Evaluate $$\int_0^{\infty}\frac{x^2\cos(mx)}{(x^2+a^2)(x^2+b^2)}\,dx$$ given that $m>0,a>b>0$.
I did the following integral and got my answer as $$\frac{\pi[\cos(bm i)-\cos(am i)]}{2(b^2-a^2)},$$ however the answer mention is $$\frac{\pi [ae^{-ma}-be^{-mb}]}{2(b^2-a^2)},$$ I have no idea how did the $cos$ vanish in the final-answer ,can someone help me on this.
Hint: Consider the complex integral
$\int_C \frac{z^2.e^{imz}}{(z^2+a^2)(z^2+b^2)}dz$
where $C$ being the closed contour oriented counterclockwise consisting of the real axis $(-\infty,\infty)$ together with a big semicircle $C_R: |z|=R$ in the upper half plane enclosing all the singularities. Use Residue theorem to see
$2πi[Res(f,ai)+Res(f,bi)]=$$ \int_{C_R} \frac{z^2.e^{imz}}{(z^2+a^2)(z^2+b^2)}dz$$+ \int_{-\infty}^{\infty} \frac{x^2.e^{imx}}{(x^2+a^2)(x^2+b^2)}dx$
and Jordan's lemma to show that the first integral goes to zero.