0
|
2710 Views
|
5 Replies
|
1 Total Likes
View groups...
Share
GROUPS:

# indefinite integral question with denominator

Posted 9 years ago
 What is this denominator appearing in all terms of indefinite integral? w120 + y1 (w121 + w122 y1) w1 = w100 + w110 x + w101 y1 + w120 x^2 + w102 y1^2 + w111 x y1 + w112 x y1^2 + w121 x^2 y1 + w122 x^2 y1^2; Expand[[Integral]0.5*(D[w1, x])^2 [DifferentialD]x] (0.0833333 w110^3)/(w120 + y1 (w121 + w122 y1)) + ( 0.5 w110^2 w120 x)/(w120 + y1 (w121 + w122 y1)) + ( 1. w110 w120^2 x^2)/(w120 + y1 (w121 + w122 y1)) + ( 0.666667 w120^3 x^3)/(w120 + y1 (w121 + w122 y1)) + ( 0.25 w110^2 w111 y1)/(w120 + y1 (w121 + w122 y1)) + ( 1. w110 w111 w120 x y1)/(w120 + y1 (w121 + w122 y1)) + ( 0.5 w110^2 w121 x y1)/(w120 + y1 (w121 + w122 y1)) + ( 1. w111 w120^2 x^2 y1)/(w120 + y1 (w121 + w122 y1)) + ( 2. w110 w120 w121 x^2 y1)/(w120 + y1 (w121 + w122 y1)) + ( 2. w120^2 w121 x^3 y1)/(w120 + y1 (w121 + w122 y1)) + ( 0.25 w110 w111^2 y1^2)/(w120 + y1 (w121 + w122 y1)) + ( 0.25 w110^2 w112 y1^2)/(w120 + y1 (w121 + w122 y1)) + ( 0.5 w111^2 w120 x y1^2)/(w120 + y1 (w121 + w122 y1)) + ( 1. w110 w112 w120 x y1^2)/(w120 + y1 (w121 + w122 y1)) + ( 1. w110 w111 w121 x y1^2)/(w120 + y1 (w121 + w122 y1)) + ( 0.5 w110^2 w122 x y1^2)/(w120 + y1 (w121 + w122 y1)) + ( 1. w112 w120^2 x^2 y1^2)/(w120 + y1 (w121 + w122 y1)) + ( 2. w111 w120 w121 x^2 y1^2)/(w120 + y1 (w121 + w122 y1)) + ( 1. w110 w121^2 x^2 y1^2)/(w120 + y1 (w121 + w122 y1)) + ( 2. w110 w120 w122 x^2 y1^2)/(w120 + y1 (w121 + w122 y1)) + ( 2. w120 w121^2 x^3 y1^2)/(w120 + y1 (w121 + w122 y1)) + ( 2. w120^2 w122 x^3 y1^2)/(w120 + y1 (w121 + w122 y1)) + ( 0.0833333 w111^3 y1^3)/(w120 + y1 (w121 + w122 y1)) + ( 0.5 w110 w111 w112 y1^3)/(w120 + y1 (w121 + w122 y1)) + ( 1. w111 w112 w120 x y1^3)/(w120 + y1 (w121 + w122 y1)) + ( 0.5 w111^2 w121 x y1^3)/(w120 + y1 (w121 + w122 y1)) + ( 1. w110 w112 w121 x y1^3)/(w120 + y1 (w121 + w122 y1)) + ( 1. w110 w111 w122 x y1^3)/(w120 + y1 (w121 + w122 y1)) + ( 2. w112 w120 w121 x^2 y1^3)/(w120 + y1 (w121 + w122 y1)) + ( 1. w111 w121^2 x^2 y1^3)/(w120 + y1 (w121 + w122 y1)) + ( 2. w111 w120 w122 x^2 y1^3)/(w120 + y1 (w121 + w122 y1)) + ( 2. w110 w121 w122 x^2 y1^3)/(w120 + y1 (w121 + w122 y1)) + ( 0.666667 w121^3 x^3 y1^3)/(w120 + y1 (w121 + w122 y1)) + ( 4. w120 w121 w122 x^3 y1^3)/(w120 + y1 (w121 + w122 y1)) + ( 0.25 w111^2 w112 y1^4)/(w120 + y1 (w121 + w122 y1)) + ( 0.25 w110 w112^2 y1^4)/(w120 + y1 (w121 + w122 y1)) + ( 0.5 w112^2 w120 x y1^4)/(w120 + y1 (w121 + w122 y1)) + ( 1. w111 w112 w121 x y1^4)/(w120 + y1 (w121 + w122 y1)) + ( 0.5 w111^2 w122 x y1^4)/(w120 + y1 (w121 + w122 y1)) + ( 1. w110 w112 w122 x y1^4)/(w120 + y1 (w121 + w122 y1)) + ( 1. w112 w121^2 x^2 y1^4)/(w120 + y1 (w121 + w122 y1)) + ( 2. w112 w120 w122 x^2 y1^4)/(w120 + y1 (w121 + w122 y1)) + ( 2. w111 w121 w122 x^2 y1^4)/(w120 + y1 (w121 + w122 y1)) + ( 1. w110 w122^2 x^2 y1^4)/(w120 + y1 (w121 + w122 y1)) + ( 2. w121^2 w122 x^3 y1^4)/(w120 + y1 (w121 + w122 y1)) + ( 2. w120 w122^2 x^3 y1^4)/(w120 + y1 (w121 + w122 y1)) + ( 0.25 w111 w112^2 y1^5)/(w120 + y1 (w121 + w122 y1)) + ( 0.5 w112^2 w121 x y1^5)/(w120 + y1 (w121 + w122 y1)) + ( 1. w111 w112 w122 x y1^5)/(w120 + y1 (w121 + w122 y1)) + ( 2. w112 w121 w122 x^2 y1^5)/(w120 + y1 (w121 + w122 y1)) + ( 1. w111 w122^2 x^2 y1^5)/(w120 + y1 (w121 + w122 y1)) + ( 2. w121 w122^2 x^3 y1^5)/(w120 + y1 (w121 + w122 y1)) + ( 0.0833333 w112^3 y1^6)/(w120 + y1 (w121 + w122 y1)) + ( 0.5 w112^2 w122 x y1^6)/(w120 + y1 (w121 + w122 y1)) + ( 1. w112 w122^2 x^2 y1^6)/(w120 + y1 (w121 + w122 y1)) + ( 0.666667 w122^3 x^3 y1^6)/(w120 + y1 (w121 + w122 y1))
