.

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

To continue, we reverse the order of integration, doing the $\mu$ integral first, and obtain

$$\pi \frac{R^5}{16} \int_1^\infty d\lambda \left . \left ( \l... ...^3 }{3} \right ) \right \vert _{-1}^{+1} e^{-\alpha \lambda R}$$

$$\pi \frac{R^5}{16} \int_1^\infty d\lambda \left ( \lambda^4 ... ...)^5 }{5} +\frac{1^3-(-1)^3 }{3} \right ) e^{-\alpha \lambda R}$$

i.e.,

$$\pi \frac{R^5}{16} \int_1^\infty d\lambda \left ( \frac{2}{3}... ...ac{2}{5}\lambda^2 +\frac{K}{3} \right ) e^{-\alpha \lambda R}$$