2
$\begingroup$

I'm integrating a positive function f(t) times sin(t) from 0 to pi/5 and get -38.

Actually f is slightly negative for a short time (smallest value ~ -0.0005), but far from enough to explain this. Here is the code for the function: 

NormFact[k_, S_] := 
Sqrt[(4 S - 2 k - 1) (4 S - k) (4 S - k + 1) (k + 
  1) (k + 2)/(4*Pi*S)];

xi[k_, j_, S_] := (-1)^j*
 Binomial[k, j]*(4*S - k - 1)!/((j + 2)!*(4*S - k - j - 1)!);

Ofunc[t_, S_, coeffs_] := 
1 - Cos[t/2]^(4*S) + 
coeffs.Table[
 NormFact[k, S]*
  Sum[xi[k, j, S]*Sin[t/2]^(2*j + 2)*Cos[t/2]^(4*S - 2*j - 2), {j,
     0, k}], {k, 0, Length[coeffs] - 1}];

cfs = {-2.6155133, 0.9614036, 0.100279, -0.432464, 
0.39887624, -0.2508507, 0.10555472, -0.0007824, -0.0526054, 
0.0703274, -0.0688152, 0.05138949, -0.03833719, 
0.02062684, -0.0018939, -0.0027808};

And here's the result when I integrate:

In:= Integrate[Ofunc[t, 58.5, cfs]*Sin[t], {t, 0, Pi/5}]
Out:= -38.132 + 0. I

Can anyone see my mistake?

Improve this question
$\endgroup$
8
  • 2
    $\begingroup$ On v9.0.1, I get 37.8958. What version are you using? Run "ReleaseID" /. ("Kernel" /. SystemInformation["Small"]) $\endgroup$
    – rcollyer
    Commented Aug 8, 2013 at 16:27
  • $\begingroup$ Hm! I get 8.0.1.0 (2063990, 2063802). Note: I tried restarting the program and get the same negative answer. $\endgroup$
    – jorgen
    Commented Aug 8, 2013 at 16:30
  • 1
    $\begingroup$ I confirmed I get that answer on 8.0.1, but I don't get it on v9. So, it is likely a bug in v8. Although, I haven't tried v8.0.4. My only advice: upgrade. $\endgroup$
    – rcollyer
    Commented Aug 8, 2013 at 16:36
  • $\begingroup$ Thanks for advice! Note: it seems to me the answer you get must also be wrong, since the function and sine are both too small on the interval to give 38 (try plotting it) $\endgroup$
    – jorgen
    Commented Aug 8, 2013 at 16:37
  • 1
    $\begingroup$ Hmm, I hadn't paid attention to that. NIntegrate gives an answer I'm comfortable with (0.13956), and it works fine on v8.0.1, too. So, I'd use that, especially considering the coefficients in your function have vastly different scales, and may be confusing things. $\endgroup$
    – rcollyer
    Commented Aug 8, 2013 at 16:48

0

Reset to default

Your Answer

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