How to integrate the expression ${(\frac{\omega \rho}{3c})}^2(\frac{1}{\gamma^2}+\theta^2)^2[K^2_{2/3}(\xi) +\frac{\theta^2}{\frac{1}{\gamma^2}+\theta^2}K^2_{1/3}(\xi)]$ w.r.t. $d\theta$ for the limit -infinity to infinity, where $\xi = (\omega \rho)/(3c)[1/\gamma^2 + \theta^2]^{3/2}$. This is the expression to obtain the formula for power distribution of synchrotron radiation per solid angle from a uniform circular motion of the electron.
The final answer is an integral of linear $K_{5/3}(\xi)d\xi$ from $\xi_{(\theta=0)}$ to infinity multiplied by $\xi_{(\theta=0)}$.
(Ref: J.D. Jackson, chapter 14)