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Question Number 55467 by peter frank last updated on 24/Feb/19

Commented by tanmay.chaudhury50@gmail.com last updated on 25/Feb/19

$$thinkingdifferentway...$$$${\int }_0^{\mathrm{\infty }}\frac{dx}{1+x^2}>{\int }_0^{\mathrm{\infty }}\frac{cos\left(lnx\right)}{1+x^2}dx>{\int }_0^{\mathrm{\infty }}\frac{-dx}{1+x^2}$$$$sincemaxvaluecos\left(lnx\right)=1$$$$minvaluecos\left(lnx\right)=-1$$$$so{\int }_0^{\mathrm{\infty }}\frac{dx}{1+x^2}>I>{\int }_0^{\mathrm{\infty }}\frac{-dx}{1+x^2}$$$$\mid {tan}^{-1}x{\mid }_0^{\mathrm{\infty }}>I>(-1)\mid {tan}^{-1}x{\mid }_0^{\mathrm{\infty }}$$$$\frac{\pi }{2}>I>\frac{-\pi }{2}$$

Commented by tanmay.chaudhury50@gmail.com last updated on 25/Feb/19

$$p={\int }_0^{\mathrm{\infty }}\frac{cos\left(lnx\right)}{1+x^2}dxq={\int }_0^{\mathrm{\infty }}\frac{sin\left(lnx\right)}{1+x^2}dx$$$$p+iq={\int }_0^{\mathrm{\infty }}\frac{e^{i\left(lnx\right)}}{1+x^2}dx$$$$p+iq={\int }_0^{\mathrm{\infty }}\frac{e^{{lnx}^i}}{1+x^2}dx\to {\int }_0^{\mathrm{\infty }}\frac{x^i}{1+x^2}dx$$$$x=tana$$$${\int }_0^{\frac{\pi }{2}}\frac{{\left(tana\right)}^i}{{sec}^{2}a}\times {sec}^{2}ada$$$${\int }_0^{\frac{\pi }{2}}{\left(tana\right)}^{i}da$$$$waitpls...$$$$$$

Commented by maxmathsup by imad last updated on 26/Feb/19

$$residusmethodbutnotsureitsonlyatry...$$$$letI={\int }_0^{\mathrm{\infty }}\frac{cos\left(ln\left(x\right)\right)}{1+x^2}dx\Rightarrow I=Re\left({\int }_0^{\mathrm{\infty }}\frac{e^{ilnx}}{1+x^2}dx\right)and$$$${\int }_0^{\mathrm{\infty }}\frac{e^{ilnx}}{1+x^2}dx={\int }_0^{\mathrm{\infty }}\frac{e^{\frac{i}{2}ln\left(x^2\right)}}{1+x^2}dx=\frac{1}{2}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{e^{\frac{i}{2}ln\left(x^2\right)}}{1+x^2}dxlet\phi \left(z\right)=\frac{e^{\frac{i}{2}ln\left(z^2\right)}}{1+z^2}$$$$thepolesof\phi areiand-iso{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi Res(\phi ,i)$$$$=2i\pi \frac{e^{\frac{i}{2}ln(-1)}}{2i}=\pi e^{\frac{i}{2}\left(i\pi \right)}=\pi e^{-\frac{\pi }{2}}\Rightarrow I=\frac{\pi }{2}{e}^{-\frac{\pi }{2}}.$$$$$$

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