10
$\begingroup$

Integrate[Cos[phi]^2*Sin[phi]^n, {phi, 0, Pi}, Assumptions -> n > 0]
on Mathematica 13.0 returns the obviously mistaken error message (Integrate::idiv):

Curiously enough, just replacing $n$ by $n-2$,
Integrate[Cos[phi]^2*Sin[phi]^(n-2), {phi, 0, Pi}, Assumptions -> n > 2]
returns the correct answer $\frac{\sqrt{\pi }\, \Gamma \left(\frac{n-1}{2}\right)}{2 \Gamma \left(\frac{n}{2}+1\right)}.$

Any idea what is going on?


Thank you for the workarounds. Still, I would be interested to know why this fails, is there a lesson here, is it a more general difficulty? The integrand seems so simple and well-behaved, why does Mathematica think it "does not converge"?

$\endgroup$
7
  • 1
    $\begingroup$ The first integral does not evaluate and has no error message for me in V13.0 (Mac M1). (The second does the same as shown above.) -- This works for me, I think: Integrate[Cos[phi]^2*Sin[phi]^n // TrigToExp, {phi, 0, Pi}, Assumptions -> n > 0] $\endgroup$
    – Michael E2
    Jan 23 at 18:54
  • 3
    $\begingroup$ I think this is a bug and should be reported to "support@wolfram.com" $\endgroup$ Jan 23 at 19:12
  • $\begingroup$ @MichaelE2: Your workaround does work, but its output (\[Pi]^(3/2) Sec[(n \[Pi])/2])/(2 Gamma[1/2 - n/2] Gamma[2 + n/2]) is not convenient (say for $n=1$). $\endgroup$
    – user64494
    Jan 23 at 19:30
  • 1
    $\begingroup$ Sorry for a mistake: someone turned of my Integrate::idiv message. I didn't do it, so I must have aborted something when it had been internally turned off. Anyway, it misled me, as did no one else contradicting me. After turning it on, I now get the same error as you. $\endgroup$
    – Michael E2
    Jan 23 at 23:59
  • 1
    $\begingroup$ Reported as a bug $\endgroup$ Jan 24 at 23:06

4 Answers 4

5
$\begingroup$

I suggest a workaround

f[n_] = Integrate[Sin[phi]^n, {phi, 0, Pi}, Assumptions -> n > 0]
FullSimplify[f[n] - f[n + 2]]

$$\frac{\sqrt{\pi } \Gamma \left(\frac{n+1}{2}\right)}{2 \Gamma \left(\frac{n}{2}+2\right)}$$

Another way of calculating the integral

FullSimplify[
 Integrate[Cos[phi]^2*Sin[phi]^n, {phi, 0, Pi/4, Pi}, 
  Assumptions -> n > 0]]

suggests that in MA way of integrating the antiderivative has a singularity at $\phi=\pi/4$. Have no idea why.

$\endgroup$
5
$\begingroup$

Here's a way to get the answer by changing the integral by symmetry. Powers of sine and cosine that depend on parameters have always been tricky for Integrate. Not sure why the trick below works and the original fails.

Integrate[
 Cos[phi]^2*Sin[phi]^n + Sin[phi]^2*Cos[phi]^n, {phi, 0, Pi/2}, 
 Assumptions -> n > 0]
(*  (Sqrt[π] Gamma[(1 + n)/2])/(2 Gamma[2 + n/2])  *)

Workarounds from my comments:

Both

Integrate[Cos[phi]^2*Sin[phi]^n, {phi, 0, Pi}, 
  GenerateConditions -> False] // FullSimplify[#, n < -4] &

Normal@Integrate[#, {phi, 0, Pi}] & /@ 
  Expand[(1 - Sin[phi]^2) Sin[phi]^n] // FullSimplify

give

(*  (Sqrt[π] Gamma[(1 + n)/2])/(2 Gamma[2 + n/2])  *)

but not in a satisfying way.

