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Question Number 29077 by abdo imad last updated on 04/Feb/18
letgiveg(x)=∫0∞arctan(x(1+t2))1+t2dtfindasimpleformofg′(x)withoutintegral.
Commented by abdo imad last updated on 10/Feb/18
g′(x)=∫0∞dt1+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+x−i)=1x2(z2+1+ix)(z2+1−ix)rootsofz2+1+ix=0?⇔z2=−1−ix=(i1+ix)2∣1+ix∣=1+1x2=1+x2∣x∣and1+ix=1+x2∣x∣(11+x2∣x∣+1x1+x2∣x∣i)=reiθ⇒cosθ=∣x∣1+x2andsinθ=∣x∣x1+x2.letsupposex>0⇒cosθ=x1+x2andsinθ=11+x2⇒tanθ=1x⇒θ=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+1−ix=0becontinued.....
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