Using complex variables to evaluate real trig integral

Blaine Jimenez

Blaine Jimenez

Answered question

2022-04-08

Using complex variables to evaluate real trig integral
02πcos(2θ)12acos(θ)+a2dθ,1<a<1
I did the substitution z=eiθ and then got cos(2θ)=12(z2+1z2) ,so
02πcos(2θ)12acos(θ)+a2dθ=1i|z|=112(z2+1z2)1azaz+a2dz.
Did I make a mistake here?

Answer & Explanation

StettyNagEragpouj

StettyNagEragpouj

Beginner2022-04-09Added 7 answers

z2+z2(az)(az1)=z4+1z(az)(az1)
Should be pretty easy to compute the residues at z=0 and z=a. (Just delete the corresponding factor from the denominator and plug in the value to compute the residues.)

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