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Question Number 33835 by prof Abdo imad last updated on 25/Apr/18

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

Commented by prof Abdo imad last updated on 29/Apr/18

$$letconsiderthecomplexfunction$$$$\phi \left(z\right)=\frac{e^{i\pi z}}{{(z^2+1+i)}^2}$$$$wehave{z}^2+1+i=z^2-(-1-i)$$$$-1-i=\sqrt{2}(\frac{-1}{\sqrt{2}}-\frac{i}{\sqrt{2}})=\sqrt{2}(cos(\pi +\frac{\pi }{4})+isin(\pi +\frac{\pi }{4}))$$$$=\sqrt{2}{e}^{i\frac{5\pi }{4}}\Rightarrow z^2-(-1-i)=z^2-\sqrt{2}{e}^{i\frac{5\pi }{4}}$$$$=(z{-}^4{\left(\sqrt{2}{e}^{i\frac{5\pi }{8}}\right)}^{2}sothepolesare{2}^{\frac{1}{4}}{e}^{i\frac{5\pi }{8}}=z_0$$$$and-2^{\frac{1}{4}}{e}^{i\frac{5\pi }{8}}=z_1\left(doubles\right)$$$$I={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{cos\left(\pi x\right)}{{(x^2+1+i)}^2}dx=Re\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{e^{i\pi x}}{{(x^2+1+i)}^2}dx\right)$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi Res(\phi ,z_0)$$$$Res(\phi ,z_0)={lim}_{z\to z_0}\frac{1}{(2-1)!}{\left({(z-z_0)}^2\phi \left(z\right)\right)}^{{}^{\prime }}$$$$={lim}_{z\to z_0}{\left(\frac{e^{i\pi z}}{{(z-z_1)}^2}\right)}^{{}^{\prime }}$$$$={lim}_{z\to z_0}\frac{i\pi e^{i\pi z}{(z-z_1)}^2-2(z-z_1)e^{i\pi z}}{{(z-z_1)}^4}$$$$={lim}_{z\to z_0}\frac{i\pi e^{i\pi z}(z-z_1)-2e^{i\pi z}}{{(x-z_1)}^3}$$$$=\frac{i\pi e^{i\pi z_0}(z_0-z_1)-2e^{i\pi z_0}}{(z_0-z_1)3}$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=(-2\pi (z_0-z_1-2)(cos\left(\pi z_0\right)+isin\left(\pi z_0\right)){(z-z_0)}^{-3}$$$$I=Re\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz\right).$$

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