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Question Number 99413 by maths mind last updated on 20/Jun/20

$${\int }_0^{+\mathrm{\infty }}\frac{sin\left(ax\right)}{e^{2\pi x}-1}dx$$

Answered by mathmax by abdo last updated on 20/Jun/20

$$\mathrm{I}={\int }_0^{\mathrm{\infty }}\frac{\sin \left(\mathrm{ax}\right)}{{\mathrm{e}}^{2\pi \mathrm{x}}-1}\mathrm{dx}\Rightarrow \mathrm{I}={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-2\pi \mathrm{x}}\sin \left(\mathrm{ax}\right)}{1-{\mathrm{e}}^{-2\pi \mathrm{x}}}={\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-2\pi \mathrm{x}}\sin \left(\mathrm{ax}\right)\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\mathrm{e}}^{-2\pi \mathrm{nx}}\mathrm{dx}$$$$=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-(2\pi +2\pi \mathrm{n})\mathrm{x}}\sin \left(\mathrm{ax}\right)\mathrm{dx}{=}_{2\pi (\mathrm{n}+1)\mathrm{x}=\mathrm{t}}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{t}}\sin \left(\mathrm{a}\frac{\mathrm{t}}{2\pi (\mathrm{n}+1)}\right)\frac{\mathrm{dt}}{2\pi (\mathrm{n}+1)}$$$$=\frac{1}{2\pi }\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{1}{\mathrm{n}+1}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{t}}\sin \left(\frac{\mathrm{at}}{2\pi (\mathrm{n}+1)}\right)\mathrm{dt}\mathrm{we}\mathrm{have}$$$${\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{t}}\sin \left(\frac{\mathrm{at}}{2\pi (\mathrm{n}+1)}\right)\mathrm{dt}=\mathrm{Im}\left({\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-(1+\frac{\mathrm{ai}}{2\pi (\mathrm{n}+1)})\mathrm{t}}\mathrm{dt}\right)\mathrm{and}$$$${\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-(1+\frac{\mathrm{ai}}{2\pi (\mathrm{n}+1)})\mathrm{t}}\mathrm{dt}={[-\frac{1}{1+\frac{\mathrm{ai}}{2\pi (\mathrm{n}+1)}}{\mathrm{e}}^{-(1+\frac{\mathrm{ai}}{2\pi (\mathrm{n}+1)})\mathrm{t}}]}_0^{\mathrm{\infty }}$$$$=\frac{1}{1+\frac{\mathrm{ai}}{2\pi (\mathrm{n}+1)}}=\frac{2\pi (\mathrm{n}+1)}{2\pi (\mathrm{n}+1)+\mathrm{ai}}=\frac{2\pi (\mathrm{n}+1)(2\pi (\mathrm{n}+1)-\mathrm{ai})}{4{\pi }^2{(\mathrm{n}+1)}^2+{\mathrm{a}}^2}\Rightarrow$$$$\mathrm{Im}\left({\int }_0^{\mathrm{\infty }}(....)\mathrm{dt}\right)=\frac{-2\pi \mathrm{a}(\mathrm{n}+1)}{4{\pi }^2{(\mathrm{n}+1)}^2+{\mathrm{a}}^2}\Rightarrow$$$$\mathrm{I}=\frac{1}{2\pi }\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{1}{(\mathrm{n}+1)}\times \frac{-2\pi \mathrm{a}(\mathrm{n}+1)}{4{\pi }^2{(\mathrm{n}+1)}^2+{\mathrm{a}}^2}=-\mathrm{a}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{1}{4{\pi }^2{(\mathrm{n}+1)}^2+{\mathrm{a}}^2}$$$$\mathrm{and}\mathrm{this}\mathrm{serie}\mathrm{can}\mathrm{be}\mathrm{found}\mathrm{by}\mathrm{fourier}....$$

Commented by maths mind last updated on 20/Jun/20

$$nicesirthankyouforthesolutionandyourtime$$

Commented by abdomathmax last updated on 20/Jun/20

$$\mathrm{you}\mathrm{are}\mathrm{welcome}\mathrm{sir}.$$

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