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Question Number 35217 by abdo.msup.com last updated on 16/May/18

$$calculate{\int }_0^{\mathrm{\infty }}\frac{xsin\left(2x\right)}{x^2+4}dx$$

Commented by prof Abdo imad last updated on 20/May/18

$$letputI={\int }_0^{\mathrm{\infty }}\frac{xsin\left(2x\right)}{x^2+4}dx$$$$2I={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{xsin\left(2x\right)}{x^2+4}dx=Im\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{xe}^{i2x}}{x^2+4}\right)dx$$$$letconsiderthecompolexfinction$$$$\phi \left(z\right)=\frac{z{e}^{i2z}}{z^2+4}$$$$\phi \left(z\right)=\frac{z{e}^{i2z}}{(z-2i)(z+2i)}sothepolesof\phi are$$$$2iand-2i$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi Res(\phi ,2i)$$$$Res(\phi ,2i)={lim}_{z\to 2i}(z-2i)\phi \left(z\right)$$$$=\frac{\left(2i\right)e^{2i\left(2i\right)}}{4i}=\frac{1}{2}{e}^{-4}\Rightarrow$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi \frac{e^{-4}}{2}=i\pi e^{-4}\Rightarrow$$$$2I=\pi e^{-4}\Rightarrow I=\frac{\pi }{2}{e}^{-4}$$$$★I=\frac{\pi }{2}{e}^{-4}★$$

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