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Question Number 33987 by abdo imad last updated on 28/Apr/18

$$find{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{cos\left(\alpha x\right)}{{(1+x^2)}^3}dxwith\alpha ⩾0.$$

Commented by abdo mathsup 649 cc last updated on 02/May/18

$$letputf\left(\alpha \right)={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{cos\left(\alpha x\right)}{{(1+x^2)}^3}dx$$$$f\left(\alpha \right)=Re\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{e^{i\alpha x}}{{(1+x^2)}^3}dx\right)letconsiderthe$$$$compolexfunction\phi \left(z\right)=\frac{e^{i\alpha z}}{{(1+z^2)}^3}thepolesof$$$$\phi areiand-i\left(triples\right)so$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi Res(\phi ,i)$$$$Res(\phi ,i)={lim}_{z\to i}\frac{1}{(3-1)!}{\left({(z-i)}^3\phi \left(z\right)\right)}^{{}^{″}}$$$$={lim}_{z\to i}\frac{1}{2}{\left(\frac{e^{i\alpha z}}{{(z+i)}^3}\right)}^{{}^{″}}but$$$${\left(\frac{e^{i\alpha z}}{{(z+i)}^3}\right)}^{{}^{\prime }}=\frac{i\alpha e^{i\alpha 2}{(z+i)}^3-3{(z+i)}^2{e}^{i\alpha z}}{{(z+i)}^6}$$$$=\frac{i\alpha e^{i\alpha z}(z+i)-3e^{i\alpha z}}{{(z+i)}^4}$$$${\left(\frac{e^{i\alpha z}}{{(z+i)}^3}\right)}^{{}^{″}}=\frac{(-{\alpha }^2{e}^{i\alpha z}(z+i)+i\alpha e^{i\alpha z}-3i\alpha e^{i\alpha z}){(z+i)}^4-4{(z+i)}^3(i\alpha e^{i\alpha z}(z+i)-3e^{i\alpha z})}{{(z+i)}^8}$$$$=\frac{(-{\alpha }^2{e}^{i\alpha z}(z+i)-2i\alpha e^{i\alpha z})(z+i)-4(i\alpha e^{i\alpha z}(z+i)-3e^{i\alpha z})}{{(z+i)}^5}$$$$Res(\overline{\phi },i)=\frac{(-2i{\alpha }^2{e}^{-i\alpha }-2i\alpha e^{-\alpha })\left(2i\right)-4(-2\alpha e^{-\alpha }-3e^{-\alpha })}{{\left(2i\right)}^5}$$$$....becontinued...$$

Commented by abdo mathsup 649 cc last updated on 03/May/18

$$Res(\phi ,i)=\frac{4{\alpha }^2{e}^{-i\alpha }+4\alpha e^{-\alpha }+8\alpha e^{-\alpha }+12e^{-\alpha }}{32i}$$$$=\frac{4{\alpha }^2{e}^{-i\alpha }+24e^{-\alpha }}{32i}$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi \frac{4{\alpha }^2{e}^{-i\alpha }+24e^{-\alpha }}{32i}$$$$=\frac{8\pi }{32}\frac{{\alpha }^2{e}^{-i\alpha }+6e^{-\alpha }}{1}=\frac{\pi }{4}({\alpha }^{2}cos\left(\alpha \right)-i{\alpha }^{2}sin\alpha +6e^{-\alpha })$$$$\Rightarrow f\left(\alpha \right)=\frac{\pi }{4}({\alpha }^{2}cos\alpha +6e^{-\alpha }).$$

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