Page:Philosophical Transactions of the Royal Society A - Volume 184.djvu/48

48 $$\int_0^\pi\frac{\cos\phi\,d\phi}{\sqrt{r^2 + c^2 - 2cr\sin\theta\,\cos\phi}}$$

$$= \frac{2}{\mathrm{R}_1}\int_0^{\frac{\pi}{2}}\frac{\left(2\cos^2\phi - 1\right)\,d\phi}{\sqrt{1 - \frac{\mathrm{R}_1^2 - \mathrm{R}^2}{\mathrm{R}_1^2}\cos^2\phi}}$$

$$= \frac{2}{\mathrm{R}_1}\int_0^{\frac{\pi}{2}}\frac{2\left(\cos^2\phi - \frac{\mathrm{R}_1^2}{\mathrm{R}_1^2 - \mathrm{R}^2}\right) - 1 + \frac{2\mathrm{R}_1^2}{\mathrm{R}_1^2 - \mathrm{R}^2}}{\sqrt{1 - \frac{\mathrm{R}_1^2 - \mathrm{R}^2}{\mathrm{R}_1^2}\cos^2\phi}}\,d\phi,$$

$$= \frac{2}{\mathrm{R}_1}\left\{\frac{2}{1 - \frac{\mathrm{R}^2}{\mathrm{R}_1^2}}\left[\int_0^{\frac{\pi}{2}}\frac{d\phi}{\sqrt{1 - \frac{\mathrm{R}_1^2 - \mathrm{R}^2}{\mathrm{R}_1^2}\sin^2\phi}} - \int_0^{\frac{\pi}{2}}\sqrt{1 - \frac{\mathrm{R}_1^2 - \mathrm{R}^2}{\mathrm{R}_1^2}\sin^2\phi}\,d\phi\right] - \int_0^{\frac{\pi}{2}}\frac{d\phi}{\sqrt{1 - \frac{\mathrm{R}_1^2 - \mathrm{R}^2}{\mathrm{R}_1^2}\sin^2\phi}}\right\}$$

$$= \frac{4}{\mathrm{R}_1\left(1 - \frac{\mathrm{R}^2}{\mathrm{R}_1^2}\right)}\left[\left\{\log\frac{4\mathrm{R}_1}{\mathrm{R}} + \frac{1^2}{2^2}\frac{\mathrm{R}^2}{\mathrm{R}_1^2}\left(\log\frac{4\mathrm{R}_1}{\mathrm{R}} - \frac{2}{1.2}\right) +\,.\,.\,.\,\right\} - \left\{1 + \frac{1}{2}\frac{\mathrm{R}^2}{\mathrm{R}_1^2}\left(\log\frac{4\mathrm{R}_1}{\mathrm{R}} - \frac{1}{1.2}\right) + \frac{1^2\cdot3}{2^2\cdot4}\frac{\mathrm{R}^4}{\mathrm{R}_1^4}\left(\log\frac{4\mathrm{R}_1}{\mathrm{R}} - \frac{2}{1.2} - \frac{2}{3.4}\right) +\,.\,.\,.\,\right\}\right]$$$$ - \frac{2}{\mathrm{R}^\prime}\left\{\log\frac{4\mathrm{R}_1}{\mathrm{R}} + \frac{1^2}{2^2}\frac{\mathrm{R}^2}{\mathrm{R}_1^2}\left(\log\frac{4\mathrm{R}_1}{\mathrm{R}} - \frac{2}{1.2}\right) + \frac{1^2\cdot3^2}{2^2\cdot4^2}\frac{\mathrm{R}^4}{\mathrm{R}_1^4}\left(\log\frac{4\mathrm{R}_1}{\mathrm{R}} - \frac{2}{1.2} - \frac{2}{3.4}\right) +\,.\,.\,.\,\right\}$$

[, 'Ell. Func.,' p. 54.]

$$= \frac{2}{\mathrm{R}_1}\left\{\log\frac{4\mathrm{R}_1}{\mathrm{R}} - 2 + \frac{\mathrm{R}^2}{\mathrm{R}_1^2}\frac{5\log\frac{4\mathrm{R}_1}{\mathrm{R}} - 7}{4} + \frac{\mathrm{R}^4}{\mathrm{R}_1^4}\frac{162\log\frac{4\mathrm{R}_1}{\mathrm{R}} - 225}{128} + etc.\right\}.$$

Now

$\mathrm{R}_1^2 = 4c^2 - 4c\mathrm{R}\cos\chi + \mathrm{R}^2$

$= 4c^2\left(1 - \frac{\mathrm{R}}{c}\cos\chi + \frac{\mathrm{R}^2}{4c^2}\right)$|undefined

$= 4c^2\left(1 - s\cos\chi + \frac{s^2}{4}\right)$

Therefore

$\log\frac{4\mathrm{R}_1}{\mathrm{R}} = \log\frac{8}{s} - \frac{s}{2}\cos\chi - \frac{s^2}{4}\frac{\cos2\chi}{2} - \frac{s^3}{8}\frac{\cos3\chi}{3} -\,$&c.|undefined

$= l + 2 - \frac{s}{2}\cos\chi - \frac{s^2}{8}\cos2\chi - \frac{s^3}{24}\cos3\chi -\,$&c.

$\frac{\mathrm{R}^2}{4\mathrm{R}_1^2} = \frac{s^2}{16} + \frac{s^3}{16}\cos\chi + \frac{s^4}{64}\left(1 + 2\cos2\chi\right) +\,$&c.