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Question Number 52683 by maxmathsup by imad last updated on 11/Jan/19

$$letf\left(\lambda \right)={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{sin\left(\lambda x\right)}{{(x^2+2\lambda x+1)}^2}dxwith\mid \lambda \mid <1$$$$1)findthevalueoff\left(\lambda \right)$$$$2)calculate{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{sin\left(\frac{x}{2}\right)}{{(x^2+x+1)}^2}dx$$$$3)findA\left(\theta \right)={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{sin\left(\left(cos\theta \right)x\right)}{{(x^2+2cos\theta x+1)}^2}thatwesuppose0<\theta <\frac{\pi }{2}$$

Commented by maxmathsup by imad last updated on 13/Jan/19

$$1)f\left(\lambda \right)=Im\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{e^{i\lambda x}}{{(x^2+2\lambda x+1)}^2}dx\right)letconsderthecomplexfunction$$$$\phi \left(z\right)=\frac{e^{i\lambda z}}{{(x^2+2\lambda x+1)}^2}polesof\phi ?letdeterminerootsof{x}^2+2\lambda x+1$$$${\mathrm{\Delta }}^{{}^{\prime }}={\lambda }^2-1<0\Rightarrow {\mathrm{\Delta }}^{{}^{\prime }}=-(1-{\lambda }^2)={\left(i\sqrt{1-{\lambda }^2}\right)}^2\Rightarrow z_1=-\lambda +i\sqrt{1-{\lambda }^2}$$$$and{z}_2=-\lambda -i\sqrt{1-{\lambda }^2}\Rightarrow \phi havedoublepoles{z}_{1}and{z}_2$$$$\phi \left(z\right)=\frac{e^{i\lambda z}}{{(z-z_1)}^2{(z-z_2)}^2}residustheoremgive$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi Res(\phi ,z_1)but$$$$Res(\phi ,z_1)={lim}_{z\to z_1}\frac{1}{(2-1)!}{\left\{{(z-z_1)}^2\phi \left(z\right)\right\}}^{\left(1\right)}$$$$={lim}_{z\to z_1}{\left\{\frac{e^{i\lambda z}}{{(z-z_2)}^2}\right\}}^{\left(1\right)}$$$$={lim}_{z\to z_1}\frac{i\lambda e^{i\lambda z}{(z-z_2)}^2-2(z-z_2)e^{i\lambda z}}{{(z-z_2)}^4}={lim}_{z\to z_1}\frac{(i\lambda (z-z_2)-2)e^{i\lambda z}}{{(z-z_2)}^3}$$$$=\frac{(i\lambda (z_1-z_2)-2)e^{{iz}_1}}{{(z_1-z_2)}^3}=\frac{(i\lambda \left(2i\sqrt{1-{\lambda }^2}\right)-2)e^{i\lambda z_1}}{{\left(2i\sqrt{1-{\lambda }^2}\right)}^3}$$$$=\frac{(-2\lambda \sqrt{1-{\lambda }^2}-2)e^{i\lambda (-\lambda +i\sqrt{1-{\lambda }^2})}}{-8i(1-{\lambda }^2)\sqrt{1-{\lambda }^2}}=\frac{(1+\lambda \sqrt{1-{\lambda }^2})e^{-i{\lambda }^2}{e}^{-\lambda \sqrt{1-{\lambda }^2}}}{4i(1-{\lambda }^2)\sqrt{1-{\lambda }^2}}$$$$\Rightarrow {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi \frac{(1+\lambda \sqrt{1-{\lambda }^2})e^{-\lambda \sqrt{1-{\lambda }^2}}}{4i(1-{\lambda }^2)\sqrt{1-{\lambda }^2}}{e}^{-i{\lambda }^2}$$$$=\frac{\pi }{2}\frac{(1+\lambda \sqrt{1-{\lambda }^2})e^{-\lambda \sqrt{1-{\lambda }^2}}}{(1-{\lambda }^2)\sqrt{1-{\lambda }^2}}(cos\left({\lambda }^2\right)-isin\left({\lambda }^2\right))\Rightarrow$$$$f\left(\lambda \right)=-\frac{\pi }{2}sin\left({\lambda }^2\right)\frac{(1+\lambda \sqrt{1-{\lambda }^2})e^{-\lambda \sqrt{1-{\lambda }^2}}}{(1-{\lambda }^2)\sqrt{1-{\lambda }^2}}.$$$$$$

Commented by maxmathsup by imad last updated on 13/Jan/19

$$2){\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{sin\left(\frac{x}{2}\right)}{{(x^2+x+1)}^2}dx=f\left(\frac{1}{2}\right)=-\frac{\pi }{2}sin\left(\frac{1}{4}\right)\frac{(1+\frac{1}{2}\sqrt{\frac{3}{4}})e^{-\frac{1}{2}\sqrt{\frac{3}{4}}}}{\frac{3}{4}\sqrt{\frac{3}{4}}}$$$$=-\frac{\pi }{2}sin\left(\frac{1}{4}\right)\frac{(1+\frac{\sqrt{3}}{4})e^{-\frac{\sqrt{3}}{4}}}{\frac{3\sqrt{3}}{8}}=-4\pi sin\left(\frac{1}{4}\right)\frac{(1+\frac{\sqrt{3}}{4})e^{-\frac{\sqrt{3}}{4}}}{3\sqrt{3}}.$$

Commented by Abdo msup. last updated on 13/Jan/19

$$3)A\left(\theta \right)=f\left(cos\theta \right)=-\frac{\pi }{2}sin\left({cos}^2\theta \right)\frac{(1+cos\theta sin\theta )e^{-cos\theta sin\theta }}{{sin}^3\theta }$$

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