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Question Number 57746 by maxmathsup by imad last updated on 10/Apr/19

$$letf\left(x\right)={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dt}{{(t^2-2xt+1)}^2}with\mid x\mid <1\left(xreal\right)$$ $$1)determineaexplicitformforf\left(x\right)$$ $$2)findalsog\left(x\right)={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{tdt}{{(t^2-2xt+1)}^3}$$ $$3)calculate{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dt}{{(t^2-\sqrt{2}t+1)}^2}and{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{tdt}{{(t^2-\sqrt{2}t+1)}^3}$$ $$4)calculateA\left(\theta \right)={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dt}{{(t^2-2cos\theta t+1)}^2}and$$ $$B\left(\theta \right)={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{tdt}{{(t^2-2cos\theta t+1)}^3}with0<\theta <\frac{\pi }{2}.$$

Commented bymaxmathsup by imad last updated on 13/Apr/19

$$1)letconsiderthecomplexfunction\phi \left(z\right)=\frac{1}{{(z^2-2xz+1)}^2}$$ $$letdeterminethepolesof\phi z^2-2xz+1=0$$ $${\mathrm{\Delta }}^{{}^{\prime }}=x^2-1={\left(i\sqrt{1-x^2}\right)}^2\Rightarrow z_1=x+i\sqrt{1-x^2}and{z}_2=x-i\sqrt{1-x^2}$$ $$\phi \left(z\right)=\frac{1}{{(z-z_1)}^2{(z-z_2)}^2}thepolesof\phi are{z}_{1}and{z}_2\left(doubles\right)$$ $$residustheoremgive{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi Res(\phi ,z_1)$$ $$Res(\phi ,z_1)={lim}_{z\to z_1}\frac{1}{(2-1)!}{\left\{{(z-z_1)}^2\phi \left(z\right)\right\}}^{\left(1\right)}$$ $$={lim}_{z\to z_1}{\left\{\frac{1}{{(z-z_2)}^2}\right\}}^{\left(1\right)}={lim}_{z\to z_1}\frac{-2(z-z_2)}{{(z-z_2)}^4}={lim}_{z\to z_2}\frac{-2}{{(z-z_2)}^3}$$ $$=\frac{-2}{{(z_1-z_2)}^3}=\frac{-2}{{\left(2i\sqrt{1-x^2}\right)}^3}=\frac{-2}{-8i(1-x^2)\sqrt{1-x^2}}=\frac{1}{4i(1-x^2)\sqrt{1-x^2}}\Rightarrow$$ $$f\left(x\right)={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi \frac{1}{4i(1-x^2)\sqrt{1-x^2}}\Rightarrow f\left(x\right)=\frac{\pi }{2(1-x^2)\sqrt{1-x^2}}$$ $$with\mid x\mid <1.$$

Commented bymaxmathsup by imad last updated on 13/Apr/19

$$2)wehave{f}^{{}^{\prime }}\left(x\right)={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\partial }{\partial x}\left(\frac{1}{{(t^2-2xt+1)}^2}\right)dt$$ $$={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{2t(t^2-2xt+1)}{{(t^2-2xt+1)}^4}dt=2{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{tdt}{{(t^2-2xt+1)}^3}=2g\left(x\right)\Rightarrow g\left(x\right)=\frac{1}{2}{f}^{{}^{\prime }}\left(x\right)$$ $$wehavef\left(x\right)=\frac{\pi }{2(1-x^2)\sqrt{1-x^2}}\Rightarrow f\left(x\right)=\frac{\pi }{2}\left\{{(1-x^2)}^{-1}{(1-x^2)}^{-\frac{1}{2}}\right\}\Rightarrow$$ $$f^{{}^{\prime }}\left(x\right)=\frac{\pi }{2}\{2x{(1-x^2)}^{-2}{(1-x^2)}^{-\frac{1}{2}}+x{(1-x^2)}^{-\frac{3}{2}}{(1-x^2)}^{-1}\}$$ $$=\frac{\pi }{2}\{\frac{2x}{{(1-x^2)}^2\sqrt{1-x^2}}+\frac{x}{(1-x^2)(1-x^2)\sqrt{1-x^2}}\}$$ $$=\frac{\pi x}{2}\{\frac{2}{{(1-x^2)}^2\sqrt{1-x^2}}+\frac{x}{{(1-x^2)}^2\sqrt{1-x^2}}\}$$ $$=\frac{\pi x(x+2)}{2{(1-x^2)}^2\sqrt{1-x^2}}\Rightarrow g\left(x\right)=\frac{\pi x(x+2)}{4{(1-x^2)}^2\sqrt{1-x^2}}.$$ $$$$

Commented bymaxmathsup by imad last updated on 13/Apr/19

$$3){\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dt}{{(t^2-\sqrt{2}t+1)}^2}=f\left(\frac{\sqrt{2}}{2}\right)=\frac{\pi }{2(1-{\left(\frac{\sqrt{2}}{2}\right)}^2\sqrt{1-{\left(\frac{\sqrt{2}}{2}\right)}^2}}$$ $$=\frac{\pi }{2(1-\frac{1}{2})\sqrt{1-\frac{1}{2}}}=\frac{\pi }{\sqrt{\frac{1}{2}}}=\pi \sqrt{2}$$ $${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{tdt}{{(t^2-\sqrt{2}t+1)}^3}=g\left(\frac{\sqrt{2}}{2}\right)=\frac{\pi \frac{\sqrt{2}}{2}(\frac{\sqrt{2}}{2}+2)}{4{(1-{\left(\frac{\sqrt{2}}{2}\right)}^2)}^2\sqrt{1-{\left(\frac{\sqrt{2}}{2}\right)}^2}}$$ $$=\frac{\pi \sqrt{2}(4+\sqrt{2})}{16{(1-\frac{1}{2})}^2\sqrt{1-\frac{1}{2}}}=\frac{\pi \sqrt{2}(4+\sqrt{2})}{4.\frac{1}{\sqrt{2}}}=\frac{2\pi (4+\sqrt{2})}{4}=\frac{\pi }{2}(4+\sqrt{2}).$$

Commented bymaxmathsup by imad last updated on 13/Apr/19

$$4)wehaveA\left(\theta \right)={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dt}{{(t^2-2cos\theta t+1)}^2}=f\left(cos\theta \right)=\frac{\pi }{2(1-{cos}^2\theta )\sqrt{1-{cos}^2\theta }}$$ $$=\frac{\pi }{2{sin}^2\theta sin\theta }=\frac{\pi }{2{sin}^3\theta }(0<\theta <\frac{\pi }{2})$$ $$alsowehaveB\left(\theta \right)={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{tdt}{{(t^2-2cos\theta t+1)}^3}\Rightarrow B\left(\theta \right)=g\left(cos\theta \right)$$ $$=\frac{\pi cos\theta (cos\theta +2)}{4{(1-{cos}^2\theta )}^2\sqrt{1-{cos}^2\theta }}=\frac{\pi cos\theta (2+cos\theta )}{4{sin}^4\theta sin\theta }=\frac{2\pi cos\theta +\pi {cos}^2\theta }{4{sin}^5\theta }.$$ $$$$

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