And

Integrate[Cos[phi]^2*Sin[phi]^n // TrigToExp, {phi, 0, Pi}, Assumptions -> n > 0]

gives

(*  (π^(3/2) Sec[(n π)/2])/(2 Gamma[1/2 - n/2] Gamma[2 + n/2])  *)

which you have to use Limit to evaluate at odd integers. Or we can expand secant and gamma according to their infinite products to get the first result above:

(π^(3/2) Sec[(n π)/2])/(2 Gamma[1/2 - n/2] Gamma[2 + n/2]) /. 
  Sec[(n π)/2]/Gamma[1/2 - n/2] -> 1/(Pi ((π - n Pi)/(2 π))*
    Product[
     (1 - (π - n π)^2 /
      (4 k^2 π^2)) ((1 + 1/k)^(1/2 (-1 - n))/(1 + (-1 - n)/(2 k))),
     {k, Infinity}]) // FullSimplify

Update:

Integrate tries a substitution $\phi = \arctan(p/2)$, which fails for reasons I don't understand, since the transformed integral, while improper, converges:

Cos[phi]^2*Sin[phi]^n*Dt[phi, p] /. phi -> ArcTan[p/2] /. 
   t : _Sin | _Cos :> TrigExpand[t] /. p -> phi // Simplify
Integrate[2 %, {phi, 0, \[Infinity]}, Assumptions -> n > 0]
(*
  (8 (phi/Sqrt[4 + phi^2])^n)/(4 + phi^2)^2
  (Sqrt[\[Pi]] Gamma[(1 + n)/2])/(2 Gamma[2 + n/2])  
*)

My guess is that there's a simple(?) bug that is not particularly interesting, but maybe someone else will have more insight. Integrate can generate answers to equivalent integrals involving things like Hypergeometric2F1[2, 1/2 + n/2, 3/2, 1], which is undefined for n>0. It evaluates to ComplexInfinity if n is given a definite, nonnegative value. I do not know if that is connected to the idiv divergence error.

$\endgroup$
3
$\begingroup$

This seems to work:

res = Integrate[Cos[phi]^m*Sin[phi]^n, {phi, 0, \[Pi]}] /. m -> 2


N[Table[{NIntegrate[Cos[phi]^2*Sin[phi]^n, {phi, 0, Pi}], 
  res}, {n, 1, 6}]]

(* {{0.666667, 0.666667}, {0.392699, 0.392699}, {0.266667, 
 0.266667}, {0.19635, 0.19635}, {0.152381, 0.152381}, {0.122718, 0.122718}} *)
$\endgroup$
3
  • $\begingroup$ true, but it does not give the symbolic answer for any $n$, does it? $\endgroup$ Jan 23 at 21:11
  • $\begingroup$ @CarloBeenakker I obtained a symbolic answer for res that claims to be valid for any $n$ with a real part greater than $-1$. (Output: ConditionalExpression[(Sqrt[\[Pi]] Gamma[(1 + n)/2])/(2 Gamma[(4 + n)/2]), Re[n] > -1]) Is it the condition (so that it is not valid for any $n$) or did you get something else? $\endgroup$
    – Michael E2
    Jan 23 at 21:19
  • $\begingroup$ apologies, I had not understood that from the answer; so somehow "Integrate[Cos[phi]^m*Sin[phi]^n, {phi, 0, [Pi]}]" works but "Integrate[Cos[phi]^2*Sin[phi]^n, {phi, 0, [Pi]}]" fails; no idea why, but at least this is a way to work around the bug. $\endgroup$ Jan 23 at 21:21
3
$\begingroup$

Two small steps are better than one big step. A simple workaround in 13.0.0 on Windows 10 is as follows.

Integrate[1/2*Cos[2*phi]*Sin[phi]^n, {phi, 0, Pi},  Assumptions -> n > 0] + 
Integrate[1/2*Sin[phi]^n, {phi, 0, Pi}, Assumptions -> n > 0]

(Sqrt[\[Pi]] Gamma[(1 + n)/2])/(2 Gamma[1 + n/2]) - ( n Sqrt[\[Pi]] Gamma[(1 + n)/2])/(4 Gamma[2 + n/2])

FullSimplify[%]

(Sqrt[\[Pi]] Gamma[(1 + n)/2])/(2 Gamma[2 + n/2])

Addition. Concerning possible explanation of the incorrect result, I share @MichaelE2's opinion since

Integrate[Cos[phi]^2*Sin[phi]^n, phi, Assumptions -> n > 0]

(1/(1 + n))Sqrt[Cos[phi]^2] