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Question Number 26125 by offrinshingal last updated on 20/Dec/17

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

$${\int }_R\frac{x{e}^{irx}}{(x^2-k^2)}dx-{\int }_R\frac{x{e}^{-irx}}{(x^2-k^2)}dx={\int }_R\frac{x(e^{irx}-e^{-irx})}{(x^2-k^2)}dx$$$$=2i\pi (Res(f.k)+Res(f.-k))withf\left(z\right)=\frac{z(e^{irx}-e^{-irx})}{(x^2-k^{2)}}$$$$Res(f.k)={lim}_{z->k}(z-k)f\left(z\right)={lim}_{z->k}z(e^{irz}-e^{-rz}){(z+k)}^{-1}$$$$$$$$=\frac{1}{2}(e^{irk}-e^{-irk})=isin\left(rk\right)$$$$Res(f.-k)=lim{}_{z->-k}(z+k)f\left(z\right)={lim}_{z->-k}\frac{z(e^{irz}-e^{-irz})}{(z-k)}$$$$=\frac{1}{2}(e^{-irk}-e^{irk})=-isin\left(rk\right)-->{\int }_R(...)dx-{\int }_R(...)dx=0$$

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