1
$\begingroup$

I am trying to generate a function that represents essentially a Beta Distribution that revolved around the Y axis. To do this I am trying to integrate only one quadrant because the expansion from that is trivial. To do this I am using the following code:

$Assumptions=0<x<1 && \[Alpha]>1 && \[Beta] >1&&x<R<1;
dy=D[Sqrt[R^2-x^2],R]
CurveFunc=dy*R^(\[Alpha]-1)*(1-R)^(\[Beta]-1)/Beta[\[Alpha],\[Beta]];
CurvePDF=Expand[Integrate[CurveFunc,{R,x,1} ]]

This produces the following Integral with primarily made up with hypergeometric functions:

enter image description here

But a quick review of that function and you will see that it is undefined or singular at Integer values of [Alpha]. Which is incorrect because if you substitute in an integer value of [Alpha] before hand it will produce a valid return ie:

CurvePDF=Expand[Integrate[CurveFunc/.\[Alpha]->6,{R,x,1} ]]

Will return a function that is continuous. I would like to find the generic integral that works for all real Alpha > 1. Does anyone have an idea how to get Mathematica to produce an integral without those singularities?

|improve this question
$\endgroup$
  • $\begingroup$ Typically, solutions of this sort are "generically correct"; that is, they are valid except on a countable subset of parameter values. Unfortunately, those values (e.g. the integers) are often the values of interest in applications. $\endgroup$ – J. M. is away♦ Apr 11 '17 at 13:44
  • $\begingroup$ I have tried including the assumption that Alpha is an integer but it still produces the same always invalid result for that version. $\endgroup$ – N D Apr 11 '17 at 13:49
0
$\begingroup$

When you calculate the integral for certain values of alpha and have a close look at the results, you can extract a general rule for Integer-alphas:

 int11[r_ /; r > 1 && EvenQ[r]] = (-1)^(r/2) 1/
      Beta[r, \[Beta]] 2^-\[Beta] Sqrt[\[Pi]] Gamma[\[Beta]] MeijerG[{{1/
      2}, {(r + \[Beta])/2, (r + 1 + \[Beta])/2}}, {{0, 
      r/2}, {(r + 1)/2}}, x^2]

 int11[r_ /; r > 1 && OddQ[r]] = (-1)^((r - 1)/2) 1/
      Beta[r, \[Beta]] 2^-\[Beta] Sqrt[\[Pi]] Gamma[\[Beta]] MeijerG[{{1/
      2}, {(r + 1)/2 + \[Beta]/2, (r + \[Beta])/2}}, {{0, (r + 1)/
      2}, {r/2}}, x^2]

Here the test

    Table[
          int11[j] == 
          Expand[Integrate[CurveFunc /. \[Alpha] -> j, {R, x, 1}]] // 
          Simplify, {j, 2, 15}]

    (*   {True, True, True, True, True, True, True, True, True, \
          True, True, True, True, True}    *)

    Plot[CurvePDF /. {\[Beta] -> 2.2, x -> .1}, {\[Alpha], 1, 15}, 
       PlotRange -> All, 
       Epilog -> 
       Point[Table[{\[Alpha], 
       int11[\[Alpha]] /. {\[Beta] -> 2.2, x -> .1}}, {\[Alpha], 2, 
       15}]], AxesLabel -> {\[Alpha], 
       "CurvePDF/.{\[Beta]\[Rule]2.2,x\[Rule].1}"}]

enter image description here

|improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.