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Question Number 53600 by maxmathsup by imad last updated on 23/Jan/19

$$calculate{A}_m={\int }_0^{\mathrm{\infty }}\frac{sin\left(mx\right)}{e^{2\pi x}-1}dxwithm>0$$

Commented bymaxmathsup by imad last updated on 24/Jan/19

$$wehave{A}_m={\int }_0^{\mathrm{\infty }}\frac{e^{-2\pi x}sin\left(mx\right)}{1-e^{-2\pi x}}dx=Im\left({\int }_0^{\mathrm{\infty }}\frac{e^{-2\pi x}{e}^{imx}}{1-e^{-2\pi x}}dx\right)but$$ $${\int }_0^{\mathrm{\infty }}\frac{e^{-(2\pi -im)x}}{1-e^{-2\pi x}}dx={\int }_0^{\mathrm{\infty }}{e}^{-(2\pi -im)x}\left(\sum _{p=0}^{\mathrm{\infty }}{e}^{-2\pi px}\right)$$ $$\text{Missing left or extra right}$$ $$=\sum _{p=0}^{\mathrm{\infty }}\frac{1}{(2+2p)\pi -im}=\sum _{p=0}^{\mathrm{\infty }}\frac{(2+2p)\pi +im}{{(2+2p)}^2{\pi }^2+m^2}\Rightarrow$$ $$A_m=\sum _{p=0}^{\mathrm{\infty }}\frac{m}{4{(p+1)}^2{\pi }^2+m^2}and{A}_{m}canbecalculatedbyfourierseries$$ $$....becontinued...$$ $$$$ $$$$

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