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Question Number 25851 by abdo imad last updated on 15/Dec/17

$$findthevalueofintegral{\int }_R{(z-a)}^{-1}dzwithafromCaplly$$$$thisresulttofindthevalueof{\int }_0^{\mathrm{\infty }}{(2+x^{4_{}})}^{-1}dx.$$

Commented by abdo imad last updated on 21/Dec/17

$$letputI\left(\xi \right)={\int }_{-\xi }^{\xi }\frac{dx}{x-a}anda=\alpha +i\beta and\beta not0$$$$I\left(\xi \right)={\int }_{-\xi }^{\xi }\frac{x-\alpha +i\beta }{{(x-\alpha )}^2+{\beta }^{2^{}}}dx=A\left(\xi \right)+B\left(\xi \right)where$$$$A\left(\xi \right)={\int }_{-\xi }^{\xi }\frac{x-\alpha }{{(x-\alpha )}^2+{\beta }^2}dxandB\left(\xi \right)=i\beta {\int }_{-\xi }^{\xi }\frac{dx}{{(x-\alpha )}^2+{\beta }^2}$$$$butA\left(\xi \right)=\frac{1}{2}ln\left(\frac{{(\xi -\alpha )}^2+{\beta }^2}{{(\xi +\alpha )}^2+{\beta }^2}\right)andlim{}_{\xi ->\propto }A\left(\xi \right)=0$$$$andbythechangementx-\alpha =\beta t$$$$B\left(\xi \right)=i\beta {\int }_{\frac{-\xi -\alpha }{\beta }}^{\frac{\xi -\alpha }{\beta }}\frac{\beta }{{\beta }^2{t}^2+{\beta }^2}dt=i(artan\left(\frac{\xi -\alpha }{\beta }\right)+artan\left(\frac{\xi +\alpha }{\beta }\right))$$$$-->if\beta >0limB{\left(\xi \right)}_{\xi ->\propto }=i\pi$$$$if\beta <0limB{\left(\xi \right)}_{\xi ->\propto }=-i\pi ...lookthat\beta =im\left(a\right)$$$$valueofI={\int }_0^{\mathrm{\infty }}\frac{dx}{2+x^4}I=\frac{1}{2}{\int }_R\frac{dx}{2+x^4}andbychangement$$$$x=2^{\frac{1}{4}}t...I=\frac{2^{\frac{1}{4}}}{4}{\int }_R\frac{dt}{t^4+1}letfindthepolesoff\left(z\right)=\frac{1}{z^4+1}$$$$therootsof{z}^4+1=0arethecomplex{z}_k=e^{i(2k+1)\frac{\pi }{4}}kfrom\left[[0.3]\right]$$$$z_0=e^{i\frac{\pi }{4}}..z_1=e^{i\frac{3\pi }{4}}..z_2=e^{i\frac{5\pi }{4}}..z_3=e^{i\frac{7{\pi }_{}}{4}}$$$$f\left(z\right)=\sum _{k=0}^{k=3}\frac{{\lambda }_k}{z-z_k}and{\lambda }_k=\frac{1}{der(1+z^4)}=-\frac{1}{4}{z}_k\left(dermeansderivative\right)$$$${\int }_{R}f\left(z\right)dz=\frac{-1}{4}({\int }_R\frac{z_0}{z-z_0}dz+{\int }_R\frac{z_1}{z-z_1}dz+{\int }_R\frac{z_2}{z-z_2}dz+{\int }_R\frac{z_3}{z-z_3}dz)$$$$=\frac{-1}{4}(i\pi z_0+i\pi z_1-i\pi z_2-i\pi z_3)becauseim\left(z_0\right)>0$$$$im\left(z_1\right)>0..im\left(z_2\right)<0im\left(z_3\right)<0$$$${\int }_{R}f\left(z\right)dz=\frac{-i\pi }{4}(z_0+z_1+z_2+z_3)=\frac{-i\pi }{4}(z_0-z_0^{-}+z_0-z_0^{-})$$$$=\frac{-i\pi }{2}\left(2i\right)im\left(z_0\right)=\frac{\pi \sqrt{2}}{2}--->{\int }_0^{\mathrm{\infty }}\frac{dx}{2+x^4}=2^{\frac{1}{4}}.4^{-1}.\frac{\pi \sqrt{2}}{2}$$$$--->{\int }_R\frac{dx}{2+x^4}=\frac{\pi \sqrt{2}.2^{\frac{1}{4}}}{8}...$$

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