Appendix A

In order to derive Eq. (7.15) we write

 

$$\mu_{kp} \frac{2}{N+1} \frac{1}{6\pi\eta b} \sqrt{\frac{6}{\pi}} \sum_{n=0}^N \cos \left( \frac{k\pi}{N+1} (n+\frac{1}{2}) \right)$$ =

$$\times \sum_{m=n-N}^n \cos \left( \frac{p\pi}{N+1} (n-m+ \frac{1}{2}) \right) \frac{1}{\sqrt{\vert m\vert}}$$

$$\frac{2}{N+1} \frac{1}{6\pi\eta b} \sqrt{\frac{6}{\pi}} \sum_{n=0... ...} (n+\frac{1}{2}) \right) \cos \left( \frac{p\pi}{ N+1} (n+\frac{1}{2}) \right)$$ =

$$\times \sum_{m=n-N}^n \cos \left( \frac{p\pi m}{N+1} \right) \fra... ...frac{6}{\pi}} \sum_{n=0}^N \cos \left( \frac{k\pi}{N+1} (n+\frac{1}{2}) \right)$$

$$\times \sin \left( \frac{p\pi}{N+1} (n+\frac{1}{2}) \right) \sum_{m=n-N}^n \sin \left( \frac{p\pi m}{N+1} \right) \frac{1}{\sqrt{ \vert m\vert}}$$ (7.30)

We now approximate

$${\sum_{m=n-N}^n \cos \left( \frac{p\pi m}{N+1} \right) \frac{1 }{\sqrt{\vert m\vert}}}$$

$$\approx \int_{-\infty}^{\infty} dm \cos \left( \frac{p\pi m}{N+1} \right) \frac{1}{\sqrt{\vert m\vert}} = \sqrt{\frac{2(N+1)}{p}}$$ (7.31)

$${ \sum_{m=n-N}^n \sin \left( \frac{p\pi m}{N+1} \right) \frac{1 }{\sqrt{\vert m\vert}}}$$

$$\approx \int_{-\infty}^{\infty} dm \sin \left( \frac{p\pi m}{N+1} \right) \frac{1}{\sqrt{\vert m\vert}} = 0$$ (7.32)

Introducing these results into Eq. (7.30) one finds Eq. (7.15). As a technical detail we notice that in principle diagonal terms in Eq. (7.14) should have been treated separately, which is clear from Eq. (5.56). Since the contributions of all other terms is proportional to N^1/2, we omit the diagonal terms.