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Question Number 38119 by maxmathsup by imad last updated on 22/Jun/18

$$calculate{\int }_1^{+\mathrm{\infty }}\frac{dx}{x^4\sqrt{x-1}}$$

Commented by math khazana by abdo last updated on 22/Jun/18

$$changement\sqrt{x-1}=tgivex-1=t^2\Rightarrow$$$$I={\int }_0^{+\mathrm{\infty }}\frac{2tdt}{{(1+t^2)}^{4}t}=2{\int }_0^{+\mathrm{\infty }}\frac{dt}{{(1+t^2)}^4}$$$$={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dt}{{(1+t^2)}^4}letconsiderthecomplex$$$$functionf\left(z\right)=\frac{1}{{(z^2+1)}^4}$$$$f\left(z\right)=\frac{1}{{(z-i)}^4{(z+i)}^4}sothepolesoffareiand-i$$$$\left(withmultiplicity4\right)$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}f\left(z\right)dz=2i\pi Res(f,i)$$$$Res(f,i)={lim}_{z\to i}\frac{1}{(4-1)!}{\left\{{(z-i)}^{4}f\left(z\right)\right\}}^{\left(3\right)}$$$$={lim}_{z\to i}\frac{1}{3!}{\left\{{(z+i)}^{-4}\right\}}^{\left(3\right)}but$$$${{(z+i)}^{-4}\}}^{\left(1\right)}=-4{(z+i)}^{-5}$$$${\left\{{(z+i)}^{-4}\right\}}^{\left(2\right)}=20{(z+i)}^{-6}$$$${\left\{{(z+i)}^{-4}\right\}}^{\left(3\right)}=-120{(z+i)}^{-7}$$$$Res(f,i)={lim}_{z\to i}\frac{1}{6}(-120){(z+i)}^{-7}$$$$=-20{\left(2i\right)}^{-7}=\frac{-20}{2^7{i}^7}=-\frac{2^2.5}{2^2.2^5(-i)}=\frac{5}{32i}$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}f\left(z\right)dz=2i\pi \frac{5}{32i}=\frac{5\pi }{16}\Rightarrow$$$$I=\frac{5\pi }{16}.$$

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