Skip to main content

calculus and analysis - Inconsistent results for equivalent converging symbolic integrals


Bug introduced in 9.0 or earlier and persisting through 11.0.1 or later



I have looked at previous questions and I'm aware that this seems to be a known bug: Mathematica giving inconsistent results for symbolic integrals done in different ways. The origins for the bugs seem to be varied.


I'm having a hard time tracking down where Mathematica is going wrong with this particular one, especially since I am explicitly assuming that all my variables ($L$ and $\rho$) are positive. I've played around with the assumptions without any luck. Here's my code:


$Assumptions = {ρ > 0, L > 0};

limits = Sequence[{ux, 0, L}, {vx, 0, L}, {uy, 0, ux}, {vy, 0, vx}]


N1 = Integrate[a1 = Exp[-ρ (ux - uy) (vx - vy)], limits];
N2 = Integrate[a2 = Exp[-ρ (ux - uy) (vx - vy)] (ρ (ux - uy) (vx - vy)), limits];
N3 = Integrate[a3 = Exp[-ρ (ux - uy) (vx - vy)] (ρ (ux - uy) (vx - vy))^2, limits];
N4 = Integrate[a4 = Exp[-ρ (ux - uy) (vx - vy)] (ρ (ux - uy) (vx - vy))^3, limits];

int1 = Integrate[4 ρ (a1 - 7/2 a2 + 2 a3 - 1/4 a4), limits]

int2 =  Simplify[4 ρ (N1 - 7/2 N2 + 2 N3 - 1/4 N4)] // Expand


Apologies for the large expressions but the particular simplifcations only happen when all four integrands are included with the correct factors. By comparing integrands it's easy to see that int1 and int2 are the same integrals. But for int1 I get


(1 - E^(-L^2 ρ)) L^2

whereas for int2 I get


L^2 - E^(-L^2 ρ) L^2 - (7 L^4 ρ)/4 + 7/2 L^4 ρ Log[L]

I know that the outcome for int1 is correct whereas that for int2 is incorrect.


Any thoughts would be very welcome.



Answer



It's a bug, of course. Mathematica gets dizzy by the "hanging" integration order (I believe).



As simple as it is, it was tough to find out but it gives the right result if you just change the order of limits:


$Assumptions = {ρ > 0, L > 0};

limits = Sequence[{ux, 0, L}, {uy, 0, ux}, {vx, 0, L}, {vy, 0, vx}];
(* instead of 
limits = Sequence[{ux, 0, L}, {vx, 0, L}, {uy, 0, ux}, {vy, 0, vx}]
*)

N1 = Integrate[a1 = Exp[-ρ (ux - uy) (vx - vy)],                            limits];
N2 = Integrate[a2 = Exp[-ρ (ux - uy) (vx - vy)] (ρ (ux - uy) (vx - vy)),    limits];

N3 = Integrate[a3 = Exp[-ρ (ux - uy) (vx - vy)] (ρ (ux - uy) (vx - vy))^2,  limits];
N4 = Integrate[a4 = Exp[-ρ (ux - uy) (vx - vy)] (ρ (ux - uy) (vx - vy))^3,  limits];

int1 = Integrate[4 ρ (a1 - 7/2 a2 + 2 a3 - 1/4 a4), limits]

int2 = FullSimplify[4 ρ (N1 - 7/2 N2 + 2 N3 - 1/4 N4)]

(*
 (1 - E^(-L^2 ρ)) L^2
 (1 - E^(-L^2 ρ)) L^2

*)

Comments

Post a Comment

Popular posts from this blog

functions - Get leading series expansion term?

Given a function f[x] , I would like to have a function leadingSeries that returns just the leading term in the series around x=0 . For example: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x)] x and leadingSeries[(1/x + 2 + (1 - 1/x^3)/4)/(4 + x)] -(1/(16 x^3)) Is there such a function in Mathematica? Or maybe one can implement it efficiently? EDIT I finally went with the following implementation, based on Carl Woll 's answer: lds[ex_,x_]:=( (ex/.x->(x+O[x]^2))/.SeriesData[U_,Z_,L_List,Mi_,Ma_,De_]:>SeriesData[U,Z,{L[[1]]},Mi,Mi+1,De]//Quiet//Normal) The advantage is, that this one also properly works with functions whose leading term is a constant: lds[Exp[x],x] 1 Answer Update 1 Updated to eliminate SeriesData and to not return additional terms Perhaps you could use: leadingSeries[expr_, x_] := Normal[expr /. x->(x+O[x]^2) /. a_List :> Take[a, 1]] Then for your examples: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x), x] leadingSeries[Exp[x], x] leadingSeries[(1/x + 2 + (1 - 1/x...
Post a Comment
Read more

How to thread a list

I have data in format data = {{a1, a2}, {b1, b2}, {c1, c2}, {d1, d2}} Tableform: I want to thread it to : tdata = {{{a1, b1}, {a2, b2}}, {{a1, c1}, {a2, c2}}, {{a1, d1}, {a2, d2}}} Tableform: And I would like to do better then pseudofunction[n_] := Transpose[{data2[[1]], data2[[n]]}]; SetAttributes[pseudofunction, Listable]; Range[2, 4] // pseudofunction Here is my benchmark data, where data3 is normal sample of real data. data3 = Drop[ExcelWorkBook[[Column1 ;; Column4]], None, 1]; data2 = {a #, b #, c #, d #} & /@ Range[1, 10^5]; data = RandomReal[{0, 1}, {10^6, 4}]; Here is my benchmark code kptnw[list_] := Transpose[{Table[First@#, {Length@# - 1}], Rest@#}, {3, 1, 2}] &@list kptnw2[list_] := Transpose[{ConstantArray[First@#, Length@# - 1], Rest@#}, {3, 1, 2}] &@list OleksandrR[list_] := Flatten[Outer[List, List@First[list], Rest[list], 1], {{2}, {1, 4}}] paradox2[list_] := Partition[Riffle[list[[1]], #], 2] & /@ Drop[list, 1] RM[list_] := FoldList[Transpose[{First@li...
Post a Comment
Read more

front end - keyboard shortcut to invoke Insert new matrix

I frequently need to type in some matrices, and the menu command Insert > Table/Matrix > New... allows matrices with lines drawn between columns and rows, which is very helpful. I would like to make a keyboard shortcut for it, but cannot find the relevant frontend token command (4209405) for it. Since the FullForm[] and InputForm[] of matrices with lines drawn between rows and columns is the same as those without lines, it's hard to do this via 3rd party system-wide text expanders (e.g. autohotkey or atext on mac). How does one assign a keyboard shortcut for the menu item Insert > Table/Matrix > New... , preferably using only mathematica? Thanks! Answer In the MenuSetup.tr (for linux located in the $InstallationDirectory/SystemFiles/FrontEnd/TextResources/X/ directory), I changed the line MenuItem["&New...", "CreateGridBoxDialog"] to read MenuItem["&New...", "CreateGridBoxDialog", MenuKey["m", Modifiers-...
Post a Comment
Read more