Question Number 102158 by mathmax by abdo last updated on 07/Jul/20

calculate ∫_(−∞) ^(+∞) (dx/((x^2 +1)^2 (x^2 +9)^2 ))

Answered by mathmax by abdo last updated on 09/Jul/20

I =∫_(−∞) ^(+∞) (dx/((x^2 +1)^2 (x^2 +9)^2 )) let ϕ(z)=(1/((z^2 +1)^2 (z^2 +9)^2 )) we have ϕ(z) =(1/((z−i)^2 (z+i)^2 (z−3i)^2 (z+3i)^2 )) the poles of ϕ are +^− i and +^− 3i ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ { Res(ϕ,i)+Res(ϕ,3i)} Res(ϕ,i) =lim_(z→i) (1/((2−1)!)){(z−i)^2 ϕ(z)}^((1)) lim_(z→i) {(1/((z+i)^2 (z^2 +9)^2 ))}^((1)) =lim_(z→i) −((2(z+i)(z^2 +9)^2 +2(2z)(z^2 +9)(z+i)^2 )/((z+i)^4 (z^2 +9)^4 )) =−lim_(z→i) ((2(z^2 +9)+4z(z+i))/((z+i)^3 (z^2 +9)^3 )) =−((2×8+4i(2i))/((2i)^3 ×8^3 )) =−(8/(−8i×8^3 )) =(1/(8^3 i)) Res(ϕ,3i) =lim_(z→3i) (1/((2−1)!)){(z−3i)^2 ϕ(z)}^((1)) =lim_(z→3i) {(1/((z^2 +1)^2 (z+3i)^2 ))}^((1)) =−lim_(z→3i) ((2(2z)(z^2 +1)(z+3i)^2 +2(z+3i)(z^2 +1)^2 )/((z^2 +1)^4 (z +3i)^4 )) =−lim_(z→3i) ((4z(z+3i) +2(z^2 +1))/((z^2 +1)^3 (z+3i)^3 )) =−((12i(6i)+2(−8))/((−8)^3 (6i)^3 )) =−((−12×6−16)/(−8^3 ×6^3 (−i))) =−((12×6 +16)/(8^3 ×6^3 i)) =−((72+16)/(48^3 i)) =−((88)/(48^3 i)) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{(1/(8^3 i))−((88)/(48^3 i))} =2π{ (1/8^3 ) −((88)/(48^3 ))} =I