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Question Number 28883 by abdo imad last updated on 31/Jan/18

$$find{\int }_0^{\mathrm{\infty }}\frac{dt}{{(1+t^2)}^4}$$

Commented by abdo imad last updated on 02/Feb/18

$$I=\frac{1}{2}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dt}{{(1+t^2)}^4}letputf\left(z\right)=\frac{1}{{(1+z^2)}^4}$$$$f\left(z\right)=\frac{1}{{(z-i)}^4{(z+i)}^4}thepolesoffareiand-iwithordr4$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}f\left(z\right)dz=2i\pi Res(f,i)but$$$$Res(f,i)={lim}_{z\to i}\frac{1}{(4-1)!}{\left({(z-i)}^{4}f\left(z\right)\right)}^{\left(3\right)}$$$$={lim}_{z\to i}\frac{1}{6}{\left({(z+i)}^{-4}\right)}^{\left(3\right)}andwehave$$$${{(z+i)}^{-4})}^{\left(1\right)}=-4{(z+i)}^{-5}$$$${\left({(z+i)}^{-4}\right)}^{\left(2\right)}=20{(z+i)}^{-6}$$$${\left({(z+i)}^{-4}\right)}^{\left(3\right)}=-120{(z+i)}^{-7}$$$$Res(f,i)=\frac{1}{6}(-120){\left(2i\right)}^{-7}=\frac{-20}{2^7{i}^7}but{i}^7=\frac{i^8}{i}=\frac{1}{i}$$$$Res(f,i)=\frac{-20i}{2^7}=-\frac{2^2.5i}{2^7}=-i\frac{5}{32}and$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}f\left(z\right)dz=2i\pi (-i\frac{5}{32})=\frac{10\pi }{32}=\frac{5\pi }{16}.finally$$$$I=\frac{1}{2}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}f\left(z\right)dz=\frac{5\pi }{32}.$$

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