7
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Bug introduced in 9.0 and fixed in 10.1


I was trying to get the following integral

Integrate[Sqrt[y/x] (Sin[t]^2 Cos[t])/(x+y+2 Sqrt[x y] Cos[t]), {t,0,Pi}, 
Assumptions -> {x > 0, y > 0, x > y}]

Mathematica answers

RFalse[x_,y_]=Pi(1/(8y)-(3y)/(8x^2))

This is wrong as the result is

RTrue[x_,y_]=-(Pi y)/(4x^2)

However, if one gives definite value for the parameters Mathematica calculates the integral correctly. Running

x1 = RandomReal[{10, 20}];
y1 = RandomReal[10];
Chop[Integrate[Sqrt[y1/x1] (Sin[t]^2 Cos[t])/(x1+y1+2 Sqrt[x1 y1] Cos[t]), 
{t,0,Pi}]-RTrue[x1,y1]]==0

returns True, while

Chop[Integrate[Sqrt[y1/x1] (Sin[t]^2 Cos[t])/(x1+y1+2 Sqrt[x1 y1] Cos[t]), 
{t,0,Pi}]-RFalse[x1,y1]]==0

returns False.

Many integrals of this type do not give the correct answer.

Any idea?

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8
  • $\begingroup$ What do you mean by Mathematica answers RFalse[x_,y_]=Pi(1/(8y)-(3y)/(8x^2))? That's quite a strange answer! $\endgroup$ Apr 24, 2015 at 13:47
  • $\begingroup$ Also, just in case, your Integrate[ ] has a syntax error. I'm downvoting until you correct it $\endgroup$ Apr 24, 2015 at 13:51
  • $\begingroup$ After correcting your integration expression to Integrate[Sqrt[y/x] (Sin[t]^2 Cos[t])/(x + y + 2 Sqrt[x y] Cos[t]), {t, 0, Pi}, Assumptions ->{x > 0, y > 0, x > y}] with version 10.1 I get the expected -((Pi*y)/(4*x^2)) $\endgroup$
    – Bob Hanlon
    Apr 24, 2015 at 13:52
  • $\begingroup$ @BobHanlon I posted that as an answer a few seconds after your comment. Deleting $\endgroup$ Apr 24, 2015 at 13:54
  • 3
    $\begingroup$ I don't think this question should be closed. The OP has found a real bug in V10.0.x, $\endgroup$
    – m_goldberg
    Apr 24, 2015 at 16:23

2 Answers 2

5
$\begingroup$
$Version

"10.0 for Mac OS X x86 (64-bit) (September 10, 2014)"

As entered Mathematica returns the wrong result.

Integrate[Sqrt[y/x] (Sin[t]^2 Cos[t])/(x + y + 2 Sqrt[x y] Cos[t]), {t, 0, 
  Pi}, Assumptions -> {x > 0, y > 0, x > y}]

Pi*(1/(8*y) - (3*y)/(8*x^2))

However, a workaround is to convert the trig functions to exponentials

Integrate[Sqrt[y/x] (Sin[t]^2 Cos[t])/(x + y + 2 Sqrt[x y] Cos[t]) // 
  TrigToExp, {t, 0, Pi}, Assumptions -> {x > 0, y > 0, x > y}]

-((Pi*y)/(4*x^2))

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3
  • $\begingroup$ So is this a bug? $\endgroup$
    – LLlAMnYP
    Apr 24, 2015 at 15:34
  • 3
    $\begingroup$ @LLlAMnYP - Looks like one to me, but I never studied entomology. $\endgroup$
    – Bob Hanlon
    Apr 24, 2015 at 15:38
  • $\begingroup$ Thank you for confirming this and for the workaround. $\endgroup$ Apr 24, 2015 at 16:17
3
$\begingroup$

This appears to be a bug in V10.0.x which was fixed in V10.1.0.

$Version
"10.1.0  for Mac OS X x86 (64-bit) (March 24, 2015)"
Integrate[Sqrt[y/x] (Sin[t]^2 Cos[t])/(x + y + 2 Sqrt[x y] Cos[t]), {t, 0, Pi},
  Assumptions -> {x > 0, y > 0, x > y}]
-((π y)/(4 x^2))
$\endgroup$
1
  • $\begingroup$ Also version 8 for the Mac was free of the bug. $\endgroup$ Apr 24, 2015 at 16:49

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