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Question Number 130994 by mnjuly1970 last updated on 31/Jan/21

$$$$$$...advancedintegral...$$$$\mathrm{\Omega }={\int }_0^{\mathrm{\infty }}\left(\frac{e^{-x}sin\left(2x\right)}{sinh\left(x\right)}\right)dx=?$$$$$$

Answered by mindispower last updated on 31/Jan/21

$$x\to 0$$$$sin\left(2x\right)=2x+o\left(x\right)$$$$e^{-x}sin\left(2x\right)=2x+o\left(x\right),sh\left(x\right)=x+o\left(x\right)$$$$\frac{e^{-x}sin\left(2x\right)}{sh\left(x\right)}\sim 1,integrabl$$$$x\to \mathrm{\infty },sh\left(x\right)>1\Rightarrow \mid \frac{e^{-x}sin\left(2x\right)}{sh\left(x\right)}\mid <e^{-x},integrabl$$$$\Rightarrow \mathrm{\Omega }existandfinit$$$$sh\left(x\right)=\frac{e^{2x}-1}{2e^x}$$$$\mathrm{\Omega }=\int \frac{2e^{-2x}sin\left(2x\right)}{1-e^{-2x}}dx={\int }_0^{\mathrm{\infty }}\frac{e^{-t}sin\left(t\right)}{1-e^{-t}}dt$$$$=\sum _{n⩾0}{\int }_0^{\mathrm{\infty }}{e}^{-(1+n)t}sin\left(t\right)dt$$$$=\sum _{n⩾0}Im{\int }_0^{\mathrm{\infty }}{e}^{t(i-(1+n))}dt$$$$=\sum _{n⩾0}Im{\left[\frac{e^{t(i-(1+n))}}{i-(1+n)}\right]}_0^{\mathrm{\infty }}$$$$=\sum _{n⩾1}Im\left(\frac{i+(1+n)}{1+n^2}\right)=\sum _{n⩾0}\frac{1}{1+n^2}..done$$

Commented by mnjuly1970 last updated on 31/Jan/21

$$bravobravobravo$$$$mrpower$$$$extraordinary...$$

Commented by mnjuly1970 last updated on 31/Jan/21

Answered by Dwaipayan Shikari last updated on 31/Jan/21

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

Commented by mnjuly1970 last updated on 31/Jan/21

$$sepasmrpayan$$$$tashakor...$$

Answered by mathmax by abdo last updated on 31/Jan/21

$$\mathrm{\Phi }={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-\mathrm{x}}\sin \left(2\mathrm{x}\right)}{\mathrm{sh}\left(\mathrm{x}\right)}\mathrm{dx}\Rightarrow \mathrm{\Phi }=2{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-\mathrm{x}}\sin \left(2\mathrm{x}\right)}{{\mathrm{e}}^{\mathrm{x}}-{\mathrm{e}}^{-\mathrm{x}}}\mathrm{dx}$$$$=2{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-2\mathrm{x}}\sin \left(2\mathrm{x}\right)}{1-{\mathrm{e}}^{-2\mathrm{x}}}\mathrm{dx}{=}_{2\mathrm{x}=\mathrm{t}}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-\mathrm{t}}\mathrm{sint}}{1-{\mathrm{e}}^{-\mathrm{t}}}\mathrm{dt}$$$$={\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{t}}\mathrm{sint}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{nt}}\mathrm{dt}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-(\mathrm{n}+1)\mathrm{t}}\mathrm{sint}\mathrm{dt}\mathrm{but}$$$${\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-(\mathrm{n}+1)\mathrm{t}}\mathrm{sint}\mathrm{dt}=\mathrm{Im}\left({\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-(\mathrm{n}+1)\mathrm{t}+\mathrm{it}}\mathrm{dt}\right)\mathrm{and}$$$${\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{(-(\mathrm{n}+1)+\mathrm{i})\mathrm{t}}\mathrm{dt}={\left[\frac{1}{-(\mathrm{n}+1)+\mathrm{i}}{\mathrm{e}}^{(-(\mathrm{n}+1)+\mathrm{i})\mathrm{t}}\right]}_0^{\mathrm{\infty }}=\frac{1}{\mathrm{n}+1-\mathrm{i}}$$$$=\frac{\mathrm{n}+1+\mathrm{i}}{{(\mathrm{n}+1)}^2+1}\Rightarrow \mathrm{\Phi }=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{1}{{(\mathrm{n}+1)}^2+1}=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{1}{{\mathrm{n}}^2+1}$$$$\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{this}\mathrm{serie}\mathrm{is}\mathrm{known}$$

Commented by mnjuly1970 last updated on 31/Jan/21

$$thanksalotmrmathmax..$$

Commented by mathmax by abdo last updated on 31/Jan/21

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

Answered by mnjuly1970 last updated on 31/Jan/21

$$note::\int e^{ax}sin\left(bx\right)dx\stackrel{alculus\left(I\right)}{=}\frac{e^{ax}(asin\left(bx\right)-bcos\left(bx\right))}{a^2+b^2}+C$$$$\mathrm{\Omega }=2{\int }_0^{\mathrm{\infty }}\frac{e^{-x}sin\left(2x\right)}{e^x-e^{-x}}=2{\int }_0^{\mathrm{\infty }}\frac{e^{-2x}sin\left(2x\right)}{1-e^{-2x}}dx$$$$=2{\int }_0^{\mathrm{\infty }}\left(e^{-2x}sin\left(2x\right)\underset{n=0}{\sum ^{\mathrm{\infty }}}{e}^{-2nx}\right)dx$$$$=2\underset{n=0}{\sum ^{\mathrm{\infty }}}\left({\int }_0^{\mathrm{\infty }}sin\left(2x\right)e^{-2x(1+n)}dx\right)$$$$=2\underset{n=0}{\sum ^{\mathrm{\infty }}}{\left[\frac{e^{-2(1+n)x}(-2(1+n)sin\left(2x\right)-2cos\left(2x\right))}{4{(1+n)}^2+4}\right]}_0^{\mathrm{\infty }}$$$$=2\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{2}{4({(1+n)}^2+1)}=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^2+1}$$$$=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^2+1}\underset{function}{\stackrel{upsilon}{=}}\frac{\pi coth\left(\pi \right)-1}{2}✓✓$$

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