0
$\begingroup$

When I numerically calculate this integral

  NIntegrate[(1/2)*Log[1 + Sin[x]]^2, {x, 0, Pi}]

The result is $0.416217488896557$ and I verified it using another tool.

However, when I use

  Integrate[(1/2)*Log[1 + Sin[x]]^2, {x, 0, Pi}]

The result is

$$\frac{1}{48} \left(-96 C \log (2)+i \left(-192 \text{Li}_3\left(\frac{1}{2}+\frac{i}{2}\right)+105 \zeta (3)+4 \log ^3(2)\right)+7 \pi ^3-24 \pi \log ^2(2)-5 i \pi ^2 \log (2)\right)$$

Numerically, this evaluates to $4.777566543250842 -2.9605947323337506\times 10^{-16} i$

Am I doing something wrong?

I am using Mathematica 14.0.0.0 on an x86, 64-bit, Windows 10 machine.

Update I tried it on Mathematica 12.2.0.0 and it left the integral unevaluated and I expected this because I couldn't imagine it would have a closed-form solution.

$\endgroup$
4
  • 1
    $\begingroup$ This is weird. What is that $C$ supposed to represent in your result? I get the correct result in v14.0 (Win). Can you please try your code again on a fresh kernel (use Quit[]) to remove any possible effects of caching? $\endgroup$
    – Domen
    Apr 2 at 13:37
  • $\begingroup$ @Domen: From a fresh kernel, resolves the problem $$\frac{1}{48} \left(2 i \left(-192 \text{Li}_3\left(\frac{1}{2}+\frac{i}{2}\right)+105 \zeta (3)+4 \log ^3(2)\right)-7 \pi ^3+12 \pi \log ^2(2)-10 i \pi ^2 \log (2)\right)$$ Using Chop, that gives the correct result. I did the same thing earlier by restarting Mathematica, ut the issue persisted. $\endgroup$
    – Moo
    Apr 2 at 13:57
  • 1
    $\begingroup$ Good! These things happen ... Please let us know if you can reproduce this issue again (ie. find a sequence of integrations that lead to the wrong result). In this case, it can then be further analyzed or reported to the WRI as a "bug". $\endgroup$
    – Domen
    Apr 2 at 14:04
  • $\begingroup$ @Domen: Just FYI - Please see the latest answer - another person was able to see what I saw and provided a nice workaround. $\endgroup$
    – Moo
    Apr 3 at 1:49

2 Answers 2

1
$\begingroup$

If you do

p = Integrate[(1/2)*Log[1 + Sin[x]]^2, x]
Limit[p, x -> \[Pi], Direction -> "FromBelow"] - 
Limit[p, x -> 0, Direction -> "FromAbove"] // FullSimplify

you get

(7*Pi^3)/24 + Pi*Log[2]^2 - 
4*I*(PolyLog[3, 1 - I] - PolyLog[3, 1 + I])]

which gives the same result as NIntegrate.

$\endgroup$
1
  • $\begingroup$ Thanks - in a new iteration, after restarting the kernel, it produces the correct results - which I had done earlier. The problem persisted but in this round it worked. $\endgroup$
    – Moo
    Apr 2 at 14:01
3
$\begingroup$

For me, like the OP's original result,

Integrate[(1/2)*Log[1 + Sin[x]]^2, {x, 0, Pi}]

yields

(*
1/48 (7 \[Pi]^3 - 96 Catalan Log[2] - 5 I \[Pi]^2 Log[2] - 
   24 \[Pi] Log[2]^2 + 
   I (4 Log[2]^3 - 192 PolyLog[3, 1/2 + I/2] + 105 Zeta[3]))
*)

But

Integrate[(1/2)*Log[1 + Sin[x]]^2, {x, 0, Pi/2, Pi}, 
 Assumptions -> 0 < x < Pi]

yields

(*
1/48 (-7 \[Pi]^3 - 10 I \[Pi]^2 Log[2] + 12 \[Pi] Log[2]^2 + 
   2 I (4 Log[2]^3 - 192 PolyLog[3, 1/2 + I/2] + 105 Zeta[3]))
*)

N[%, 80]
(*
0.416217488896552119403646053625517279786339361782973377550142 + 0.*10^-61 I
*)

which agrees with the numerical integration.

I've found that sometimes specifying the integration path helps, as it does in this case.

$\endgroup$
1
  • $\begingroup$ Thanks for showing that you got the same thing as I did as I felt I had not done anything to cause an errant result. What is even stranger is that using a fresh kernel then generated the correct result. I like your approach of providing as much input as possible to try and avoid these things. +1. $\endgroup$
    – Moo
    Apr 3 at 1:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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