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Problem with Integrate

Posted 10 years ago

Hello.

I was trying to do the following integral:

Integrate[1/Sqrt[2 (1 + x)^3 + 1], {x, 0, 1}] // N

The output in Mathematica 9 is a complex number that is not correct

-2.96267 - 1.92762 I

When I try to solve by using NIntegrate, the output is the correct one

NIntegrate[1/Sqrt[2 (1 + x)^3 + 1], {x, 0, 1}]

0.376065

Is this a bug in Integrate or I am missing something? Can someone test in Mathematica 10?

Thanks

POSTED BY: otsocol
24 Replies

Let me add two remarks

1) the exact value of the jump position seems to be xj = 2^(2/3) - 1

N[2^(2/3) - 1, 20]
(* 0.58740105196819947475 *)

2) an antiderivative without a jump can be calculated as follows

ad2F1[x_] = Integrate[1/Sqrt[1 + 2 x^3], {x, 0, t}, Assumptions -> t > 0] /. t -> 1 + x

(* Out[116]= (1 + x) Hypergeometric2F1[1/3, 1/2, 4/3, -2 (1 + x)^3] *)

Hence we find a hypergeometric function instead of the "bizarre" elliptic integral.

Proof

D[ad2F1[x], x] // Simplify

(* Out[109]= 1/Sqrt[1 + 2 (1 + x)^3] *)

Did somebody notice already that the simple substitution x+1->y, dx->dy, {x,0,1} -> {y,1,2} leads to the correct result?

In fact:

sol2 = Integrate[1/Sqrt[2 y^3 + 1], {y, 1, 2}]

(* Out[27]= Sqrt[2] Hypergeometric2F1[1/6, 1/2, 7/6, -(1/2)] -  Hypergeometric2F1[1/6, 1/2, 7/6, -(1/16)] *)

sol2 // N

(* Out[28]= 0.376065 *)

solN = NIntegrate[1/Sqrt[2 (1 + x)^3 + 1], {x, 0, 1}]

(* Out[39]= 0.376065 *)


$Version

(* Out[40]= "8.0 for Microsoft Windows (64-bit) (October 7, 2011)" *)

Best regards, Wolfgang

Yes, for the indefinite integral you're right; it is still in Version 10.0.0 the case. The difference of valued indefinite integrals does not convert into the Hypergeometric2F1[] form and the bug reappears

In[11]:= N[(ReplaceAll[#, y -> 2] - ReplaceAll[#, y -> 1]) &[Integrate[1/Sqrt[2 y^3 + 1], y]]]
Out[11]= -2.96267 - 1.92762 I

In[15]:= N[FullSimplify[(ReplaceAll[#, y -> 2] - ReplaceAll[#, y -> 1])] &[Integrate[1/Sqrt[2 y^3 + 1], y]]]
Out[15]= -2.96267 - 1.92762 I

In[17]:= (Integrate[1/Sqrt[2 y^3 + 1], y] /. y -> x + 1) === Integrate[1/Sqrt[2 (1 + x)^3 + 1], x]
Out[17]= True

that means this is a bug not about a path singularity, because the indefinite integral does not go along a path. It's something about turning back to the right leaf. For example, look at this, the derivative does not turn back to the original integrand, but presents an equation of six-th degree instead:

otsocol6Degree

POSTED BY: Udo Krause

One gets the intended result with a simple shift of the integration variable of the definite integral

In[87]:= NIntegrate[1/Sqrt[2 (1 + x)^3 + 1], {x, 0, 1}]
Out[87]= 0.376065

In[88]:= $Version
Out[88]= "10.0 for Microsoft Windows (64-bit) (June 29, 2014)"

Let y = x + 1, dy = dx and

In[90]:= NIntegrate[1/Sqrt[2 y^3 + 1], {y, 1, 2}]
Out[90]= 0.376065

In[91]:= Integrate[1/Sqrt[2 y^3 + 1], {y, 1, 2}]
Out[91]= Sqrt[2]*Hypergeometric2F1[1/6, 1/2, 7/6, -1/2] - Hypergeometric2F1[1/6, 1/2, 7/6, -1/16]

In[92]:= % // N
Out[92]= 0.376065

With other words, from 1 to 2 one does not encounter a singularity like the one Daniel mentioned, because

In[117]:= (Integrate[1/Sqrt[2 y^3 + 1], y] /. y -> x + 1) ===  Integrate[1/Sqrt[2 (1 + x)^3 + 1], x]
Out[117]= True

of course.

POSTED BY: Udo Krause
Posted 10 years ago

Dear Udo, thanks for your comments.

