nice calculus... evaluate ::::↷ Ω=∫_0 ^( ∞) (((√x) tan^(−1) (x))/(1+x^2 ))dx=???


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Question Number 125390 by mnjuly1970 last updated on 10/Dec/20

$n i c e c a l c u l u s . . .$ $e v a l u a t e ::::↷$ $Ω = \int_{0}^{\infty} \frac{\sqrt{x} {t a n}^{- 1} (x)}{1 + x^{2}} d x = ? ? ?$
	Answered by mathmax by abdo last updated on 11/Dec/20
	$I =∫_0 ^∞ (((√x)arctan(x))/(1+x^2 ))dx changement (√x)=t give x=t^2 ⇒ I=∫_0 ^∞ ((t arctan(t^2 ))/(1+t^4 ))(2t)dt =2 ∫_0 ^∞ (t^2 /(1+t^4 ))arctan(t^2 )dt =∫_(−∞) ^(+∞) ((t^2 arctan(t^2 ))/(t^4 +1))dt its clear that I is convergent let ϕ(z)=((z^2 arctan(z^2 ))/(z^4 +1)) we have ϕ(z)=((z^2 arctan(z^2 ))/((z^2 −i)(z^2 +i))) =((z^2 arctan(z^2 ))/((z−e^((iπ)/4) )(z+e^((iπ)/4) )(z−e^(−((iπ)/4)) )(z+e^(−((iπ)/4)) ))) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{Res(ϕ,e^((iπ)/4) )+Res(ϕ,−e^(−((iπ)/4)) )} Res(ϕ,e^((iπ)/4) ) =((iarctan(i))/(2e^((iπ)/4) (2i))) =(1/4)e^(−((iπ)/4)) arctan(i) Res(ϕ,−e^(−((iπ)/4)) ) =((−iarctan(−i))/((−2i)(−2e^(−((iπ)/4)) ))) =(1/4)e^((iπ)/4) arctan(i) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =((2iπ)/4) arctan(i){e^((iπ)/4) +e^(−((iπ)/4)) } =((iπ)/2)arctan(i)×2cos((π/4)) =iπarctan(i)×((√2)/2) ⇒I=((iπ)/( (√2))) arctan(i) rest to determine the value of arctan(i)... be continued...$
	$I = \int_{0}^{\infty} \frac{\sqrt{x} \arctan (x)}{1 + x^{2}} dx changement \sqrt{x} = t give x = t^{2} \Rightarrow$ $I = \int_{0}^{\infty} \frac{t \arctan (t^{2})}{1 + t^{4}} (2 t) dt = 2 \int_{0}^{\infty} \frac{t^{2}}{1 + t^{4}} \arctan (t^{2}) dt$ $= \int_{- \infty}^{+ \infty} \frac{t^{2} \arctan (t^{2})}{t^{4} + 1} dt its clear that I is convergent let$ $φ (z) = \frac{z^{2} \arctan (z^{2})}{z^{4} + 1} we have φ (z) = \frac{z^{2} \arctan (z^{2})}{(z^{2} - i) (z^{2} + i)}$ $= \frac{z^{2} \arctan (z^{2})}{(z - e^{\frac{i π}{4}}) (z + e^{\frac{i π}{4}}) (z - e^{- \frac{i π}{4}}) (z + e^{- \frac{i π}{4}})}$ $\int_{- \infty}^{+ \infty} φ (z) dz = 2 i π {Res (φ, e^{\frac{i π}{4}}) + Res (φ, - e^{- \frac{i π}{4}})}$ $Res (φ, e^{\frac{i π}{4}}) = \frac{iarctan (i)}{{2 e}^{\frac{i π}{4}} (2 i)} = \frac{1}{4} e^{- \frac{i π}{4}} \arctan (i)$ $Res (φ, - e^{- \frac{i π}{4}}) = \frac{- iarctan (- i)}{(- 2 i) (- {2 e}^{- \frac{i π}{4}})} = \frac{1}{4} e^{\frac{i π}{4}} \arctan (i) \Rightarrow$ $\int_{- \infty}^{+ \infty} φ (z) dz = \frac{2 i π}{4} \arctan (i) {e^{\frac{i π}{4}} + e^{- \frac{i π}{4}}}$ $= \frac{i π}{2} \arctan (i) \times 2 \cos (\frac{π}{4}) = i π \arctan (i) \times \frac{\sqrt{2}}{2}$ $\Rightarrow I = \frac{i π}{\sqrt{2}} \arctan (i) rest to determine the value of \arctan (i) . . .$ $be continued . . .$
	Answered by mnjuly1970 last updated on 11/Dec/20

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