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Question Number 29077 by abdo imad last updated on 04/Feb/18

let give g(x)=∫_0 ^∞   ((arctan(x(1+t^2 )))/(1+t^2 ))dt  find a simple  form of  g^′ (x) without integral.

letgiveg(x)=0arctan(x(1+t2))1+t2dtfindasimpleformofg(x)withoutintegral.

Commented by abdo imad last updated on 10/Feb/18

g^′ (x)= ∫_0 ^∞     (dt/(1+x^2 (1+t^2 )^2 ))=(1/2) ∫_(−∞) ^(+∞)      (dt/(1+x^2 (1+t^2 )^2 ))  let introduce the complex function  ϕ(z) =(1/(1+x^2 (1+z^2 )^2 ))  poles ofϕ?  we have  ϕ(z)=    (1/((x(1+z^2 )+i)(x(1+z^2 )−i)))  =       (1/((xz^2 +x+i)( xz^2  +x−i)))= (1/(x^2 (z^2  +1 +(i/x))(z^2  +1−(i/x))))  roots of z^2  +1 +(i/x)=0?⇔ z^2  =−1−(i/x)=(i(√(1+(i/x))))^2   ∣1+(i/x)∣=(√(1+(1/x^2 ) ))  =((√(1+x^2 ))/(∣x∣))and 1+(i/x)  =((√(1+x^2 ))/(∣x∣))((1/((√(1+x^2 ))/(∣x∣))) + (1/(x((√(1+x^2 ))/(∣x∣))))i)=r e^(iθ)   ⇒cosθ=((∣x∣)/(√(1+x^2 )))  and sinθ =((∣x∣)/(x(√(1+x^2 )))) .let suppose x>0⇒cosθ= (x/(√(1+x^2 )))  and sinθ= (1/(√(1+x^2 )))⇒tanθ= (1/x) ⇒θ=arctan((1/x))and  1+(i/x)=((√(1+x^2 ))/x) e^(iarctan((1/x)))  ⇒(√(1+(i/x)))  =(((1+x^2 )^(1/4) )/(√x)) e^((i/2)arctan((1/x)))   ⇒  z= +−i (((1+x^2 )^(1/4) )/(√x)) e^((i/2)arctan((1/x)))  the poles are  z_0 =i(((1+x^2 )^(1/4) )/(√x))(cos((1/2)arctan((1/x)) +isin((1/2)arctan((1/x)))  z_1 =−i(((1+x^2 )^(1/4) )/(√x))(cos((1/2)arctan((1/x))+isin((1/2)arctan((1/x)))  roots of z^2  +1 −(i/x)=0   be continued.....

g(x)=0dt1+x2(1+t2)2=12+dt1+x2(1+t2)2letintroducethecomplexfunctionφ(z)=11+x2(1+z2)2polesofφ?wehaveφ(z)=1(x(1+z2)+i)(x(1+z2)i)=1(xz2+x+i)(xz2+xi)=1x2(z2+1+ix)(z2+1ix)rootsofz2+1+ix=0?z2=1ix=(i1+ix)21+ix∣=1+1x2=1+x2xand1+ix=1+x2x(11+x2x+1x1+x2xi)=reiθcosθ=x1+x2andsinθ=xx1+x2.letsupposex>0cosθ=x1+x2andsinθ=11+x2tanθ=1xθ=arctan(1x)and1+ix=1+x2xeiarctan(1x)1+ix=(1+x2)14xei2arctan(1x)z=+i(1+x2)14xei2arctan(1x)thepolesarez0=i(1+x2)14x(cos(12arctan(1x)+isin(12arctan(1x))z1=i(1+x2)14x(cos(12arctan(1x)+isin(12arctan(1x))rootsofz2+1ix=0becontinued.....

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