But if you try (I am using Mathematica 9)

In[1]:= sol = Integrate[1/Sqrt[2 y^3 + 1], y]

Out[1]= (1/(3^(1/4) Sqrt[1 + 2 y^3]))(-1)^(1/6) 2^( 2/3) Sqrt[(-1)^(5/6) (-1 + (-2)^(1/3) y)] Sqrt[ 1 + (-2)^(1/3) y + (-2)^(2/3) y^2] EllipticF[ArcSin[Sqrt[-(-1)^(5/6) - I (-2)^(1/3) y]/3^(1/4)], (-1)^( 1/3)]

In[2]:= (sol /. y -> 2) - (sol /. y -> 1) // N

Out[2]= -2.96267 - 1.92762 I

You still have a wrong answer. So, in my opnion, there is a bug in Integrate function.

POSTED BY: otsocol

Although possible (but you commented that you think not) it's unlikely that they are related--there are a variety of crashing bugs that one will come across in Mathematica (not frequently, but they happen) and I am guessing that they happen because of things deep in the works.

POSTED BY: David Reiss

Regarding the crash associated with plotting a long expression, I wonder if the bug is related to this post: http://community.wolfram.com/groups/-/m/t/314527?p_p_auth=3uJCatzr

Postscript: I tried the plot in http://community.wolfram.com/groups/-/m/t/314527?p_p_auth=3uJCatzr, but using a shorter expression for the first argument to Plot. The plot in that post still crashes (randomly, it seems). So, perhaps the bugs are not relatives.

POSTED BY: W. Craig Carter
Posted 10 years ago

No crash in Mathematica 9

POSTED BY: otsocol

I filed a bug report.

POSTED BY: David Reiss

Yes, that's what happens on my computer also. Weird.

POSTED BY: Frank Kampas

Plotting theDerivative using

Plot[theDerivative,{x,0,1}] 

does not crash my Kernel, but plotting the full expression for theDerivative as written in my post does crash it reproducibly.

POSTED BY: David Reiss

If I calculate

sln = DSolve[{y'[x] == 1/Sqrt[2 (1 + x)^3 + 1], y[0] == 0}, y, x]

yp[x_] = D[y[x] /. sln, x]

Plot[yp[x] - 1/Sqrt[2 (1 + x)^3 + 1], {x, 0, 1}]

works giving me a plot that stays with 1.5 x 10^-15

Also, plotting "theDerivative" does not crash my computer.

POSTED BY: Frank Kampas

A side observation...

If one computes

theIntegral = Integrate[1/Sqrt[2 (1 + x)^3 + 1], x]

and then computes

theDerivative = D[theIntegral, x] // FullSimplify

which yields

-(((-1)^(5/6) Sqrt[3] Sqrt[
     1 + 2 (1 + x)^3])/(Sqrt[-(-1)^(5/6) (1 + 2^(1/3) + 2^(1/3) x)]
      Sqrt[(-1)^(5/6) (-1 + (-2)^(1/3) (1 + x))] Sqrt[
     1 + (-2)^(1/3) (1 + x) + (-2)^(2/3) (1 + x)^2] Sqrt[
     3 - (-1)^(1/6) Sqrt[3] + (1 + x) Root[108 + #1^6 &, 1]] Sqrt[
     3 + (-1)^(5/6) Sqrt[3] + (1 + x) Root[108 + #1^6 &, 2]])) 

and then attempts to Plot this using

Plot[-(((-1)^(5/6) Sqrt[3] Sqrt[
      1 + 2 (1 + x)^3])/(Sqrt[-(-1)^(5/6) (1 + 2^(1/3) + 2^(1/3) x)]
       Sqrt[(-1)^(5/6) (-1 + (-2)^(1/3) (1 + x))] Sqrt[
      1 + (-2)^(1/3) (1 + x) + (-2)^(2/3) (1 + x)^2] Sqrt[
      3 - (-1)^(1/6) Sqrt[3] + (1 + x) Root[108 + #1^6 &, 1]] Sqrt[
      3 + (-1)^(5/6) Sqrt[3] + (1 + x) Root[108 + #1^6 &, 2]])) , {x, 0, 1}]

the result is that it crashes the Mathematica Kernel. (I will report this.)

In[1]:= SystemInformation["Small"]

Out[1]= {"Kernel" -> {"SystemID" -> "MacOSX-x86-64", 
   "ReleaseID" -> "10.0.0.0 (5098698, 5098537)", 
   "CreationDate" -> 
    DateObject[{2014, 6, 29}, TimeObject[{20, 38, 32}]]}, 
 "FrontEnd" -> {"OperatingSystem" -> "MacOSX", 
   "ReleaseID" -> "10.0.0.0 (5098698, 2014062704)", 
   "CreationDate" -> 
    DateObject[{2014, 6, 27}, TimeObject[{23, 17, 40.}]]}}
POSTED BY: David Reiss

Slipping in Evaluate prevents the crash. fyi.

