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Question Number 129684 by mnjuly1970 last updated on 17/Jan/21

$$...advancedcalculus...$$$$provethat:$$$${\int }_0^{\pi }cos(tan\left(x\right)-cot\left(x\right))dx=\frac{\pi }{e^2}$$$$$$

Answered by mindispower last updated on 17/Jan/21

$$=2{\int }_0^{\frac{\pi }{2}}cos(tg\left(x\right)-cot\left(x\right))dx$$$$=2{\int }_0^{\mathrm{\infty }}\frac{cos(t-\frac{1}{t})}{1+t^2}dt$$$$=Re{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{e^{i(t-\frac{1}{t})}}{1+t^2}$$$$t={re}^{iϵ},ϵ\in [0,\pi [$$$$e^{i({re}^{iϵ}-\frac{e^{-iϵ}}{r})}=e^{i(t-\frac{1}{t})},$$$$e^i({re}^{iϵ}-\frac{e^{-iϵ}}{r})\to 0,r\to \mathrm{\infty }$$$$wehnr\to 0$$$$r\to 0^{+}$$$$e^{i({re}^{iϵ}+i\frac{sin\left(ϵ\right)-cos\left(ϵ\right)}{r})}\to 0$$$$\Rightarrow {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{e^{i(t-\frac{1}{t})}}{1+t^2}dt=2i\pi .(Res(\frac{e^{i(t-\frac{1}{t})}}{1+t^2},t=i)$$$$=2i\pi Re\left(\frac{e^{i(i-\frac{1}{i})}}{2i}\right)=2i\pi .\left(\frac{e^{-2}}{2i}\right)=\frac{\pi }{e^2}$$$$2{\int }_0^{\mathrm{\infty }}\frac{cos(t-\frac{1}{t})}{1+t^2}dt=Re\left(\frac{\pi }{e^2}\right)=\frac{\pi }{e^2}$$$$$$$$$$

Answered by mathmax by abdo last updated on 17/Jan/21

$$\mathrm{I}={\int }_0^{\frac{\pi }{2}}\cos (\mathrm{tanx}-\frac{1}{\mathrm{tanx}})\mathrm{dx}+{\int }_{\frac{\pi }{2}}^{\pi }\cos (\mathrm{tanx}-\frac{1}{\mathrm{tanx}})\mathrm{dx}(\to \mathrm{x}=\frac{\pi }{2}+\mathrm{t})$$$$=2{\int }_0^{\frac{\pi }{2}}\cos (\mathrm{tanx}-\frac{1}{\mathrm{tanx}})\mathrm{dx}{=}_{\mathrm{tanx}=\mathrm{t}}2{\int }_0^{\mathrm{\infty }}\cos (\mathrm{t}-\frac{1}{\mathrm{t}})\frac{\mathrm{dt}}{{\mathrm{t}}^2+1}$$$$={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\cos (\mathrm{t}-\frac{1}{\mathrm{t}})}{{\mathrm{t}}^2+1}\mathrm{dt}=\mathrm{Re}\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{\mathrm{e}}^{\mathrm{i}(\mathrm{t}-\frac{1}{\mathrm{t}})}}{{\mathrm{t}}^2+1}\mathrm{dt}\right)\mathrm{let}$$$$\phi \left(\mathrm{z}\right)=\frac{{\mathrm{e}}^{\mathrm{i}(\mathrm{z}-\frac{1}{\mathrm{z}})}}{{\mathrm{z}}^2+1}\Rightarrow \phi \left(\mathrm{z}\right)=\frac{{\mathrm{e}}^{\mathrm{i}(\mathrm{z}-\frac{1}{\mathrm{z}})}}{(\mathrm{z}-\mathrm{i})(\mathrm{z}+\mathrm{i})}\mathrm{residus}[\mathrm{theorem}\mathrm{give}$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(\mathrm{z}\right)\mathrm{dz}=2\mathrm{i}\pi \mathrm{Res}(\phi ,\mathrm{i})=2\mathrm{i}\pi \times \frac{{\mathrm{e}}^{\mathrm{i}(\mathrm{i}-\frac{1}{\mathrm{i}})}}{2\mathrm{i}}=\pi {\mathrm{e}}^{-1-1}=\frac{\pi }{{\mathrm{e}}^2}\Rightarrow$$$$★\mathrm{I}=\frac{\pi }{{\mathrm{e}}^2}★$$

Answered by mnjuly1970 last updated on 20/Jan/21

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