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Bug introduced in 9.0 or earlier, and fixed in 10.2


I noticed a bug in Mathematica. It computes incorrectly a definite integral

Integrate[y* Sqrt[(l1^2 - y^2) (l2^2 - y^2)]/(a^2 - y^2), {y, 0, l1}, 
  Assumptions -> l2 > a > l1 > 0]

and the response was (l1 l2)/2.

The result of integration is obviously incorrect. I sent them an e-mail and reminder. They do not respond. So, I believe the users should know about the bug.

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migrated from mathematica.meta.stackexchange.com Apr 27 '16 at 22:52

This question came from our discussion, support, and feature requests site for users of Wolfram Mathematica.

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    $\begingroup$ I'm fairly certain this should be posted on the main site if it is truly a bug. $\endgroup$ – William Apr 27 '16 at 20:03
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    $\begingroup$ On Mathematica 10.4.1 I get (2*l1*l2 - 2*Sqrt[(a^2 - l1^2)*(-a^2 + l2^2)]*Pi - (4*I)*Sqrt[(a^2 - l1^2)*(-a^2 + l2^2)]*Log[a] + 2*(-2*a^2 + l1^2 + l2^2)*Log[l1 + l2] + 2*a^2*Log[-l1^2 + l2^2] - l1^2*Log[-l1^2 + l2^2] - l2^2*Log[-l1^2 + l2^2] - (2*I)*Sqrt[(a^2 - l1^2)*(-a^2 + l2^2)]*Log[-l1^2 + l2^2] + (2*I)*Sqrt[-((a^2 - l1^2)*(a^2 - l2^2))]* Log[-(a^2*(l1^2 + l2^2)) + 2*l1*l2*(l1*l2 - I*Sqrt[(a^2 - l1^2)*(-a^2 + l2^2)])])/4 $\endgroup$ – Rolf Mertig Apr 27 '16 at 20:11
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    $\begingroup$ @RolfMertig So it does appear to be fixed I'm on 10.0.0.0 with (l1 l2)/2. $\endgroup$ – William Apr 27 '16 at 20:12
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    $\begingroup$ @William Yep. People really should always say which Mathematica version they use. And for the notebook FrontEnd also the operating system. $\endgroup$ – Rolf Mertig Apr 27 '16 at 20:15
  • $\begingroup$ With 10.4.1 for Mac OS X x86 (64-bit) (April 11, 2016), Assuming[{l2 > a > l1 > 0}, Integrate[y*Sqrt[(l1^2 - y^2) (l2^2 - y^2)]/(a^2 - y^2), {y, 0, l1}] // FullSimplify] evaluates to (1/2)*(l1*l2 - Sqrt[(a - l1)*(a + l1)* (-a^2 + l2^2)]*Pi + (-2*a^2 + l1^2 + l2^2)* ArcTanh[l1/l2] - I*Sqrt[(-(a - l1))*(a + l1)* (a - l2)*(a + l2)]* (2*Log[a] + Log[-l1^2 + l2^2] - Log[(-a^2)*(l1^2 + l2^2) + 2*l1*l2*(l1*l2 - I*Sqrt[(a - l1)*(a + l1)* (-a^2 + l2^2)])])) $\endgroup$ – Bob Hanlon Apr 28 '16 at 0:31

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