Time For The Missing Term...

From the post "Electron Orbit B Field II" dated 17 Oct 2014,

$\Pi =\cfrac { \partial \,  }{ \partial t } \int _{ 0 }^{ r }{ xtan(\theta )e^{ -\cfrac { x }{ a_{ e }cos(\theta ) }  } } d\, r$

Consider the differentiation wrt $t$ first,

$III=\left( xtan(\theta )e^{ -\cfrac { x }{ a_{ e }cos(\theta ) }  } \right) ^{ ' }=\cfrac { dx }{ dt } tan(\theta )e^{ -\cfrac { x }{ a_{ e }cos(\theta ) }  }+xsec^{ 2 }(\theta )e^{ -\cfrac { x }{ a_{ e }cos(\theta ) }  }\cfrac { d\theta  }{ dt } -xtan(\theta ).e^{ -\cfrac { x }{ a_{ e }cos(\theta ) }  }.\cfrac { 1 }{ a_{ e }cos(\theta ) } \cfrac { dx }{ dt } -xtan(\theta ).e^{ -\cfrac { x }{ a_{ e }cos(\theta ) }  }.\cfrac { x }{ a_{ e } } tan(\theta )sec(\theta )\cfrac { d\theta  }{ dt }$

$=e^{ -\cfrac { x }{ a_{ e }cos(\theta ) }  }\left\{ tan(\theta )\cfrac { dx }{ dt } +xsec^{ 2 }(\theta )\cfrac { d\theta  }{ dt } -\cfrac { x }{ a_{ e } } tan(\theta )sec(\theta )\cfrac { dx }{ dt } -\cfrac { x^{ 2 } }{ a_{ e } } tan^{ 2 }(\theta )sec(\theta )\cfrac { d\theta  }{ dt }  \right\}$

Since,

$\cfrac { r }{ x } =tan(\theta )$

$-\cfrac { r }{ x^{ 2 } } \cfrac { d\, x }{ d\, t } =sec^{ 2 }(\theta )\cfrac { d\, \theta  }{ d\, t }$

$xsec^{ 2 }(\theta )\cfrac { d\, \theta  }{ d\, t } =-\cfrac { r }{ x } \cfrac { d\, x }{ d\, t } =-\cfrac { d\, x }{ d\, t } tan(\theta )$

So,

$III=e^{ -\cfrac { x }{ a_{ e }cos(\theta ) }  }\left\{ tan(\theta )\cfrac { dx }{ dt } -\cfrac { d\, x }{ d\, t } tan(\theta )-\cfrac { x }{ a_{ e } } tan(\theta )sec(\theta )\cfrac { dx }{ dt } +\cfrac { x }{ a_{ e } } \cfrac { tan^{ 2 }(\theta ) }{ sec(\theta ) } \cfrac { dx }{ dt }  \right\}$

$III=e^{ -\cfrac { x }{ a_{ e }cos(\theta ) }  }\cfrac { x }{ a_{ e } } \left\{ \cfrac { tan^{ 2 }(\theta ) }{ sec(\theta ) } -tan(\theta )sec(\theta ) \right\} \cfrac { dx }{ dt }$

Substitute back into $\Pi$,

$\Pi =\int _{ 0 }^{ r }{ e^{ -\cfrac { x }{ a_{ e }cos(\theta ) }  }\cfrac { x }{ a_{ e } } \left\{ \cfrac { tan^{ 2 }(\theta ) }{ sec(\theta ) } -tan(\theta )sec(\theta ) \right\} \cfrac { dx }{ dt }  } d\, r$

But,

$\cfrac { r }{ x } =tan(\theta )$

$\cfrac { 1 }{ x } =sec^{ 2 }(\theta )\cfrac { d\, \theta  }{ dr }$

$dr=xsec^{ 2 }(\theta ){ d\, \theta  }$

So,

$\Pi =\cfrac { x^{ 2 } }{ a_{ e } } \cfrac { dx }{ dt } \int _{ 0 }^{ \theta_0 }{ e^{ -\cfrac { x }{ a_{ e }cos(\theta ) }  }\left\{ \cfrac { tan^{ 2 }(\theta ) }{ sec(\theta ) } -tan(\theta )sec(\theta ) \right\}  } sec^{ 2 }(\theta ){ d\, \theta  }$

