.

$$\pi \frac{R^5}{16} \int_{-1}^{+1}d\mu \int_1^\infty d\lambda ... ...\mu^2) \left( \lambda^2 \mu^2-1 \right) e^{-\alpha \lambda R}$$

which yields, expanding,

$$\pi \frac{R^5}{16} \int_{-1}^{+1}d\mu \int_1^\infty d\lambda ... ...left( \lambda^2 \mu^2-1 \right) \right ) e^{-\alpha \lambda R}$$

$$\pi \frac{R^5}{16} \int_{-1}^{+1}d\mu \int_1^\infty d\lambda ... ...mbda^2 - \lambda^2 K^4 +\mu^2 \right ) e^{-\alpha \lambda R}$$