5 Replies
Sort By:
Posted 9 years ago
 Here is the integration: \[Integral]((w110 + 2 w120 x + y1 (w111 + 2 w121 x + w112 y1 + 2 w122 x y1))^3/(12 (w120 + y1 (w121 + w122 y1)))) \[DifferentialD]y1 which gives: (1/(720 w122^6))(60 w122 (w112^3 (w121^4 - 3 w120 w121^2 w122 + w120^2 w122^2) + 3 w112 w122^2 (-2 w110 w111 w121 w122 + w110^2 w122^2 + w111^2 (w121^2 - w120 w122)) - 3 w112^2 w122 (w110 w122 (-w121^2 + w120 w122) + w111 (w121^3 - 2 w120 w121 w122)) + w122^3 (-w111^3 w121 + 3 w110 w111^2 w122 + 2 w122^2 x (3 w110^2 + 6 w110 w120 x + 4 w120^2 x^2))) y1 + 30 w122^2 (-3 w111 w112 w122^2 (w111 w121 - 2 w110 w122) - 3 w112^2 w122 (-w111 w121^2 + w111 w120 w122 + w110 w121 w122) - w112^3 (w121^3 - 2 w120 w121 w122) + w122^3 (w111^3 + 12 w111 w122 x (w110 + w120 x) + 4 w121 w122 x^2 (3 w110 + 4 w120 x))) y1^2 + 20 w122^3 (w112 + 2 w122 x) (w112^2 (w121^2 - w120 w122) + w112 w122 (-3 w111 w121 + 3 w110 w122 - 2 w121^2 x + 2 w120 w122 x) + w122^2 (3 w111^2 + 6 w111 w121 x + 2 x (3 w110 w122 + 2 w121^2 x + 4 w120 w122 x))) y1^3 + 15 w122^4 (w112 + 2 w122 x)^2 (-w112 w121 + 3 w111 w122 + 4 w121 w122 x) y1^4 + 12 w122^5 (w112 + 2 w122 x)^3 y1^5 + (1/ Sqrt[-w121^2 + 4 w120 w122]) 60 (w112^3 (w121^6 - 6 w120 w121^4 w122 + 9 w120^2 w121^2 w122^2 - 2 w120^3 w122^3) - w122^3 (w111 w121 - 2 w110 w122) (-w110 w111 w121 w122 + w110^2 w122^2 + w111^2 (w121^2 - 3 w120 w122)) + 3 w112 w122^2 (-2 w110 w111 w121 w122 (w121^2 - 3 w120 w122) + w110^2 w122^2 (w121^2 - 2 w120 w122) + w111^2 (w121^4 - 4 w120 w121^2 w122 + 2 w120^2 w122^2)) - 3 w112^2 w122 (-w110 w122 (w121^4 - 4 w120 w121^2 w122 + 2 w120^2 w122^2) + w111 (w121^5 - 5 w120 w121^3 w122 + 5 w120^2 w121 w122^2))) ArcTan[(w121 + 2 w122 y1)/ Sqrt[-w121^2 + 4 w120 w122]] - 30 (w112 w121 - w111 w122) (w112^2 (w121^4 - 4 w120 w121^2 w122 + 3 w120^2 w122^2) + w122^2 (-3 w110 w111 w121 w122 + 3 w110^2 w122^2 + w111^2 (w121^2 - w120 w122)) + w112 w122 (3 w110 w122 (w121^2 - 2 w120 w122) + w111 (-2 w121^3 + 5 w120 w121 w122))) Log[ w120 + y1 (w121 + w122 y1)]) 
Posted 9 years ago
 What do you exactly mean by 'change of variables'? Integrating the last expression 'out' with respect to y1 produces a very complicated expression with an 'ArcTan' function. The result is correct (I checked it with taking derivatives) but it is not a polynomial. How can I get an exact polynomial as a result?
Posted 9 years ago
 Well, there are many ways to write a given expression. The integral weather done by hand or with Mathematica is the one that I indicated above. As I mentioned there, the denominator arise as naturally as a consequence of the change of variable needed in order to do the integration naturally.
Posted 9 years ago
 It comes from the change of variables when doing the integration with respect to x. This, of course, could be easily done by hand. Note: Collect[D[w1, x], x] gives w110 + w111 y1 + w112 y1^2 + x (2 w120 + 2 w121 y1 + 2 w122 y1^2) I am not sure why you have an Expand in your computation --and also why you have a 0.5 rather than a 1/2. But here is the integral more exposed In:= Integrate[1/2 (D[w1, x])^2, x] Out= (w110 + 2 w120 x + y1 (w111 + 2 w121 x + w112 y1 + 2 w122 x y1))^3/(12 (w120 + y1 (w121 + w122 y1))) 
Posted 9 years ago
 How can I avoid this denominator? If I integrate a polynomial by hand for example no such a thing is produced.