$\Pi =\cfrac { x^{ 2 } }{ a_{ e } } \cfrac { dx }{ dt } \int _{ 0 }^{ \theta_0 }{ e^{ -\cfrac { x }{ a_{ e }cos(\theta ) }  }\left\{ tan^{ 2 }(\theta )sec(\theta )-tan(\theta )sec^{ 3 }(\theta ) \right\}  } { d\, \theta  }$

$\Pi =\cfrac { x^{ 2 } }{ a_{ e } } \cfrac { dx }{ dt } \int _{ 0 }^{ \theta_0 }{ e^{ -\cfrac { x }{ a_{ e }cos(\theta ) }  }\left\{ tan^{ 2 }(\theta )sec(\theta )-\cfrac { sin(\theta ) }{ cos^{ 4 }(\theta ) }  \right\}  } { d\, \theta  }$

Since,

$tan(\theta)d(sec(\theta))=tan^{ 2 }(\theta)sec(\theta)d(\theta)=\sqrt { sec^{ 2 }(\theta)-1 } d(sec(\theta))$

and

$d(cos(\theta))=-sin(\theta)$

We break $\Pi$ into two parts,

$\Pi _{ 1 }=\cfrac { x^{ 2 } }{ a_{ e } } \cfrac { dx }{ dt } \int _{ 0 }^{ \theta_0 }{ e^{ -\cfrac { x }{ a_{ e }cos(\theta ) }  }\left\{ tan^{ 2 }(\theta )sec(\theta ) \right\}  } { d\, \theta  }$

$\Pi _{ 1 }=\cfrac { x^{ 2 } }{ a_{ e } } \cfrac { dx }{ dt } \int _{ 0 }^{ sec(\theta_0) }{ e^{ -\cfrac { x.sec(\theta ) }{ a_{ e } }  }\left\{ \sqrt { sec^{ 2 }(\theta)-1 }  \right\}  } { d\, sec(\theta ) }$

$\Pi _{ 1 }=\cfrac { x^{ 2 } }{ a_{ e } } \cfrac { dx }{ dt } \int _{ 0 }^{y_0 }{ e^{ -\cfrac { x.y }{ a_{ e } }  }\left\{ \sqrt { y^{ 2 }-1 }  \right\}  } { d\, y }$

where   $y_0=sec(\theta_0)$

and

$\Pi _{ 2 }=\cfrac { x^{ 2 } }{ a_{ e } } \cfrac { dx }{ dt } \int _{ 0 }^{ \theta_0 }{ e^{ -\cfrac { x }{ a_{ e }cos(\theta ) }  }\left\{ -\cfrac { sin(\theta ) }{ cos^{ 4 }(\theta ) }  \right\}  } { d\, \theta  }$

$\Pi _{ 2 }=\cfrac { x^{ 2 } }{ a_{ e } } \cfrac { dx }{ dt } \int _{ 1 }^{ cos(\theta_0) }{ e^{ -\cfrac { x }{ a_{ e }cos(\theta ) }  }\left\{ \cfrac { 1 }{ cos^{ 4 }(\theta ) }  \right\}  } { d\, cos(\theta )  }$

$\Pi _{ 2 }=\cfrac { x^{ 2 } }{ a_{ e } } \cfrac { dx }{ dt } \int _{ 1 }^{ y_0 }{ e^{ -\cfrac { x }{ a_{ e }y }  }\left\{ \cfrac { 1 }{ y^{ 4 } }  \right\}  } { d\,y  }$

where $y_0=cos(\theta_0)$

Combining, just to look at it,

$$\Pi=\cfrac { x^{ 2 } }{ a_{ e } }\cfrac { dx }{ dt } \left\{ \int _{ 1 }^{sec(\theta_0) }{ e^{ -\cfrac { x.y }{ a_{ e } }  }\left\{ \sqrt { y^{ 2 }-1 }  \right\}  } { d\, y }+  \int _{ 1 }^{ cos(\theta_0) }{ e^{ -\cfrac { x }{ a_{ e }y }  }\left\{ \cfrac { 1 }{ y^{ 4 } }  \right\}  } { d\,y  }\right\}$$

At every point along $x$, the integration is done from zero to $\theta_0$.  When $x$ changes, $\theta_0$ changes, the limit of integration changes but not $\theta$.  The variable $\theta$ is independent of $x$.

This corrects the post "Electron Orbit B Field II" dated 17 Oct 2014.

 I apologize, I have half a brain and it is infected...