Plot[ Evaluate[-(((-1)^(5/6) Sqrt[3] Sqrt[
        1 + 2 (1 + x)^3])/(Sqrt[-(-1)^(5/6) (1 + 2^(1/3) + 
           2^(1/3) x)] Sqrt[(-1)^(5/
            6) (-1 + (-2)^(1/3) (1 + x))] Sqrt[
        1 + (-2)^(1/3) (1 + x) + (-2)^(2/3) (1 + x)^2] Sqrt[
        3 - (-1)^(1/6) Sqrt[3] + (1 + x) Root[108 + #1^6 &, 1]] Sqrt[
        3 + (-1)^(5/6) Sqrt[3] + (1 + x) Root[108 + #1^6 &, 
           2]]))], {x, 0, 1}]

Ditto pre-processing the expression with FullSimplify.

expr2 = FullSimplify[  expr , x >= 0] 

I don't want to discourage your reporting the crash.

POSTED BY: Bruce Miller

Interesting. I've reported the crash and it's in the bug system. And I just added your example to the report.

POSTED BY: David Reiss

Much simpler example that causes the crash:

Plot[Root[108 + #1^6 &, 2], {x, 0, 1}]
POSTED BY: David Reiss

@otsocol, Yes, I was looking at that antiderivative.

As to whether the antiderivative itself is correct, About all I can say is it gives a numerical result consistent with NIntegrate if one only integrates from 0 to 1/2 (which excludes that singularity point). This might or might not answer your question.

POSTED BY: Daniel Lichtblau
Posted 10 years ago

Hi Daniel. Thanks for your answer.

When you say that the antiderivative has a jump discontinuity near x=0.59, I think you are looking at the Mathematica answer

In[6]:= Integrate[1/Sqrt[2 (1 + x)^3 + 1], x]

Out[6]= ((-1)^(1/6) 2^(2/3) Sqrt[(-1)^(5/6) (-1 + (-2)^(1/3) (1 + x))] Sqrt[1 + (-2)^(1/3) (1 + x) + (-2)^(2/3) (1 + x)^2] EllipticF[ ArcSin[Sqrt[-(-1)^(5/6) - I (-2)^(1/3) (1 + x)]/3^(1/4)], (-1)^( 1/3)])/(3^(1/4) Sqrt[1 + 2 (1 + x)^3])

Is that correct? But is it possible to trust in this result in this case?

POSTED BY: otsocol

I see what you mean

sln = DSolve[{y'[x] == 1/Sqrt[2 (1 + x)^3 + 1], y[0] == 0}, y]

Plot [Im[y[x] /. sln], {x, 0, 1}, PlotRange -> All ] shows that y[x] develops an imaginary part at 0.59

POSTED BY: Frank Kampas
sln = DSolve[{y'[x] == 1/Sqrt[2 (1 + x)^3 + 1], y[0] == 0}, y]

sln represents a syntax error as such. If one is concerned about newbies and rookies not sticking to the syntax, one should stick to it at least oneself.

POSTED BY: Udo Krause

Frank: In brief: Mathematica is using the Newton-Leibniz method to do the definite integral. It computes the antiderivative. It looks for points on the path of integration where the antiderivative might have discontinuities. It evaluates by taking limits at endpoints and all these points.

The antiderivative has a jump discontinuity near x=0.59, which of course is between 0 and 1. The code that looks for discontinuities is failing to find it. Not entirely a surprise (finding them involves some hard-core equation solving, often transcendental). But it's worth further investigation.

POSTED BY: Daniel Lichtblau

Daniel, I don't understand your post.

POSTED BY: Frank Kampas

There is a path singularity around .59 that is not found. Will investigate.

POSTED BY: Daniel Lichtblau

Using Mathematica 10

In[2]:= Integrate[1/Sqrt[2 (1 + x)^3 + 1], {x, 0, 1}]

Out[2]= (1/(3^(1/4)))(-1)^(1/12) 2^( 2/3) (EllipticF[ ArcSin[((-1)^(11/12) Sqrt[1 + 2^(1/3)])/3^(1/4)], (-1)^(1/3)] - EllipticF[ ArcSin[((-1)^(11/12) Sqrt[1 + 2 2^(1/3)])/3^(1/4)], (-1)^(1/3)])

In[3]:= N[%]

Out[3]= -2.96267 - 1.92762 I

Seeing -1 raised to fractional powers makes a branch cut problem a possibility.

POSTED BY: Frank Kampas
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