Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 119462 by mnjuly1970 last updated on 24/Oct/20

$$...advancedcalculus...$$$$evaluate::$$$$\mathrm{\Omega }={\int }_0^{\mathrm{\infty }}\frac{{tan}^{-1}\left(x\right)}{e^{2\pi x}-1}dx=?$$$$m.n.1970$$

Answered by mathmax by abdo last updated on 25/Oct/20

$$\mathrm{let}\mathrm{take}\mathrm{a}\mathrm{try}\mathrm{with}\mathrm{series}$$$$\mathrm{A}={\int }_0^{\mathrm{\infty }}\frac{\mathrm{arctanx}}{{\mathrm{e}}^{2\pi \mathrm{x}}-1}\mathrm{dx}\Rightarrow \mathrm{A}={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-2\pi \mathrm{x}}\mathrm{arctanx}}{1-{\mathrm{e}}^{-2\pi \mathrm{x}}}\mathrm{dx}$$$$={\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-2\pi \mathrm{x}}\mathrm{arctanx}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\mathrm{e}}^{-2\mathrm{n}\pi \mathrm{x}}\mathrm{dx}$$$$=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-2\pi (\mathrm{n}+1)\mathrm{x}}\mathrm{arctanx}\mathrm{dx}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\mathrm{A}}_{\mathrm{n}}$$$${\mathrm{A}}_{\mathrm{n}}={\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-2\pi (\mathrm{n}+1)\mathrm{x}}\mathrm{arctanx}\mathrm{dx}$$$${=}_{\mathrm{by}\mathrm{parts}}{[-\frac{{\mathrm{e}}^{-2\pi (\mathrm{n}+1)\mathrm{x}}}{2\pi (\mathrm{n}+1)}\mathrm{arctanx}]}_0^{\mathrm{\infty }}+{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-2\pi (\mathrm{n}+1)\mathrm{x}}}{2\pi (\mathrm{n}+1)}\times \frac{\mathrm{dx}}{1+{\mathrm{x}}^2}$$$$=\frac{1}{2\pi (\mathrm{n}+1)}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-2\pi (\mathrm{n}+1)\mathrm{x}}}{1+{\mathrm{x}}^2}\mathrm{dx}{=}_{2\pi (\mathrm{n}+1)\mathrm{x}=\mathrm{t}}\frac{1}{2\pi (\mathrm{n}+1)}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-\mathrm{t}}}{1+\frac{{\mathrm{t}}^2}{4{\pi }^2{(\mathrm{n}+1)}^2}}\times \frac{\mathrm{dt}}{2\pi (\mathrm{n}+1)}$$$$=\frac{1}{4{\pi }^2{(\mathrm{n}+1)}^2}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-\mathrm{t}}}{{\mathrm{t}}^2+4{\pi }^2{(\mathrm{n}+1)}^2}\times 4{\pi }^2{(\mathrm{n}+1)}^2\mathrm{dt}$$$$={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-\mathrm{t}}}{{\mathrm{t}}^2+4{\pi }^2{(\mathrm{n}+1)}^2}\mathrm{dt}={\int }_0^{\mathrm{\infty }}\frac{1}{{\mathrm{t}}^2+4{\pi }^2{(\mathrm{n}+1)}^2}\left(\sum _{\mathrm{p}=0}^{\mathrm{\infty }}\frac{{(-\mathrm{t})}^{\mathrm{p}}}{\mathrm{p}!}\right)\mathrm{dt}$$$$=\sum _{\mathrm{p}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{p}}}{\mathrm{p}!}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{t}}^{\mathrm{p}}}{{\mathrm{t}}^2+4{\pi }^2{(\mathrm{n}+1)}^2}\mathrm{dt}....\mathrm{be}\mathrm{continued}...$$

