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Question Number 162525 by mathlove last updated on 30/Dec/21

$$\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{\sqrt{x}}{(x^2+4x+4)}=?$$

Answered by Ar Brandon last updated on 24/Mar/22

$$I={\int }_0^{\mathrm{\infty }}\frac{\sqrt{x}}{x^2+4x+4}dx,x=t^2\Rightarrow dx=2tdt$$$$=2{\int }_0^{\mathrm{\infty }}\frac{t^2}{t^4+4t^2+4}dt=2{\int }_0^{\mathrm{\infty }}\frac{t^2}{{(t^2+2)}^2}dt={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{t^2}{{(t^2+2)}^2}dt$$$$\phi \left(z\right)=\frac{z^2}{{(z^2+2)}^2},\mathrm{poles}:z_1=-i\sqrt{2},z_2=i\sqrt{2}$$$$I=2i\pi Res(\phi ,z_2)$$$$Res(\phi ,z_2)=\underset{z\to z_2}{\lim }\frac{d}{dz}\left\{\frac{z^2}{{(z-z_1)}^2}\right\}=\underset{z\to z_2}{\lim }\left\{\frac{2z{(z-z_1)}^2-2z^2(z-z_1)}{{(z-z_1)}^4}\right\}$$$$=\frac{2z_2{(z_2-z_1)}^2-2z_2^2(z_2-z_1)}{{(z_2-z_1)}^4}=\frac{2\sqrt{2}i{\left(2\sqrt{2}i\right)}^2-2{\left(i\sqrt{2}\right)}^2\left(2\sqrt{2}i\right)}{64}$$$$=\frac{-16\sqrt{2}i+8\sqrt{2}i}{64}=-\frac{\sqrt{2}}{8}i\Rightarrow I=2i\pi (-\frac{\sqrt{2}}{8}i)=\frac{\pi }{2\sqrt{2}}$$

Answered by cortano last updated on 30/Dec/21

$${\int }_0^{\mathrm{\infty }}\frac{\sqrt{x}}{x^2+4x+4}dx={\int }_0^{\mathrm{\infty }}\frac{2u^2}{u^4+4u^2+4}du$$$$[u=\sqrt{x}]$$$$={\int }_0^{\mathrm{\infty }}\frac{2u^2}{{(u^2+2)}^2}du=-\frac{u}{u^2+2}{\mid }_0^{\mathrm{\infty }}+\int \frac{du}{u^2+2}$$$$=0+\frac{1}{\sqrt{2}}{\left[{\tan }^{-1}\left(\frac{u}{\sqrt{2}}\right)\right]}_0^{\mathrm{\infty }}=\frac{\pi }{2\sqrt{2}}=\frac{\pi \sqrt{2}}{4}$$

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