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Mathematics Equation Solving Wolfram Language

Is this definite integral solvable?

Jan Krepela

Jan Krepela, I am retired

Posted 8 years ago

I have been looking at the eccentricity in induction motor. Through using the Maxwell's equations I have arrived at the equation (1) in the attached file. However, I cannot find a solution for the equation (2) in a closed form using Mathematica. I believe, that the solution can be greatly simplified as described in the attached file. Am I correct or not? Attachments:

POSTED BY: Jan Krepela

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Mike Luntz

Posted 8 years ago

This may be useless information by now, but I looked at the integral and found that there is a relatively simple equation for this integral, at least for integer values of p. See the attached notebook. Attachments:

POSTED BY: Mike Luntz

Kay Herbert

Kay Herbert, Chief scientist, Sloan Valve Co.

Posted 8 years ago

Hyper geometric functions aren't uncommon. if p=2 it greatly simplifies: In[19]:= -(( 2 \[Pi] HypergeometricPFQRegularized[{1/2, 1, 1}, {1 - p, 1 + p}, ( 2 e)/(-1 + e)])/(-1 + e)) /. p -> 2 Out[19]= (2 (-2 + 2 Sqrt[(-1 - e)/(-1 + e)] - 2 Sqrt[(-1 - e)/(-1 + e)] e + e^2) \[Pi])/(Sqrt[(-1 - e)/(-1 + e)] (-1 + e) e^2) In[20]:= Simplify[( 2 (-2 + 2 Sqrt[(-1 - e)/(-1 + e)] - 2 Sqrt[(-1 - e)/(-1 + e)] e + e^2) \[Pi])/(Sqrt[(-1 - e)/(-1 + e)] (-1 + e) e^2)] Out[20]= -(( 2 (2 - e^2 - 2 Sqrt[(1 + e)/(1 - e)] + 2 e Sqrt[(1 + e)/(1 - e)]) \[Pi])/((-1 + e) e^2 Sqrt[(1 + e)/( 1 - e)]))

Hyper geometric functions aren't uncommon. if p=2 it greatly simplifies:

In[19]:= -((
  2 \[Pi] HypergeometricPFQRegularized[{1/2, 1, 1}, {1 - p, 1 + p}, (
    2 e)/(-1 + e)])/(-1 + e)) /. p -> 2

Out[19]= (2 (-2 + 2 Sqrt[(-1 - e)/(-1 + e)] - 
   2 Sqrt[(-1 - e)/(-1 + e)] e + e^2) \[Pi])/(Sqrt[(-1 - e)/(-1 + 
  e)] (-1 + e) e^2)

In[20]:= Simplify[(
 2 (-2 + 2 Sqrt[(-1 - e)/(-1 + e)] - 2 Sqrt[(-1 - e)/(-1 + e)] e + 
    e^2) \[Pi])/(Sqrt[(-1 - e)/(-1 + e)] (-1 + e) e^2)]

Out[20]= -((
 2 (2 - e^2 - 2 Sqrt[(1 + e)/(1 - e)] + 
    2 e Sqrt[(1 + e)/(1 - e)]) \[Pi])/((-1 + e) e^2 Sqrt[(1 + e)/(
  1 - e)]))

enter image description here

POSTED BY: Kay Herbert

Jan Krepela

Jan Krepela, I am retired

Posted 8 years ago

Hello Herbert, At this moment I am kind of overwhelmed. If the function you have arrived at were called sin or arctg I would know what you are talking about. But the Out[34] is something I have never dreamed existed in mathematical language. I have even doubted I have joined the correct community after the e-mail from the moderator. I have a limited knowledge of mathematics, that is why I have bought Mathematica two month ago to help me solve the equations Maxwell's theory led me to. I have no doubt that your solution is absolutely brilliant. However, I would like to bring the solution of the problem to a usable stage. Hence, I would like anybody that works seriously with induction motors to be able to calculate "one directional pull" for a 4 pole motor (p=2) with eccentricity 30%. In the meantime, I have to work hard to learn the Wolfram language to be acceptable. Being just barely tolerated is no fun.

POSTED BY: Jan Krepela

Kay Herbert

Kay Herbert, Chief scientist, Sloan Valve Co.

Posted 8 years ago

If I use symmetry take 2 x integral from 0 to Pi and define the constraints on the parameters, Mathematica gives me an analytical answer to the integral: In[34]:= Integrate[2 Cos[p b]/ (1 + e Cos[b]), {b, 0, Pi}, Assumptions -> e > 0 && p > 0 && p \[Element] Integers && e <= 1] Out[34]= -(( 2 \[Pi] HypergeometricPFQRegularized[{1/2, 1, 1}, {1 - p, 1 + p}, ( 2 e)/(-1 + e)])/(-1 + e))

If I use symmetry take 2 x integral from 0 to Pi and define the constraints on the parameters, Mathematica gives me an analytical answer to the integral:

In[34]:= Integrate[2 Cos[p b]/ (1 + e Cos[b]), {b, 0, Pi}, 
 Assumptions -> e > 0 && p > 0 && p \[Element] Integers && e <= 1]

Out[34]= -((
 2 \[Pi] HypergeometricPFQRegularized[{1/2, 1, 1}, {1 - p, 1 + p}, (
   2 e)/(-1 + e)])/(-1 + e))

POSTED BY: Kay Herbert

Daniel Lichtblau

Daniel Lichtblau, Wolfram Research

Posted 8 years ago

If you post actual Mathematica code people are more likely to try it out.

POSTED BY: Daniel Lichtblau

EDITORIAL BOARD

EDITORIAL BOARD, WOLFRAM

Posted 8 years ago

Welcome to Wolfram Community! Please make sure you know the rules: https://wolfr.am/READ-1ST Please post some code that you tried, this is an essential forum guideline.