Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

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 by maxmathsup 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 by maxmathsup 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 by maxmathsup 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 by maxmathsup by imad last updated on 13/Apr/19

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com