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Question Number 47064 by maxmathsup by imad last updated on 04/Nov/18

$$1)calculate{u}_n={\int }_0^{\mathrm{\infty }}\frac{sin\left(nx\right)}{sh\left(2x\right)}dxwithnintegrnatural$$$$2)calculate\sum _{n=0}^{\mathrm{\infty }}{u}_n.$$

Commented by maxmathsup by imad last updated on 07/Nov/18

$$1)wehave{u}_n={\int }_0^{\mathrm{\infty }}\frac{2sin\left(nx\right)}{e^{2x}-e^{-2x}}dx=2{\int }_0^{\mathrm{\infty }}\frac{e^{-2x}sin\left(nx\right)}{1-e^{-4x}}dx$$$$=2{\int }_0^{\mathrm{\infty }}{e}^{-2x}sin\left(nx\right)\left(\sum _{p=0}^{\mathrm{\infty }}{e}^{-4px}\right)ex$$$$=2\sum _{p=0}^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}{e}^{-(2+4p)x}sin\left(nx\right)dx$$$${=}_{(2+4p)x=u}2\sum _{p=0}^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}{e}^{-u}sin\left(n\frac{u}{2+4p}\right)\frac{du}{2+4p}$$$$=\sum _{p=0}^{\mathrm{\infty }}\frac{1}{2p+1}{\int }_0^{\mathrm{\infty }}{e}^{-u}sin\left(\frac{nu}{4p+2}\right)duletdetermine{I}_{\lambda }={\int }_0^{\mathrm{\infty }}{e}^{-u}sin\left(\lambda u\right)du(\lambda >0)$$$$I_{\lambda }=Im\left({\int }_0^{\mathrm{\infty }}{e}^{-u+i\lambda u}du\right)=Im\left({\int }_0^{\mathrm{\infty }}{e}^{(-1+i\lambda )u}du\right)but$$$${\int }_0^{\mathrm{\infty }}{e}^{(-1+i\lambda )u}du={\left[\frac{1}{-1+i\lambda }{e}^{(-1+i\lambda )u}\right]}_0^{+\mathrm{\infty }}=\frac{-1}{-1+i\lambda }=\frac{1}{1-i\lambda }=\frac{1+i\lambda }{1+{\lambda }^2}\Rightarrow$$$$I_{\lambda }=\frac{\lambda }{1+{\lambda }^2}\Rightarrow u_n=\sum _{p=0}^{\mathrm{\infty }}\frac{1}{2p+1}\left(\frac{n}{(4p+2)(1+{\left(\frac{n}{4p+2}\right)}^2)}\right)$$$$=\sum _{p=0}^{\mathrm{\infty }}\frac{n}{(2p+1)(4p+2+\frac{n^2}{4p+2})}=\sum _{p=0}^{\mathrm{\infty }}\frac{n(4p+2)}{(2p+1)({(4p+2)}^2+n^2)}=\sum _{p=0}^{\mathrm{\infty }}\frac{2n}{{(4p+2)}^2+n^2}$$$$u_{n}canbecalculatedbyfourierserie....becontinued....$$$$$$

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