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Question Number 33735 by prof Abdo imad last updated on 22/Apr/18

$$calculate{\int }_0^{\mathrm{\infty }}\frac{cos\left(2x\right)dx}{(x^2+1)(2x^2+3)}.$$

Commented by prof Abdo imad last updated on 25/Apr/18

$$letputI={\int }_0^{\mathrm{\infty }}\frac{cos\left(2x\right)}{(x^2+1)(2x^2+3)}dx$$$$2I={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{cos\left(2x\right)}{(x^2+1)(2x^2+3)}dx$$$$=Re\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{e^{i2x}}{(x^2+1)(2x^2+3)}dx\right)letconsiderthe$$$$complexfunction\phi \left(z\right)=\frac{e^{i2z}}{(z^2+1)(2z^2+3)}$$$$\phi \left(z\right)=\frac{e^{i2z}}{2(z^2+1)(z^2+\frac{3}{2})}=\frac{e^{i2z}}{2(z-i)(z+i)(z-i\sqrt{\frac{3}{2})(z+i\frac{\sqrt{3}}{2})}}$$$$sothepolesof\phi arei,-i,i\sqrt{\frac{3}{2}},-i\sqrt{\frac{3}{2}}$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi (Res(\phi ,i)+Res(\phi ,i\sqrt{\frac{3}{2}}))$$$$Res(\phi ,i)=\frac{e^{-2}}{2\left(2i\right)(-1+\frac{3}{2})}=\frac{e^{-2}}{4i.\frac{1}{2}}=\frac{e^{-2}}{2i}$$$$Res(\phi ,i\sqrt{\frac{3}{2}})=\frac{e^{-2\sqrt{\frac{3}{2}}}}{2(-\frac{3}{2}+1)\left(2i\sqrt{\frac{3}{2}}\right)}$$$$=\frac{e^{-\sqrt{6}}}{-2i\frac{\sqrt{3}}{\sqrt{2}}}=-\frac{e^{-\sqrt{6}}}{i\sqrt{6}}\Rightarrow$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi (\frac{e^{-2}}{2i}-\frac{e^{-\sqrt{6}}}{i\sqrt{6}})$$$$=\pi e^{-2}-\frac{2\pi }{\sqrt{6}}{e}^{-\sqrt{6}}\Rightarrow$$$$I=\frac{\pi }{2}{e}^{-2}-\frac{\pi }{\sqrt{6}}{e}^{-\sqrt{6}}.$$

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