\( \int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 4 }(\theta ) } e^{ -\cfrac { x }{ a_{ e }cos(\theta ) } } } { d\, (cos(\theta )) }\)
is an interesting one. And has solution,
\(z=cos(\theta)\)
\(\int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ z^{ 4 } } e^{ -\cfrac { x }{ a_{ e }z } } } { d\, z}=\cfrac { 1 }{ z^{ 4 } } \left\{ \cfrac { x }{ { a }_{ e } } { E }_{ i }(-\cfrac { x }{ { a }_{ e }z } )+xe^{ -\cfrac { x }{ { a }_{ e }z } } \right\} +c\)
where \(E_i(x)\) is the polyexponential function,
OR
\(\cfrac { a_{ e }(x^{ 2 }+2a_{ e }xz+2a_{ e }^{ 2 }z^{ 2 }) }{ x^{ 3 }z^{ 2 } } e^{ (-\cfrac { x }{ a_{ e }z } ) }\)
And so,
\( \int _{ 0 }^{ \theta _{ o } }{ \cfrac { 1 }{ cos^{ 4 }(\theta ) } e^{ -\cfrac { x }{ a_{ e }cos(\theta ) } } } { d\, (cos(\theta )) }=\left[\cfrac { a_{ e } }{ x^{ 3 }cos^{ 2 }(\theta ) } \left\{x^{ 2 }+2a_{ e }xcos(\theta )+2a_{ e }^{ 2 }cos^{ 2 }(\theta )\right\}e^{ (-\cfrac { x }{ a_{ e }cos(\theta ) } ) }\right]^{\theta_o}_{0}\)
Which has a DC term,
\(\cfrac { a_{ e } }{ x^{ 3 } } \left\{ x^{ 2 }+2a_{ e }x+2a_{ e }^{ 2 } \right\} e^{ (-\cfrac { x }{ a_{ e } } ) }\)
This term suggests that the \(B\) field is always present and the positively charged nucleuii are held permanently in the \(B\) orbits.
1/x^4*e^(-a/(2*x)) for 0.5≤a≤5, 1/x^4*e^(-2.5/(2*x)), and 1/x^5*e^(-2.5/(2*x)) are plotted below.
Since, \(x=cos(\theta)\), \(x\le1\) the graph \(\cfrac{1}{x^5}\gt\cfrac{1}{x^4}\). And
\(\cfrac{1}{x^4}e^{a/(2x)}\) increases with decreasing \(a\) which in our original expression is \(x\), the distance between the center of the loop and the orbiting negative charge.
