0-x-x-2-4x-4-

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

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

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

I=∫_0 ^∞ ((√x)/(x^2 +4x+4))dx, x=t^2 ⇒dx=2tdt =2∫_0 ^∞ (t^2 /(t^4 +4t^2 +4))dt=2∫_0 ^∞ (t^2 /((t^2 +2)^2 ))dt=∫_(−∞) ^(+∞) (t^2 /((t^2 +2)^2 ))dt ϕ(z)=(z^2 /((z^2 +2)^2 )), poles: z_1 =−i(√2) , z_2 =i(√2) I=2iπRes(ϕ, z_2 ) Res (ϕ, z_2 )=lim_(z→z_2 ) (d/dz){(z^2 /((z−z_1 )^2 ))}=lim_(z→z_2 ) {((2z(z−z_1 )^2 −2z^2 (z−z_1 ))/((z−z_1 )^4 ))} =((2z_2 (z_2 −z_1 )^2 −2z_2 ^2 (z_2 −z_1 ))/((z_2 −z_1 )^4 ))=((2(√2)i(2(√2)i)^2 −2(i(√2))^2 (2(√2)i))/(64)) =((−16(√2)i+8(√2)i)/(64))=−((√2)/8)i⇒I=2iπ(−((√2)/8)i)=(π/( 2(√2)))

$${I}=\int_{\mathrm{0}} ^{\infty} \frac{\sqrt{{x}}}{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{4}}{dx},\:{x}={t}^{\mathrm{2}} \Rightarrow{dx}=\mathrm{2}{tdt} \\ $$$$\:\:\:=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{{t}^{\mathrm{2}} }{{t}^{\mathrm{4}} +\mathrm{4}{t}^{\mathrm{2}} +\mathrm{4}}{dt}=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{{t}^{\mathrm{2}} }{\left({t}^{\mathrm{2}} +\mathrm{2}\right)^{\mathrm{2}} }{dt}=\int_{−\infty} ^{+\infty} \frac{{t}^{\mathrm{2}} }{\left({t}^{\mathrm{2}} +\mathrm{2}\right)^{\mathrm{2}} }{dt} \\ $$$$\varphi\left({z}\right)=\frac{{z}^{\mathrm{2}} }{\left({z}^{\mathrm{2}} +\mathrm{2}\right)^{\mathrm{2}} },\:\mathrm{poles}:\:{z}_{\mathrm{1}} =−{i}\sqrt{\mathrm{2}}\:,\:{z}_{\mathrm{2}} ={i}\sqrt{\mathrm{2}} \\ $$$${I}=\mathrm{2}{i}\pi{Res}\left(\varphi,\:{z}_{\mathrm{2}} \right) \\ $$$${Res}\:\left(\varphi,\:{z}_{\mathrm{2}} \right)=\underset{{z}\rightarrow{z}_{\mathrm{2}} } {\mathrm{lim}}\frac{{d}}{{dz}}\left\{\frac{{z}^{\mathrm{2}} }{\left({z}−{z}_{\mathrm{1}} \right)^{\mathrm{2}} }\right\}=\underset{{z}\rightarrow{z}_{\mathrm{2}} } {\mathrm{lim}}\left\{\frac{\mathrm{2}{z}\left({z}−{z}_{\mathrm{1}} \right)^{\mathrm{2}} −\mathrm{2}{z}^{\mathrm{2}} \left({z}−{z}_{\mathrm{1}} \right)}{\left({z}−{z}_{\mathrm{1}} \right)^{\mathrm{4}} }\right\} \\ $$$$=\frac{\mathrm{2}{z}_{\mathrm{2}} \left({z}_{\mathrm{2}} −{z}_{\mathrm{1}} \right)^{\mathrm{2}} −\mathrm{2}{z}_{\mathrm{2}} ^{\mathrm{2}} \left({z}_{\mathrm{2}} −{z}_{\mathrm{1}} \right)}{\left({z}_{\mathrm{2}} −{z}_{\mathrm{1}} \right)^{\mathrm{4}} }=\frac{\mathrm{2}\sqrt{\mathrm{2}}{i}\left(\mathrm{2}\sqrt{\mathrm{2}}{i}\right)^{\mathrm{2}} −\mathrm{2}\left({i}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} \left(\mathrm{2}\sqrt{\mathrm{2}}{i}\right)}{\mathrm{64}} \\ $$$$=\frac{−\mathrm{16}\sqrt{\mathrm{2}}{i}+\mathrm{8}\sqrt{\mathrm{2}}{i}}{\mathrm{64}}=−\frac{\sqrt{\mathrm{2}}}{\mathrm{8}}{i}\Rightarrow{I}=\mathrm{2}{i}\pi\left(−\frac{\sqrt{\mathrm{2}}}{\mathrm{8}}{i}\right)=\frac{\pi}{\:\mathrm{2}\sqrt{\mathrm{2}}} \\ $$

Answered by cortano last updated on 30/Dec/21

∫_0 ^( ∞) ((√x)/(x^2 +4x+4)) dx= ∫_0 ^( ∞) ((2u^2 )/(u^4 +4u^2 +4)) du [ u=(√x) ] = ∫_0 ^( ∞) ((2u^2 )/((u^2 +2)^2 )) du = −(u/(u^2 +2 )) ∣_0 ^∞ +∫ (du/(u^2 +2)) = 0 + (1/( (√2)))[ tan^(−1) ((u/( (√2))))]_0 ^∞ = (π/(2(√2))) = ((π(√2))/4)

$$\:\int_{\mathrm{0}} ^{\:\infty} \:\frac{\sqrt{{x}}}{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{4}}\:{dx}=\:\int_{\mathrm{0}} ^{\:\infty} \:\frac{\mathrm{2}{u}^{\mathrm{2}} }{{u}^{\mathrm{4}} +\mathrm{4}{u}^{\mathrm{2}} +\mathrm{4}}\:{du}\: \\ $$$$\:\left[\:{u}=\sqrt{{x}}\:\right]\: \\ $$$$\:=\:\int_{\mathrm{0}} ^{\:\infty} \:\frac{\mathrm{2}{u}^{\mathrm{2}} }{\left({u}^{\mathrm{2}} +\mathrm{2}\right)^{\mathrm{2}} }\:{du}\:=\:−\frac{{u}}{{u}^{\mathrm{2}} +\mathrm{2}\:}\:\mid_{\mathrm{0}} ^{\infty} \:+\int\:\frac{{du}}{{u}^{\mathrm{2}} +\mathrm{2}} \\ $$$$\:=\:\mathrm{0}\:+\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left[\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{{u}}{\:\sqrt{\mathrm{2}}}\right)\right]_{\mathrm{0}} ^{\infty} =\:\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}}\:=\:\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{4}} \\ $$

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