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Question Number 36202 by prof Abdo imad last updated on 30/May/18

$$calculate{\int }_0^{\mathrm{\infty }}\frac{x^{2}dx}{{(x^2+1)}^3}$$

Commented by math khazana by abdo last updated on 18/Aug/18

$$letA={\int }_0^{\mathrm{\infty }}\frac{x^2}{{(x^2+1)}^3}dx$$$$2A={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{x^2}{{(x^2+1)}^3}dxletconsiderthecomplex$$$$function\phi \left(z\right)=\frac{z^2}{{(z^2+1)}^3}wehave$$$$\phi \left(z\right)=\frac{z^2}{{(z-i)}^3{(z+i)}^3}sothepolesof\phi areiand-i$$$$\left(triples\right)\Rightarrow {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi Res(\phi ,i)$$$$Res(\phi ,i)={lim}_{z\to i}\frac{1}{(3-1)!}{\left\{{(z-i)}^3\phi \left(z\right)\right\}}^{\left(2\right)}$$$$=\frac{1}{2}{lim}_{z\to i}{\left\{\frac{z^2}{{(z+i)}^3}\right\}}^{\left(2\right)}$$$$=\frac{1}{2}{lim}_{z\to i}{\left\{\frac{2z{(z+i)}^3-3{(z+i)}^2{z}^2}{{(z+i)}^6}\right\}}^{\left(1\right)}$$$$=\frac{1}{2}{lim}_{z\to i}{\left\{\frac{2z(z+i)-3z^2}{{(z+i)}^4}\right\}}^{\left(1\right)}$$$$=\frac{1}{2}{lim}_{z\to i}{\left\{\frac{-z^2+2iz}{{(z+i)}^4}\right\}}^{\left(1\right)}$$$$=\frac{1}{2}{lim}_{z\to i}\left\{\frac{(-2z+2i){(z+i)}^4-4{(z+i)}^3(-z^2+2iz)}{{(z+i)}^8}\right\}$$$$=\frac{1}{2}{lim}_{z\to i}\frac{(-2z+2i)(z+i)-4(-z^2+2iz)}{{(z+i)}^5}$$$$=\frac{1}{2}\frac{-4(1-2)}{{\left(2i\right)}^5}=\frac{2}{2^{5}i}=\frac{1}{i{2}^4}=\frac{1}{16i}\Rightarrow$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(z\right)dz=2i\pi \frac{1}{16i}=\frac{\pi }{8}=2A\Rightarrow$$$$A=\frac{\pi }{16}.$$$$$$$$$$

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