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Question Number 36188 by prof Abdo imad last updated on 30/May/18
find the value of ∫_0 ^∞ ((√t)/(1+t^2 ))dt
$${find}\:{the}\:{value}\:{of}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\sqrt{{t}}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$
Commented by maxmathsup by imad last updated on 15/Aug/18
let I = ∫_0 ^∞ ((√t)/(1+t^2 )) dt changement (√t) =x give I = ∫_0 ^∞ (x/(1+x^4 ))(2x)dx = 2 ∫_0 ^∞ (x^2 /(1+x^4 )) dx =∫_(−∞) ^(+∞) (x^2 /(x^4 +1))dx let consider ϕ(z) =(z^2 /(z^4 +1)) we have ϕ(z) = (z^2 /((z^2 −i)(z^2 +i))) =(z^2 /((z−(√i))(z+(√i))(z−(√(−i)))(z+(√(−i))))) =(z^2 /((z −e^((iπ)/4) )(z+e^((−iπ)/4) )(z−e^(−((iπ)/4)) )(z+e^(−((iπ)/4)) ))) the poles of ϕ are +^− e^((iπ)/4) and +^− e^(−((iπ)/4)) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ { Res(ϕ, e^((iπ)/4) ) +Res(ϕ,−e^(−((iπ)/4)) )} Res(ϕ,z_i ) =(z_i ^2 /(4z_i ^3 )) =(z_i ^3 /(−4)) =−(1/4) z_i ^3 ⇒Res(ϕ,e^((iπ)/4) ) =−(1/4) e^(i((3π)/4)) Res(ϕ,−e^(−((iπ)/4)) ) = (((−e^(−((iπ)/4)) )^3 )/(−4)) =(1/4) e^(−((i3π)/4)) ∫_(−∞) ^(+∞) ϕ(z)dz =((2iπ)/4){− e^((i3π)/4) +e^(−((i3π)/4)) } =−((iπ)/2) ( e^((i3π)/4) −e^(−((i3π)/4)) ) =−((iπ)/2)(2i sin(((3π)/4))) = ((π(√2))/2) ⇒ I = (π/( (√2))) .
$${let}\:{I}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\sqrt{{t}}}{\mathrm{1}+{t}^{\mathrm{2}} }\:{dt}\:\:{changement}\:\sqrt{{t}}\:={x}\:{give} \\ $$$${I}\:\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{x}}{\mathrm{1}+{x}^{\mathrm{4}} }\left(\mathrm{2}{x}\right){dx}\:=\:\mathrm{2}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{4}} }\:{dx}\:=\int_{−\infty} ^{+\infty} \:\:\:\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{4}} \:+\mathrm{1}}{dx}\:{let}\:{consider} \\ $$$$\varphi\left({z}\right)\:=\frac{{z}^{\mathrm{2}} }{{z}^{\mathrm{4}} +\mathrm{1}}\:\:{we}\:{have}\:\varphi\left({z}\right)\:=\:\frac{{z}^{\mathrm{2}} }{\left({z}^{\mathrm{2}} −{i}\right)\left({z}^{\mathrm{2}} \:+{i}\right)} \\ $$$$=\frac{{z}^{\mathrm{2}} }{\left({z}−\sqrt{{i}}\right)\left({z}+\sqrt{{i}}\right)\left({z}−\sqrt{−{i}}\right)\left({z}+\sqrt{−{i}}\right)}\:=\frac{{z}^{\mathrm{2}} }{\left({z}\:−{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}+{e}^{\frac{−{i}\pi}{\mathrm{4}}} \right)\left({z}−{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}+{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)} \\ $$$${the}\:{poles}\:{of}\:\varphi\:{are}\:\overset{−} {+}{e}^{\frac{{i}\pi}{\mathrm{4}}} \:\:{and}\:\:\overset{−} {+}\:{e}^{−\frac{{i}\pi}{\mathrm{4}}} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\left\{\:{Res}\left(\varphi,\:{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\:+{Res}\left(\varphi,−{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\right\} \\ $$$${Res}\left(\varphi,{z}_{{i}} \right)\:=\frac{{z}_{{i}} ^{\mathrm{2}} }{\mathrm{4}{z}_{{i}} ^{\mathrm{3}} }\:=\frac{{z}_{{i}} ^{\mathrm{3}} }{−\mathrm{4}}\:=−\frac{\mathrm{1}}{\mathrm{4}}\:{z}_{{i}} ^{\mathrm{3}} \:\Rightarrow{Res}\left(\varphi,{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\:=−\frac{\mathrm{1}}{\mathrm{4}}\:{e}^{{i}\frac{\mathrm{3}\pi}{\mathrm{4}}} \\ $$$${Res}\left(\varphi,−{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\:=\:\frac{\left(−{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)^{\mathrm{3}} }{−\mathrm{4}}\:=\frac{\mathrm{1}}{\mathrm{4}}\:{e}^{−\frac{{i}\mathrm{3}\pi}{\mathrm{4}}} \\ $$$$\int_{−\infty} ^{+\infty} \:\:\:\varphi\left({z}\right){dz}\:=\frac{\mathrm{2}{i}\pi}{\mathrm{4}}\left\{−\:\:{e}^{\frac{{i}\mathrm{3}\pi}{\mathrm{4}}} \:\:+{e}^{−\frac{{i}\mathrm{3}\pi}{\mathrm{4}}} \right\}\:=−\frac{{i}\pi}{\mathrm{2}}\:\left(\:{e}^{\frac{{i}\mathrm{3}\pi}{\mathrm{4}}} \:−{e}^{−\frac{{i}\mathrm{3}\pi}{\mathrm{4}}} \right) \\ $$$$=−\frac{{i}\pi}{\mathrm{2}}\left(\mathrm{2}{i}\:{sin}\left(\frac{\mathrm{3}\pi}{\mathrm{4}}\right)\right)\:=\:\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{2}}\:\:\Rightarrow\:{I}\:\:=\:\frac{\pi}{\:\sqrt{\mathrm{2}}}\:\:. \\ $$

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