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Question Number 130693 by mnjuly1970 last updated on 28/Jan/21

$$...nicecalculus...$$$$evaluate::$$$$={\int }_0^{\mathrm{\infty }}\frac{sin\left(x\right)}{e^x-1}dx=??$$$$....................$$

Answered by mnjuly1970 last updated on 28/Jan/21

$${\int }_0^{\mathrm{\infty }}\frac{sin\left(x\right)}{e^x-1}dx=\frac{1}{2i}{\int }_0^{\mathrm{\infty }}\frac{e^{ix}-e^{-ix}}{e^x-1}dx$$$$=\frac{1}{2i}{\int }_0^{\mathrm{\infty }}\frac{e^{ix-x}-e^{-ix-x}}{1-e^{-x}}dx$$$$=\frac{1}{2i}{\int }_0^{\mathrm{\infty }}\{\underset{n=0}{\sum ^{\mathrm{\infty }}}{e}^{ix-x-nx}-e^{-ix-x-nx}\}dx$$$$=\frac{1}{2i}\underset{n=0}{\sum ^{\mathrm{\infty }}}{\{\frac{e^{x(i-1-n)}}{i-n-1}+\frac{e^{-x(i+1+n)}}{i+1+n}\}}_0^{\mathrm{\infty }}$$$$=\frac{-1}{2i}\underset{n=0}{\sum ^{\mathrm{\infty }}}\{\frac{1}{i-(n+1)}+\frac{1}{i+(n+1)}\}$$$$=\frac{-1}{2i}\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{i-n}+\frac{1}{i+n}=\frac{-1}{2i}\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{2i}{i^2-n^2}$$$$=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^2+1}\stackrel{⟨upsilonfunction⟩}{=}\frac{\pi coth\left(\pi \right)-1}{3}$$

Answered by Dwaipayan Shikari last updated on 28/Jan/21

$$\frac{1}{2i}\underset{n=1}{\sum ^{\mathrm{\infty }}}{\int }_0^{\mathrm{\infty }}{e}^{-nx+ix}-e^{-nx-ix}dx$$$$=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^2+1}=\frac{1}{2}(\pi coth\left(\pi \right)-1)$$

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

$$\mathrm{\Phi }={\int }_0^{\mathrm{\infty }}\frac{\mathrm{sinx}}{{\mathrm{e}}^{\mathrm{x}}-1}\mathrm{dx}\Rightarrow \mathrm{\Phi }={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-\mathrm{x}}\mathrm{sinx}}{1-{\mathrm{e}}^{-\mathrm{x}}}\mathrm{dx}={\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{x}}\mathrm{sinx}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{nx}}\mathrm{dx}$$$$=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-(\mathrm{n}+1)\mathrm{x}}\mathrm{sinx}\mathrm{dx}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\mathrm{u}}_{\mathrm{n}}$$$${\mathrm{u}}_{\mathrm{n}}=\mathrm{Im}\left({\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-(\mathrm{n}+1)\mathrm{x}}{\mathrm{e}}^{\mathrm{ix}}\mathrm{dx}\right)\mathrm{but}$$$${{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{(-(\mathrm{n}+1)+\mathrm{i})\mathrm{x}}\mathrm{dx}=\frac{1}{-(\mathrm{n}+1)+\mathrm{i}}{\mathrm{e}}^{-(\mathrm{n}+1)+\mathrm{i})\mathrm{x}}]}_0^{\mathrm{\infty }}$$$$=\frac{1}{\mathrm{n}+1-\mathrm{i}}=\frac{\mathrm{n}+1+\mathrm{i}}{{(\mathrm{n}+1)}^2+1}\Rightarrow {\mathrm{u}}_{\mathrm{n}}=\frac{1}{{(\mathrm{n}+1)}^2+1}\Rightarrow \mathrm{\Phi }=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{1}{{(\mathrm{n}+1)}^2+1}$$$$=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{1}{{\mathrm{n}}^2+1},\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{this}\mathrm{serie}\mathrm{is}\mathrm{known}\mathrm{see}\mathrm{the}\mathrm{platform}$$

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