find the value of ∫_0 ^∞ ((√t)/(1+t^2 ))dt


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Question Number 36188 by prof Abdo imad last updated on 30/May/18

$f i n d t h e v a l u e o f \int_{0}^{\infty} \frac{\sqrt{t}}{1 + t^{2}} d t$
Commented by maxmathsup by imad last updated on 15/Aug/18
$let I = ∫_0 ^∞ ((√t)/(1+t^2 )) dt changement (√t) =x give I = ∫_0 ^∞ (x/(1+x^4 ))(2x)dx = 2 ∫_0 ^∞ (x^2 /(1+x^4 )) dx =∫_(−∞) ^(+∞) (x^2 /(x^4 +1))dx let consider ϕ(z) =(z^2 /(z^4 +1)) we have ϕ(z) = (z^2 /((z^2 −i)(z^2 +i))) =(z^2 /((z−(√i))(z+(√i))(z−(√(−i)))(z+(√(−i))))) =(z^2 /((z −e^((iπ)/4) )(z+e^((−iπ)/4) )(z−e^(−((iπ)/4)) )(z+e^(−((iπ)/4)) ))) the poles of ϕ are +^− e^((iπ)/4) and +^− e^(−((iπ)/4)) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ { Res(ϕ, e^((iπ)/4) ) +Res(ϕ,−e^(−((iπ)/4)) )} Res(ϕ,z_i ) =(z_i ^2 /(4z_i ^3 )) =(z_i ^3 /(−4)) =−(1/4) z_i ^3 ⇒Res(ϕ,e^((iπ)/4) ) =−(1/4) e^(i((3π)/4)) Res(ϕ,−e^(−((iπ)/4)) ) = (((−e^(−((iπ)/4)) )^3 )/(−4)) =(1/4) e^(−((i3π)/4)) ∫_(−∞) ^(+∞) ϕ(z)dz =((2iπ)/4){− e^((i3π)/4) +e^(−((i3π)/4)) } =−((iπ)/2) ( e^((i3π)/4) −e^(−((i3π)/4)) ) =−((iπ)/2)(2i sin(((3π)/4))) = ((π(√2))/2) ⇒ I = (π/(√2)) .$
$l e t I = \int_{0}^{\infty} \frac{\sqrt{t}}{1 + t^{2}} d t c h a n g e m e n t \sqrt{t} = x g i v e$ $I = \int_{0}^{\infty} \frac{x}{1 + x^{4}} (2 x) d x = 2 \int_{0}^{\infty} \frac{x^{2}}{1 + x^{4}} d x = \int_{- \infty}^{+ \infty} \frac{x^{2}}{x^{4} + 1} d x l e t c o n s i d e r$ $φ (z) = \frac{z^{2}}{z^{4} + 1} w e h a v e φ (z) = \frac{z^{2}}{(z^{2} - i) (z^{2} + i)}$ $= \frac{z^{2}}{(z - \sqrt{i}) (z + \sqrt{i}) (z - \sqrt{- i}) (z + \sqrt{- i})} = \frac{z^{2}}{(z - e^{\frac{i π}{4}}) (z + e^{\frac{- i π}{4}}) (z - e^{- \frac{i π}{4}}) (z + e^{- \frac{i π}{4}})}$ $t h e p o l e s o f φ a r e \bar{+} e^{\frac{i π}{4}} a n d \bar{+} e^{- \frac{i π}{4}}$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {R e s (φ, e^{\frac{i π}{4}}) + R e s (φ, - e^{- \frac{i π}{4}})}$ $R e s (φ, z_{i}) = \frac{z_{i}^{2}}{4 z_{i}^{3}} = \frac{z_{i}^{3}}{- 4} = - \frac{1}{4} z_{i}^{3} \Rightarrow R e s (φ, e^{\frac{i π}{4}}) = - \frac{1}{4} e^{i \frac{3 π}{4}}$ $R e s (φ, - e^{- \frac{i π}{4}}) = \frac{{(- e^{- \frac{i π}{4}})}^{3}}{- 4} = \frac{1}{4} e^{- \frac{i 3 π}{4}}$ $\int_{- \infty}^{+ \infty} φ (z) d z = \frac{2 i π}{4} {- e^{\frac{i 3 π}{4}} + e^{- \frac{i 3 π}{4}}} = - \frac{i π}{2} (e^{\frac{i 3 π}{4}} - e^{- \frac{i 3 π}{4}})$ $= - \frac{i π}{2} (2 i s i n (\frac{3 π}{4})) = \frac{π \sqrt{2}}{2} \Rightarrow I = \frac{π}{\sqrt{2}} .$
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