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Question Number 173424 by mnjuly1970 last updated on 11/Jul/22

$$$$$$\mathrm{prove}\mathrm{that}$$$$$$$${\int }_0^{\mathrm{\infty }}\frac{\sin \left(\mathrm{x}\right)}{\sinh \left(\mathrm{x}\right)}\mathrm{dx}=\frac{\pi }{2}\tanh \left(\frac{\pi }{2}\right)$$$$$$

Answered by Mathspace last updated on 11/Jul/22

$$\mathrm{\Psi }={\int }_0^{\mathrm{\infty }}\frac{sinx}{sh\left(x\right)}dx=2{\int }_0^{\mathrm{\infty }}\frac{sinx}{e^x-e^{-x}}dx$$$$=2{\int }_0^{\mathrm{\infty }}\frac{e^{-x}sinx}{1-e^{-2x}}dx$$$$=2{\int }_0^{\mathrm{\infty }}{e}^{-x}sinx\sum _{n=0}^{\mathrm{\infty }}{e}^{-2nx}dx$$$$=2\sum _{n=0}^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}{e}^{-(2n+1)x}sinxdx$$$$but{\int }_0^{\mathrm{\infty }}{e}^{-(2n+1)x}sinxdx$$$$=Im\left({\int }_0^{\mathrm{\infty }}{e}^{-(n+1)x+ix}dx\right)$$$$and{\int }_0^{\mathrm{\infty }}{e}^{-(n+1)+i)x}dx$$$${=\frac{1}{-(n+1)+i}{e}^{-(n+1)+i)x}]}_0^{\mathrm{\infty }}$$$$=\frac{1}{n+1-i}=\frac{n+1+i}{{(n+1)}^2+1}\Rightarrow$$$$Im({\int }_0^{\mathrm{\infty }}....)=\frac{1}{{(n+1)}^2+1}\Rightarrow$$$$\mathrm{\Psi }=2\sum _{n=0}^{\mathrm{\infty }}\frac{1}{{(n+1)}^2+1}$$$$=2\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2+1}$$$$\sum _{n=0}^{\mathrm{\infty }}\frac{1}{n^2+1}=\sum _{n=0}^{\mathrm{\infty }}\frac{1}{(n+i)(n-i)}$$$$=\frac{\mathrm{\Psi }\left(i\right)-\mathrm{\Psi }(-i)}{2i}afterweuse$$$$\mathrm{\Psi }\left(z\right)-\mathrm{\Psi }(1-z)=\pi cotan\left(\pi z\right)$$$$orwecandeveloppcos\left(\alpha x\right)$$$$atfourierserietofindthevalue...$$

Commented by mnjuly1970 last updated on 11/Jul/22

$$\mathrm{grateful}\mathrm{sir}$$

Commented by Tawa11 last updated on 13/Jul/22

$$\mathrm{Great}\mathrm{sir}$$

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