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Question Number 92078 by mathmax by abdo last updated on 04/May/20

$$calculate{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{cos\left(arctan(2x+1)\right)}{x^2+x+1}dx$$

Commented by abdomathmax last updated on 09/May/20

$$I={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{cos\left(arctan(2x+1)\right)dx}{x^2+x+1}$$$$I=Re\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{e^{iarctan(2x+1)}}{x^2+x+1}dx\right)let\phi \left(z\right)=\frac{e^{iarctan(2z+1)}}{z^2+z+1}$$$$polesof\phi ?z^2+z+1=0\to \mathrm{\Delta }=-3\Rightarrow$$$$z_1=\frac{-1+i\sqrt{3}}{2}=e^{i\frac{2\pi }{3}}and{z}_2=\frac{-1-i\sqrt{3}}{2}=e^{-\frac{i2\pi }{3}}$$$$residustheoremgive{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi Res(\phi ,e^{\frac{i2\pi }{3}})$$$$Res(\phi ,e^{\frac{i2\pi }{3}})=\frac{e^{iarctan(2z_1+1)}}{(z_1-z_2)}=\frac{e^{iarctan(2e^{\frac{i2\pi }{3}}+1)}}{2i\times \frac{\sqrt{3}}{2}}\Rightarrow$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=\frac{2\pi }{\sqrt{3}}{e}^{iarctan(2e^{\frac{i2\pi }{3}}+1)}$$$$weknowarctan\left(z\right)=\frac{1}{2i}ln\left(\frac{1+iz}{1-iz}\right)$$$$2e^{\frac{i2\pi }{3}}+1=2(-\frac{1}{2}+i\frac{\sqrt{3}}{2})+1=i\sqrt{3}\Rightarrow$$$$arctan(...)=arctan\left(i\sqrt{3}\right)=\frac{1}{2i}ln\left(\frac{1+i\left(i\sqrt{3}\right)}{1-i\left(i\sqrt{3}\right)}\right)$$$$=\frac{1}{2i}ln\left(\frac{1-\sqrt{3}}{1+\sqrt{3}}\right)=\frac{1}{2i}ln(-1)+\frac{1}{2i}ln\left(\frac{\sqrt{3}-1}{\sqrt{3}+1}\right)$$$$=\frac{i\pi }{2i}+\frac{1}{2i}ln\left(\frac{\sqrt{3}-1}{\sqrt{3}+1}\right)\Rightarrow$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=\frac{2\pi }{\sqrt{3}}(\frac{\pi }{2}+\frac{1}{2i}ln\left(\frac{\sqrt{3}-1}{\sqrt{3}+1}\right))$$$$=\frac{{\pi }^2}{\sqrt{3}}+-\frac{i\pi }{\sqrt{3}}ln\left(\frac{\sqrt{3}-1}{\sqrt{3}+1}\right)$$$$I=Re\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz\right)=\frac{{\pi }^2}{\sqrt{3}}$$

Commented by Ar Brandon last updated on 09/May/20

nice

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