calculate-0-cos-abx-x-2-ax-1-x-2-bx-1-with-a-and-b-real-and-a-lt-2-b-lt-2-

Question Number 130387 by Bird last updated on 25/Jan/21

c a l c u l a t e \int_{0}^{\infty} \frac{c o s (a b x)}{(x^{2} + a x + 1) (x^{2} + b x + 1)}

w i t h a a n d b r e a l a n d ∣ a ∣< 2, ∣ b ∣< 2

Commented by mathmax by abdo last updated on 25/Jan/21

the q . is \int_{0}^{\infty} \frac{\cos (abx)}{(x^{2} + ax + 1) (x^{2} + bx + 1)} dx

Commented by mathmax by abdo last updated on 25/Jan/21

I = \int_{0}^{\infty} \frac{\cos (abx)}{(x^{2} + ax + 1) (x^{2} + bx + 1)} dx \Rightarrow

2 I = \int_{- \infty}^{+ \infty} \frac{\cos (abx)}{(x^{2} + ax + 1) (x^{2} + bx + 1)} dx = Re (\int_{- \infty}^{+ \infty} \frac{e^{iabx}}{(x^{2} + ax + 1) (x^{2} + bx + 1)} dx)

let φ (z) = \frac{e^{iabz}}{(z^{2} + az + 1) (z^{2} + bz + 1)} poles of φ ?

z^{2} + az + 1 = 0 \to Δ = a^{2} - 4 < 0 \Rightarrow z_{1} = \frac{- a + i \sqrt{4 - a^{2}}}{2} and

z_{2} = \frac{- a - i \sqrt{4 - a^{2}}}{2} also the roots of x^{2} + bx + 1 are

t_{1} = \frac{- b + i \sqrt{4 - b^{2}}}{2} and t_{2} = \frac{- b - i \sqrt{4 - b^{2}}}{2} \Rightarrow

φ (z) = \frac{e^{iabz}}{(z - z_{1}) (z - z_{2}) (z - t_{1}) (z - t_{2})} residus theorem give

\int_{R} φ (z) dz = 2 i π {Res (φ, z_{1}) + Res (φ, t_{1})}

Res (φ, z_{1}) = \frac{e^{{iabz}_{1}}}{(z_{1} - z_{2}) (z_{1} - t_{1}) (z_{1} - t_{2})}

= \frac{e^{{iabz}_{1}}}{- i \sqrt{4 - a^{2}} (z_{1}^{2} + {bz}_{1} + 1)} = \dots .

Res (φ, z_{2}) = \frac{e^{{iabz}_{2}}}{(z_{2} - z_{1}) (z_{2}^{2} + {bz}_{2} + 1)} = \dots .

rest to collect the results \dots .