Answered by mindispower last updated on 25/Oct/20

$$Abelplanaformula$$$$\sum _{m⩾0}f\left(m\right)={\int }_0^{\mathrm{\infty }}f\left(x\right)dx+\frac{f\left(0\right)}{2}+i{\int }_0^{\mathrm{\infty }}\frac{f\left(it\right)-f(-it)}{e^{2\pi t}-1}dt$$$$f\left(x\right)=\frac{1}{{(x+z)}^2}\Rightarrow$$$$\sum _{n⩾0}\frac{1}{{(n+z)}^2}={\underset{0}{\int }}^{\mathrm{\infty }}\frac{1}{{(x+z)}^2}dx+\frac{1}{2z^2}+i{\int }_0^{\mathrm{\infty }}\frac{\frac{1}{{(it+z)}^2}-\frac{1}{{(-it+z)}^2}}{e^{2\pi t}-1}dt$$$$=\frac{1}{z}+\frac{1}{2z^2}+i{\int }_0^{\mathrm{\infty }}\frac{-4itz}{{(z^2+t^2)}^2(e^{2\pi t}-1)}dt$$$$=\frac{1}{z}+\frac{1}{2z^2}+2{\int }_0^{\mathrm{\infty }}\frac{2tz}{{(z^2+t^2)}^2(e^{2\pi t}-1)}dt$$$$\sum _{n⩾0}\frac{1}{{(n+z)}^2}={\mathrm{\Psi }}^{\prime }\left(z\right)={\left(ln\mathrm{\Gamma }\left(z\right)\right)}^{″}$$$$wehave$$$$ln{\left(\mathrm{\Gamma }\left(z\right)\right)}^{″}=\frac{1}{z}+\frac{1}{2z^2}+2{\int }_0^{\mathrm{\infty }}\frac{2tz}{{(z^2+t^2)}^2(e^{2\pi t}-1)}dt$$$$byintegrate$$$$\Rightarrow ln{\left(\mathrm{\Gamma }\left(z\right)\right)}^{\prime }=ln\left(z\right)-\frac{1}{2z}+2{\int }_0^{\mathrm{\infty }}-\frac{t}{(z^2+t^2)}.\frac{dt}{(e^{2\pi t}-1)}+c$$$$ln\left(\mathrm{\Gamma }\left(z\right)\right)=zln\left(z\right)-z-\frac{1}{2}ln\left(z\right)-2{\int }_0^{\mathrm{\infty }}\int \frac{t}{z^2+t^2}dz.\frac{dt}{e^{2\pi t}-1}+cz+c^{\prime }$$$$=(z-\frac{1}{2})ln\left(z\right)+(c-1)z+2{\int }_0^{\mathrm{\infty }}\frac{arctan\left(\frac{t}{z}\right)}{e^{2\pi t}-1}dt+c^{\prime }$$$$\Rightarrow [log\left(\mathrm{\Gamma }\left(z\right)\right)-(z-\frac{1}{2})ln\left(z\right)-(c-1)z-c^{\prime }]=2\int \frac{arctan\left(\frac{t}{z}\right)}{e^{2\pi t}-1}$$$$whenz\to \mathrm{\infty }?wemustehave0botheside$$$$\Rightarrow c{=}^{\prime }\frac{log\left(2\pi \right)}{2},c=0‘‘strilingformulafor\mathrm{\Gamma }{\left(z\right)}^{″}$$$$\iff$$$$log\left(\mathrm{\Gamma }\left(z\right)\right)=(z-\frac{1}{2})ln\left(z\right)-z+\frac{log\left(2\pi \right)}{2}+2{\int }_0^{\mathrm{\infty }}\frac{{\tan }^{-1}\left(\frac{t}{z}\right)}{e^{2\pi t}-1}dt$$$$forz=1$$$$\Rightarrow log\left(\mathrm{\Gamma }\left(1\right)\right)=-1+\frac{log\left(2\pi \right)}{2}+2{\int }_0^{\mathrm{\infty }}\frac{{\tan }^{-1}\left(t\right)}{e^{2\pi t}-1}dt$$$$\Rightarrow {\int }_0^{\mathrm{\infty }}\frac{{\tan }^{-1}\left(t\right)}{e^{2\pi t}-1}dt=\frac{2-log\left(2\pi \right)}{4}$$$$$$$$$$$$$$$$$$

Commented by mnjuly1970 last updated on 25/Oct/20

$$peacebeuponyoumrpower$$$$.supernice$$$$goodforyou...$$

Commented by mindispower last updated on 25/Oct/20

$$withepleasur$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com