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Integrate[(1 + 16 Tan[2 x - y]^2)/(1 + 4 Tan[2 x - y]^2), {x, 0, 2 π}]

Mathematica (wrong) output is (tested under versions 8 and 10.0, took ~ 1 minute of CPU time)

10 π

The correct value of this integral is 6 π (not 10 π).

Integrate[(1 + 16 Tan[2 x - 2]^2)/(1 + 4 Tan[2 x - 2]^2), {x, 0, 2 π}]
(* 6 π *)
Integrate[(1 + 16 Tan[2 x - 4]^2)/(1 + 4 Tan[2 x - 4]^2), {x, 0, 2 π}]
(* 6 π *)

Version 8 and 10.0.0.0 Version 8 and 10 Version 7.0.1.0 Version 7

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    $\begingroup$ @Vaclav I noticed that most of your posts here are bug reports. Please do not do this. This site is for questions only. If you wish to report bugs, contact Wolfram directly. Wolfram will not see bugs reported here. If you only report here, they won't get fixed. This site is not affiliated with Wolfram. $\endgroup$ – Szabolcs May 22 '15 at 12:39
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    $\begingroup$ But I already reported one bug in 2012, but still not was fixed. Today I found on this website a lot of posts, where users reported a bugs in Mathematica. See for example mathematica.stackexchange.com/questions/19322 with an answer "This has been confirmed and reported as a bug". This was motivation for me add a several bugs, which I found. $\endgroup$ – Vaclav Kotesovec May 22 '15 at 12:53
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    $\begingroup$ @Szabolcs: It would make sense if m.se had a separate section for bug reports. WRI definitely WILL notice them and deal with them, simply because they are on public display and they'll want that list to shrink. With their internal bug reporting mechanism, they pick and choose what they want to fix and what they want to leave as is. I've reported two bugs in 2004, that was 5.1 iirc, and they're still not fixed. Sometimes public display is the only way to get improvements. $\endgroup$ – Andreas Lauschke May 22 '15 at 13:37
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    $\begingroup$ It is the common type of error in Integrate[] which has been discussed many times in this forum: the indefinite integral (the antiderivative) is not continuous. Hence the definite integral is not the difference of the antiderivative at both ends of the interval. MMA is not able to detect the discontinuity correctly so the remedy is: the user should check continuity and take apropriate measures to adjust the result. Try the simpler example Integrate[2/(2 + Cos[x - y]), {x, 0, 2*Pi}] which gives 0 (Version 8). $\endgroup$ – Dr. Wolfgang Hintze May 22 '15 at 15:41
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    $\begingroup$ Reporting a miscalculation of Integrate on this site is a bit like adding a grain sand to beach. Still, as the OP notes, such "questions" have been welcomed here generally, although occasionally there really is a question about the result. $\endgroup$ – Michael E2 May 22 '15 at 15:56
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I've got a theory....

Let's try to get the antiderivative:

Integrate[(1 + 16*Tan[2*x - y]^2)/(1 + 4*Tan[2*x - y]^2), x,
 Assumptions -> y \[Element] Reals]

(*returns 5 x - (5 y)/2 - ArcTan[2 Tan[2 x - y]] *)

Seems legit. You can test with D[5 x - (5 y)/2 - ArcTan[2 Tan[2 x - y]], x] and plot or rearrange. This antiderivative is technically correct, but there is a latent +k Pi missing where k is an integer.

Using the antiderivative and the fundamental theorem of calculus you can evaluate at 2 Pi and 0 and subtract the first from last giving 10 Pi. Yet if you use NIntegrate (for some random numerical y), it converges to 6 Pi as expected... so a mystery 4 Pi has been added that vanished when the singularities in both the numerator and denominator of the integrand became removable as a fraction.

I think what's happening is Mathematica is not accommodating ArcTan and Tan's branch cuts at these singularities which will result in an extra +k Pi being added on, in this case it's an extra 4 Pi on top of the true answer. Those branch cut singularities in the top and bottom of the fraction are supposed to disappear (become removable) if we're just dealing with real numbers - yet Mathematica fails to realize this. Why 4 Pi? Well probably because if you plot the denominator or numerator you'll always see exactly 4 singularities for any choice of y. For instance try setting y to 5 and doing:

Show[Plot[(1 + 16*Tan[2*x - y]^2) /. y -> 5, {x, 0, 2 Pi}], 
     Plot[(1 + 4*Tan[2*x - y]^2) /. y -> 5, {x, 0, 2 Pi}]]

If we now force y and x to only use the real part we get the answer without the mystery +4 Pi

Integrate[(1 + 16*Tan[2*Re[x] - Re[y]]^2)/(1 + 4*Tan[2*Re[x] -
 Re[y]]^2), {x, 0, 2 Pi}] 

(*Answer 6 Pi *)

It's just a theory of mine; I've only recently been studying complex analysis so I may not be right... I definitely think it's a bug now.

EXTRA: to show that my D[...antiderivative...,x] and the integrand are equivalent you can do:

TrigReduce[5 - (4 Sec[2 x - y]^2)/(1 + 4 Tan[2 x - y]^2)] == 
TrigReduce@FullSimplify[(1 + 16 Tan[2 x - y]^2)/(1 + 4 Tan[2 x - y]^2)]
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[Edit notice: I'll put the gist up front.]

10 π is not wrong

With proper assumptions given, the integral evaluates as desired by the OP, to 6 π. Without them, it gives one of the correct values of the integral, 10 π, the one that in some sense is more likely, but without the correct conditions attached. (One may well argue that is a bug. However, Mathematica's exact solvers usually give only generically correct solutions. And beyond that, Integrate is a little looser than the others, and that is, I suspect, so that it can reach answers within a reasonable amount of time, when they can be reached at all. No doubt everyone wishes for more accurate and reliable algorithms.)

