aus Wikisource, der freien Quellensammlung
Anziehung eines homogenen elliptischen Cylinders.
Nun ist aber durch Differentiation leicht zu beweisen, dass
1
t
t
−
x
2
⋅
∂
t
∂
s
=
2
x
2
⋅
∂
∂
s
(
arcsin
1
−
x
2
t
)
,
{\displaystyle {\frac {1}{t{\sqrt {t-x^{2}}}}}\cdot {\frac {\partial t}{\partial s}}={\frac {2}{\sqrt {x^{2}}}}\cdot {\frac {\partial }{\partial s}}\left(\arcsin {\sqrt {1-{\frac {x^{2}}{t}}}}\right),}
∫
s
+
γ
2
(
1
+
s
β
2
)
(
1
+
s
γ
2
)
⋅
∂
t
∂
s
t
t
−
x
2
d
s
{\displaystyle \int {\frac {s+\gamma ^{2}}{\sqrt {\left(1+{\frac {s}{\beta ^{2}}}\right)\left(1+{\frac {s}{\gamma ^{2}}}\right)}}}\cdot {\frac {\frac {\partial t}{\partial s}}{t{\sqrt {t-x^{2}}}}}ds}
=
2
γ
2
x
2
1
+
s
γ
2
1
+
s
β
2
⋅
arcsin
1
−
x
2
t
{\displaystyle ={\frac {2\gamma ^{2}}{\sqrt {x^{2}}}}{\sqrt {\frac {1+{\frac {s}{\gamma ^{2}}}}{1+{\frac {s}{\beta ^{2}}}}}}\cdot \arcsin {\sqrt {1-{\frac {x^{2}}{t}}}}}
−
2
γ
2
x
2
∫
arcsin
1
−
x
2
t
⋅
d
1
+
s
γ
2
1
+
s
β
2
.
{\displaystyle -{\frac {2\gamma ^{2}}{\sqrt {x^{2}}}}\int \arcsin {\sqrt {1-{\frac {x^{2}}{t}}}}\cdot d{\sqrt {\frac {1+{\frac {s}{\gamma ^{2}}}}{1+{\frac {s}{\beta ^{2}}}}}}.}
s
=
σ
{\displaystyle s=\sigma \,}
β
γ
π
x
2
{\displaystyle {\frac {\beta \,\gamma \,\pi }{\sqrt {x^{2}}}}}
s
=
∞
{\displaystyle s=\infty \,}
(15)
Y
=
−
2
ε
β
γ
π
ρ
y
β
2
−
γ
2
+
Φ
(
y
,
z
)
−
4
ε
ρ
y
1
−
β
2
γ
2
∫
s
=
σ
∞
arcsin
1
−
x
2
t
⋅
d
1
+
s
γ
2
1
+
s
β
2
,
{\displaystyle {\begin{aligned}Y=&-{\frac {2\,\varepsilon \,\beta \,\gamma \,\pi \,\rho \,y}{\beta ^{2}-\gamma ^{2}}}+\Phi (y,\,z)\\&-{\frac {4\,\varepsilon \,\rho \,y}{1-{\frac {\beta ^{2}}{\gamma ^{2}}}}}\int \limits _{s=\sigma }^{\infty }\arcsin {\sqrt {1-{\frac {x^{2}}{t}}}}\cdot d{\sqrt {\frac {1+{\frac {s}{\gamma ^{2}}}}{1+{\frac {s}{\beta ^{2}}}}}},\\\end{aligned}}}
ε
=
+
1
{\displaystyle \varepsilon =+1\,}
x
>
0
,
{\displaystyle x>0,\,}
ε
=
−
1
{\displaystyle \varepsilon =-1\,}
x
<
0
{\displaystyle x<0\,}
Wenn wir in (15) partiell nach
x
{\displaystyle x\,}
(16)
∂
Y
∂
x
=
4
ρ
y
1
−
β
2
γ
2
∫
s
=
σ
∞
1
t
−
x
2
⋅
d
1
+
s
γ
2
1
+
s
β
2
.
{\displaystyle {\frac {\partial Y}{\partial x}}={\frac {4\,\rho \,y}{1-{\frac {\beta ^{2}}{\gamma ^{2}}}}}\int \limits _{s=\sigma }^{\infty }{\frac {1}{\sqrt {t-x^{2}}}}\cdot d{\sqrt {\frac {1+{\frac {s}{\gamma ^{2}}}}{1+{\frac {s}{\beta ^{2}}}}}}.}
