1
$\begingroup$

I have a long set of replacement rules that I utilize to solve some longish integrals over trigonometric functions. Just look at this snippet (from this answer):

validQ[coeff___, arg_, v_] := 
  FreeQ[{coeff}, v] && Exponent[arg, v] == 1;
testInt[v_, a_, b_] := {
   coeff___*Cos[arg_] /; validQ[coeff, arg, v] :> coeff/Coefficient[arg, v]*Simplify[(First@Differences[Sin[arg] /. {{v -> a}, {v -> b}}]), Trig -> False],
   (*failure rule*)
   integrand_ :> Inactive[Integrate][integrand, {v, a, b}]
   };

See, that e.g. 2*Cos[t] /. testInt[t, 0, \[Pi]/2] returns 2 as expected. However, I am struggling with terms that are like Cos[t]:

Cos[t] /. testInt[t, 0, \[Pi]/2]
(* Inactive[Integrate][Cos[t], {t, 0, \[Pi]/2}] *)

Same for 1*Cos[t]. However, applying the rules to 1.*Cos[t] yields 2.. In principle, I could just define an additional rule for Cos with the negative side-effect that I would have to be very cautious since this would also replace the Cos in terms where it may be wrong. How can I change the patterns so that ideally coeff___*Cos also accepts 1 as a coefficient?

$\endgroup$
3
$\begingroup$

Seems that I was just able to answer my own question. I came across Optional which seems to do exactly what I want. Changing the rule according to

validQ[coeff___, arg_, v_] := FreeQ[{coeff}, v]  && Exponent[arg, v] == 1;
testInt[v_, a_, b_] := {
   (coeff___ : 1)*Cos[arg_] /; validQ[coeff, arg, v] :> coeff/Coefficient[arg, v]*Simplify[(First@Differences[Sin[arg] /. {{v -> a}, {v -> b}}]), Trig -> False],
   (*failure rule*)
   integrand_ :> Inactive[Integrate][integrand, {v, a, b}]
   };

does the job. By this, it is possible to provide a default value (here 1) if coeff is absent in the expression. To test the rules:

Cos[t] /. testInt[t, a, b]
(* -Sin[a] + Sin[b] *)
2*Cos[t] /. testInt[t, a, b]
(* 2 (-Sin[a] + Sin[b]) *)
t*Cos[t] /. testInt[t, a, b]
(* Inactive[Integrate][t Cos[t], {t, a, b}] *)

Update

As pointed out by @Karsten in a comment, it is also possible to use coeff_. instead of coeff___:1.

$\endgroup$
  • 1
    $\begingroup$ Using coeff_. should do the trick, too. $\endgroup$ – Karsten 7. Dec 4 '15 at 12:42
  • $\begingroup$ @Karsten7. True, also works! Thanks for this. $\endgroup$ – Lukas Dec 4 '15 at 12:44
  • $\begingroup$ You are using the syntax for a custom default value, but as you specify it to be the built-in default value for multiplication you might as well use that. $\endgroup$ – Karsten 7. Dec 4 '15 at 12:57

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.