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fixed in 10.1 (windows)

I'm running Mathematica 10.0.0 and encountered a disturbing error in the symbolic integration of a rather simple function

Integrate[(1 - x)*(1 + 2*x)^6/Sqrt[1 - x^2], {x, -1, 1}]/Pi

The correct value for this integral is 15 (and NIntegrate gives that correctly) but Mathematica evaluates it symbolically as 1/π+29/2. I tried Wolfram Alpha, and it also gives the wrong answer. Any idea what is going on?

incorrect Wolfram Alpha symbolic evaluation of this integral

the correct answer is 15π=47.1239

splitting the integrand into two parts does give the correct answer 15,

Integrate[(1 + 2*x)^6/Sqrt[1 - x^2], {x, -1, 1}]/Pi - 
 Integrate[x*(1 + 2*x)^6/Sqrt[1 - x^2], {x, -1, 1}]/Pi

somehow Mathematica has difficulty with square root singularities in the integrand?

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    $\begingroup$ Are you that Carlo Beenakker ? Anyway this seems better int = Integrate[(1 - x)*(1 + 2*x)^6/Sqrt[1 - x^2], x] and then Limit[int, x -> 1] - Limit[int, x -> -1]. $\endgroup$ – b.gates.you.know.what Sep 5 '14 at 12:11
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    $\begingroup$ thanks, actually splitting the integrand in two parts also gives the correct answer, but I do find it worrisome that I need to take these precautions for what is a simple square-root singularity. $\endgroup$ – Carlo Beenakker Sep 5 '14 at 12:18
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    $\begingroup$ This works, too: Integrate[(1 - x)*(1 + 2*x)^6/Sqrt[1 - x^2], {x, -1, 1}, GenerateConditions -> False] $\endgroup$ – Michael E2 Sep 5 '14 at 12:44
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    $\begingroup$ Will investigate.. $\endgroup$ – Daniel Lichtblau Sep 5 '14 at 15:11
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    $\begingroup$ @xslittlegrass I am not able to figure out what might be the issue you are reporting. If it was this NIntegrate[FEx[w]*Conjugate[FEx[w]], {w, -Infinity, Infinity}] (or same, but with Integrate) then I do not see an incorrect result. In any case, sending a report via comment, without proper formatting of the input and without full explanation of the issue, is not all that sensible in terms of getting it addressed. $\endgroup$ – Daniel Lichtblau Sep 5 '14 at 17:51
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Having experienced similar problematic issues with Mathematica I instantly thought that expanding the fraction in the integrand i.e. applying Appart could resolve the problem, and indeed it does:

Integrate[ Apart[(1 - x)(1 + 2x)^6/Sqrt[1 - x^2]], {x, -1, 1}]/Pi

15

These arguments apply to this case as well Bug in mathematica analytic integration? i.e.

... underlying complex functions behind the symbolic result are not defined in the whole complex plane...

Whenever an indetermined integral involves ArcSin, ArcCos, ArcTan etc. (here we have ArcSin[x]) one should carefully compare numerical and indetermined integrals remembering that Mathematica functions may have some arbitrary branch cuts in the complex plane. For a related issue see also How to calculate contour integrals with Mathematica? where various user defined branch cuts of analytic functions are compared.

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  • $\begingroup$ oddly it gets it correct if we split up the integral as Integrate[ .. ,{x,-1,1/2}]+Integrate[..,{x,1/2,1}] (depending on where you make the cut you get either the right or the wrong result.. ) $\endgroup$ – george2079 Sep 5 '14 at 18:35
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Main

The bug cannot only be attributed to the Sqrt in the integrand. It is trickier.

In fact, define for t=0,1,2,...

f[t_] := Integrate[(1 - x)*(1 + 2*x)^t/Sqrt[1 - x^2], {x, -1, 1}]/Pi

Then

{#, f[#]} & /@ {0, 1, 2, 3, 4, 5, 6, 7, 8}

(* Out[33]= {{0, 1}, {1, 0}, {2, 1}, {3, 1}, {4, 3}, {5, 6}, {6, (
  1 + (29 \[Pi])/2)/\[Pi]}, {7, (1 + (71 \[Pi])/2)/\[Pi]}, {8, (
  1 + (181 \[Pi])/2)/\[Pi]}} *)

Which is correct for t=0..5 but becomes wrong for t=6, 7, ... but in all cases the branch cut problem of the Sqrt in the integrand should appear. So what is so special about t=6, 7, ...?

I don't have an answer but have continued the study: writing in the integrand (for |x|<1)

Simplify[(1 - x)/Sqrt[1 - x^2 ] == Sqrt[(1 - x)/(1 + x)], -1 <= x <= 1]

(* Out[40]= True *)

the integral becomes

f1[t_] := Integrate[Sqrt[(1 - x)/(1 + x)] (1 + 2*x)^t, {x, -1, 1}]/Pi

At integer values we have

{#, f1[#]} & /@ {0, 1, 2, 3, 4, 5, 6, 7, 8}

(* Out[37]= {{0, 1}, {1, 0}, {2, 1}, {3, 1}, {4, 3}, {5, 6}, {6, 15}, {7, 36}, {8, 91}} *)

and no problem is encountered.

The integral can even be solved for real t (do the indefinite integral, then insert the limits x=1, x=-1), with the analytical result

f2[t_] := -3^t (-2 Hypergeometric2F1[1/2, -t, 1, 4/3] + Hypergeometric2F1[1/2, -t, 2, 4/3])

Except for non negative integers t this function is complex as can be seen by plotting it:

Plot[{Re[#], Im[#]} &[f2[t]], {t, -7, 7}]

(* graph now shown here *)

Best regards, Wolfgang

EDIT 07.09.14 00:55

Even stranger:

Integrate[(1 + 2 x)^6/Sqrt[1 - x^2] (1 - x), {x, -1, 1}]/Pi

(* Out[88]= (1 + (29 \[Pi])/2)/\[Pi] *)

Decomposing trivially (1-x) = 1 - x gives two integer parts

Integrate[(1 + 2 x)^6/Sqrt[1 - x^2] (1), {x, -1, 1}]/Pi

(* Out[91]= 141 *)

Integrate[(1 + 2 x)^6/Sqrt[1 - x^2] (-x), {x, -1, 1}]/Pi

(* Out[92]= -126 *)

and added = 15, as it should be

% + %%

(* Out[93]= 15 *)

Regards, Wolfgang

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fixed in 10.1 (windows):

Mathematica graphics

Mathematica graphics

code:

Clear[x]
Integrate[(1 - x)*(1 + 2*x)^6/Sqrt[1 - x^2], {x, -1, 1}]/Pi
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