By default Mathematica's functions treat variables as complex numbers. These include functions called by Integrate. Clearly in the OP's call

Integrate[(1 + 16 Tan[2 x - y]^2)/(1 + 4 Tan[2 x - y]^2), {x, 0, 2 π}]

the variable y should be treated as complex. What's less clear is whether in the line integral from x == 0 to x = 2 Pi, the variable x should be considered real. In fact, it need not be. Sometimes it can help to explicitly specify that x is real or even 0 < x < 2 Pi.

Integrate[(1 + 16 Tan[2 x - y]^2)/(1 + 4 Tan[2 x - y]^2),
 {x, 0, 2 π}, Assumptions -> -(π/2) <= y <= π/2 && 0 < x < 2 Pi]
(*  6 π  *)

In fact this result is valid only when Abs[Im[y]] < Log[3]/2. When Abs[Im[y]] > Log[3]/2, the integral is 10 π. (So on the basis of the length of the intervals, for most values of y the integral is 10 π.)

Interestingly this is one of those cases where Assuming and Assumptions give different results:

Assuming[-(π/2) <= y <= π/2 && 0 < x < 2 Pi,
 Integrate[(1 + 16 Tan[2 x - y]^2)/(1 + 4 Tan[2 x - y]^2), {x, 0, 2 π}]]
(*  10 π  *)

One can see the difference in the graph of the integrand. Just by estimating the the average height of the graph, one can see the results 6 π, 10 π are probably correct in their respective cases.

Manipulate[
 Column[{
   Plot[Evaluate@
     Through[{Re, Im}[(1 + 16*Tan[2*x - y]^2)/(1 + 4*Tan[2*x - y]^2) /. y -> a + b I]],
    {x, 0, 2 Pi}, AxesOrigin -> {0, 0}, 
    PlotRange -> {-5, 10}, Filling -> Axis, ImageSize -> 225],
   Chop[N@
     NIntegrate[(1 + 16 Tan[2 x - y]^2)/(1 + 4 Tan[2 x - y]^2) /. y -> SetPrecision[a + b I, Infinity],
      {x, 0, 2 π}, 
      PrecisionGoal -> 10, WorkingPrecision -> 50, 
      MaxRecursion -> 16], 1*^-15]
   }],
 {a, 0, 3, Appearance -> "Labeled", ImageSize -> 200},
 {b, -2, 2, Appearance -> "Labeled", ImageSize -> 200}]

Mathematica graphics Mathematica graphics

Mathematica graphics Mathematica graphics

Checking with NIntegrate

I get 10 Pi with y = I:

Block[{y = I},
 N@NIntegrate[(1 + 16 Tan[2 x - y]^2)/(1 + 4 Tan[2 x - y]^2),
    {x, 0, 2 π}, WorkingPrecision -> 25]]
% == 10 Pi
(*
  31.4159 + 0. I
  True
*)

Other times, I get 8 Pi:

Block[{y = 1 + I},
 N@NIntegrate[(1 + 16 Tan[2 x - y]^2)/(1 + 4 Tan[2 x - y]^2),
    {x, 0, I, 2 π}, WorkingPrecision -> 25]]
% == 8 Pi
(*
  25.1327 + 0. I
  True
*)

And sometimes I can even get 9 Pi:

Block[{y = 10},
 N@NIntegrate[(1 + 16 Tan[2 x - y]^2)/(1 + 4 Tan[2 x - y]^2),
    {x, 0, I, 2 π}, WorkingPrecision -> 25]]
% == 9 Pi
(*
  28.2743 + 0. I
  True
*)

However, this is interesting:

Integrate[(1 + 16 Tan[2 x - y]^2)/(1 + 4 Tan[2 x - y]^2),
  {x, 0, y, 2 π}, Assumptions -> y ∈ Reals]
(*  ConditionalExpression[(23 π)/3, y == π/2]  *)

I think this is certainly wrong. NIntegrate gives 6 Pi for y = π/2.

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Oddly, if you follow the hints given in the ConditionalExpressions you can get pointed to the right answer although constrained to an overly restrictive region.

$Version

"10.1.0 for Mac OS X x86 (64-bit) (March 24, 2015)"

expr = (1 + 16 Tan[2 x - y]^2)/(1 + 4 Tan[2 x - y]^2);

Integrate[expr, {x, 0, 2 Pi}]

ConditionalExpression[9*Pi, -(Pi/2) <= y <= (7*Pi)/2]

The wrong answer; however, using the stated conditions

Assuming[{-Pi/2 <= y <= 7 Pi/2},
 Integrate[expr, {x, 0, 2 Pi}]]

ConditionalExpression[6*Pi, 2*y < Pi]

Applying the refined condition

Assuming[{-Pi/2 <= y <= Pi/2},
 Integrate[expr, {x, 0, 2 \[Pi]}]]

6*Pi

Integrate[expr /. y -> 100 Pi, {x, 0, 2 Pi}]

6*Pi

NIntegrate[expr /. y -> 100 Pi, {x, 0, 2 Pi}]

18.8496

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Integrate and NIntegrate agree on this matter:

Table[Integrate[(1+16 Tan[2 x-y]^2)/(1+4 Tan[2 x-y]^2),{x,0,2Pi}],{y,0,10,2}]
(*==> {6π,6π,6π,6π,6π,6π}*)

Table[NIntegrate[(1+16 Tan[2 x-y]^2)/(1+4 Tan[2 x-y]^2),{x,0,2Pi}],{y,0,10,2}]
(*==> {18.8496,18.8496,18.8496,18.8496,18.8496,18.8496}*)

N[6Pi]
(*==> 18.8496*)
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