{n}}$$$$={(-1)}^{\mathrm{n}}\{\mathrm{n}+1+\mathrm{n}-1-\mathrm{n}+1+\mathrm{n}+1\}$$$$=(2\mathrm{n}+2){(-1)}^{\mathrm{n}}\Rightarrow {\mathrm{nA}}_{\mathrm{n}}=-\pi (2\mathrm{n}+2){(-1)}^{\mathrm{n}}\Rightarrow$$$${\mathrm{A}}_{\mathrm{n}}=\frac{2\pi (\mathrm{n}+1){(-1)}^{\mathrm{n}-1}}{\mathrm{n}}(\mathrm{n}⩾1)$$

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