Schwere, Elektricität und Magnetismus:153

Für ${\displaystyle V'\!}$ nehmen wir am besten den Ausdruck (7) des vorigen Paragraphen. Dann findet sich

(2)

${\displaystyle {\begin{aligned}2{\frac {\partial V'}{\partial r'}}=&\mp {\frac {a}{r'^{2}}}F(\pi )-{\frac {a}{r'^{2}}}F(0)\\&\pm \left(1+{\frac {a^{2}}{r'^{2}}}\right)\int \limits _{0}^{\pi }{\frac {F(\theta )d\theta }{(a^{2}+r'^{2}-2\,a\,r'\cos \theta )^{\frac {1}{2}}}}\\&\pm {\frac {r'^{2}-a^{2}}{r'}}\int \limits _{0}^{\pi }{\frac {(-r'+a\cos \theta )}{a^{2}+r'^{2}-2\,a\,r'\cos \theta )^{\frac {3}{2}}}}F'(\theta )d\theta ,\\\end{aligned}}}$

und daraus wird für ${\displaystyle r'=a\pm 0\!}$

(3)

${\displaystyle {\begin{aligned}2\left({\frac {\partial V'}{\partial r'}}\right)_{a\pm 0}=&\mp {\frac {1}{a}}F(\pi )-{\frac {1}{a}}F(0)\pm 2\int \limits _{0}^{\pi }{\frac {F'(\theta )d\theta }{2a\sin {\frac {1}{2}}\theta }}\\&\pm \lim {\frac {r'^{2}-a^{2}}{r'}}\int \limits _{0}^{\pi }{\frac {(-r'+a\cos \theta )F'(\theta )d\theta }{(a^{2}+r'^{2}-2\,a\,r'\cos \theta )^{\frac {3}{2}}}}.\\\end{aligned}}}$

Das letzte Integral ist noch zu transformiren. Wir schreiben

${\displaystyle {\begin{aligned}a\cos \theta -r'&=(a-r')-a(1-\cos \theta )\\&=(a-r')-2\,a\sin {\frac {1}{2}}\theta ^{2}.\\\end{aligned}}}$

Demnach ist

${\displaystyle {\begin{aligned}&\lim {\frac {r'^{2}-a^{2}}{r'}}\int \limits _{0}^{\pi }{\frac {(-r'+a\cos \theta )F'(\theta )d\theta }{a^{2}+r'^{2}-2\,a\,r'\cos \theta )^{\frac {3}{2}}}}\\=&\lim {\frac {(r'+a)}{r'}}\int \limits _{0}^{\pi }{\frac {-(r'-a)^{2}F'(\theta )d\theta }{(a^{2}+r'^{2}-2\,a\,r'\cos \theta )^{\frac {3}{2}}}}\\-&\lim {\frac {(r'^{2}-a^{2})}{r'}}\int \limits _{0}^{\pi }{\frac {2\,a\sin {\frac {1}{2}}\theta ^{2}}{\left(2\,a\sin {\frac {1}{2}}\theta \right)^{3}}}F'(\theta )d\theta .\\\end{aligned}}}$

Der letzte Bestandtheil der rechten Seite verschwindet, wenn das Integral einen endlichen Werth hat, d. h. wenn ${\displaystyle F'(0)=0\!}$ ist. Den ersten Bestandtheil zerlegen wir weiter. Es ist nemlich