2
$\begingroup$

I'm trying to integrate a trigonometric expression as

Integrate[Cos[2 t] Cos[3 t] Cos[4 t] Cos[5 t] Sin[2 t] Sin[3 t], {t, 0, 2 Pi}]

The final results is correct (which is 0), but it takes a very long time (about 1100 seconds), and it returns some output messages such as:

N::meprec :  Internal precision limit $MaxExtraPrecision = 50.` reached while evaluating -\[Pi]+10 ArcTan[Sqrt[10-2 Power[<<2>>]]/(5+Sqrt[5])].
N::meprec :  Internal precision limit $MaxExtraPrecision = 50.` reached while evaluating -\[Pi]+10 ArcTan[Sqrt[10-2 Power[<<2>>]]/(5+Sqrt[5])].
LessEqual::meprec :  Internal precision limit $MaxExtraPrecision = 50.` reached while evaluating -(1/5) Abs[-\[Pi]+10 ArcTan[Sqrt[Plus[<<2>>]]/Plus[<<2>>]]].
PossibleZeroQ::ztest1 :  Unable to decide whether numeric quantity -(\[Pi]/5)+I (Log[1-(I Sqrt[Plus[<<2>>]])/Plus[<<2>>]]-Log[1+I Power[<<2>>] Power[<<2>>]]) is equal to zero. Assuming it is.

In the end, I used TrigReduce to process this expression first then Integrate it (and got the result very fast).

But I am curious why the original version takes so long.

$\endgroup$
2
  • 2
    $\begingroup$ You're asking about internal workings. Try Trace[Integrate[ Cos[2 t] Cos[3 t] Cos[4 t] Cos[5 t] Sin[2 t] Sin[3 t], {t, 0, 2 Pi}], _PossibleZeroQ, TraceInternal -> True] to see what's causing the warning. Determining whether something is zero can be difficult. It may or may not have to do with the problem. $\endgroup$
    – Michael E2
    Aug 9 at 14:59
  • $\begingroup$ @Michael E2 the TraceInternal output is really hard to read, and I'm not sure I can find the reason from it, anyway thanks for the help (^_^) $\endgroup$
    – Ji'an Li
    Aug 9 at 15:56

1 Answer 1

4
$\begingroup$

Simplification of the integrand leads to (removable) singularities:

Cos[2 t] Cos[3 t] Cos[4 t] Cos[5 t] Sin[2 t] Sin[3 t] // Simplify
(*  1/16 Csc[5 t] Sin[6 t] Sin[8 t] Sin[10 t]  *)

That leads to checking convergence, which I guess takes a long time.

You can turn off some of the checking, and Integrate[] takes somewhere between 0.2 and 4 seconds, depending on what's been loaded and computed already.

Integrate[
 Cos[2 t] Cos[3 t] Cos[4 t] Cos[5 t] Sin[2 t] Sin[3 t],
 {t, 0, 2 Pi}, 
 GenerateConditions -> False]
(*  0  *)

You can see that the original Integrate[] obtains the antiderivative fairly quickly and then output stops, from which I inferred it was dealing with the singularities (see How much time should one give Mathematica for an integral evaluation? for some debugging techniques):

Block[{Integrate`QuickLookUpDump`dbgPrintQT = Print},
 Integrate[
  Cos[2 t] Cos[3 t] Cos[4 t] Cos[5 t] Sin[2 t] Sin[3 t],
   {t, 0, 2 Pi}]
 ]
(* output contains Csc[5 t] Sin[6 t] Sin[8 t] Sin[10 t] *)

I aborted the computation, so I don't know what happens if you wait 1100 seconds....

$\endgroup$
3
  • $\begingroup$ Thanks for help (^_^) ` Integrate[ Cos[2 t] Cos[3 t] Cos[4 t] Cos[5 t] Sin[2 t] Sin[3 t], {t, 0, 2 Pi}, GenerateConditions -> False] ` this code still needs a long time, and I aborted it after a minute wait. The debugging techniques is really helpfull, I think I need to take a good look at it, so that to figure out what happened during the points process $\endgroup$
    – Ji'an Li
    Aug 9 at 15:45
  • $\begingroup$ @Ji'anLi You're welcome. I would say that anything that takes 1100 sec. on a modern computer is doing a lot. The answer I linked shows how to get more information, but I would expect the output to be hundreds of pages. Most of the output will be impossible to understand without being familiar with the internal code. I'm not, and I wouldn't look into this any farther. I'd be satisfied with finding the Csc[5 t] factor above and thinking that's leading to difficulties. $\endgroup$
    – Michael E2
    Aug 9 at 15:59
  • $\begingroup$ @Ji'anLi Also, I'm using V13.1. Integrate[] is updated probably in every new version of Mathematica. If you're using a different version, it might explain why the GenerateConditions -> False version takes more than a few seconds. $\endgroup$
    – Michael E2
    Aug 9 at 16:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.