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Question Number 43939 by abdo.msup.com last updated on 18/Sep/18
find f(ξ) =∫_0 ^∞ (dt/(1+(t−iξ)^2 )) and calculate f^′ (ξ)
findf(ξ)=∫0∞dt1+(t−iξ)2andcalculatef′(ξ)
Commented by maxmathsup by imad last updated on 22/Sep/18
f(ξ) =∫_0 ^∞ (dt/((t−ξ)^2 −i^2 )) =∫_0 ^∞ (dt/(((t−ξ)−i)(t−ξ +i))) =(1/(2i))∫_0 ^∞ { (1/(t−ξ−i)) −(1/(t−ξ +i))}dt ⇒2i f(ξ) =∫_0 ^∞ (dt/(t−ξ−i)) −∫_0 ^∞ (dt/(t−ξ +i)) but ∫ (dt/(t−ξ−i)) =∫ ((t−ξ +i)/((t−ξ)^2 +1)) dt =∫ ((t−ξ)/((t−ξ)^2 +1))dt +i ∫ (dt/((t−ξ)^2 +1)) ∫ ((t−ξ)/((t−ξ)^2 +1))dt =(1/2)ln{ (t−ξ)^2 +1} +c_1 ∫ (dt/((t−ξ)^2 +1)) =_(t−ξ =u) ∫ (du/(1+u^2 )) =arctan(t−ξ) +c_2 ⇒ ∫ (dt/(t−ξ−i)) =(1/2)ln{(t−ξ)^2 +1} +i arctan(t−ξ) +c also ∫ (dt/(t−ξ +i)) = (1/2)ln{(t−ξ)^2 +1}−i arctan(t−ξ) ⇒ 2if(ξ)= 2i [arctan(t−ξ)]_0 ^(+∞) = 2i{ (π/2) +arctan(ξ)} ⇒ f(ξ) =(π/2) +arctan(ξ) and we have f^′ (ξ) = (1/(1+ξ^2 )) .
f(ξ)=∫0∞dt(t−ξ)2−i2=∫0∞dt((t−ξ)−i)(t−ξ+i)=12i∫0∞{1t−ξ−i−1t−ξ+i}dt⇒2if(ξ)=∫0∞dtt−ξ−i−∫0∞dtt−ξ+ibut∫dtt−ξ−i=∫t−ξ+i(t−ξ)2+1dt=∫t−ξ(t−ξ)2+1dt+i∫dt(t−ξ)2+1∫t−ξ(t−ξ)2+1dt=12ln{(t−ξ)2+1}+c1∫dt(t−ξ)2+1=t−ξ=u∫du1+u2=arctan(t−ξ)+c2⇒∫dtt−ξ−i=12ln{(t−ξ)2+1}+iarctan(t−ξ)+calso∫dtt−ξ+i=12ln{(t−ξ)2+1}−iarctan(t−ξ)⇒2if(ξ)=2i[arctan(t−ξ)]0+∞=2i{π2+arctan(ξ)}⇒f(ξ)=π2+arctan(ξ)andwehavef′(ξ)=11+ξ2.

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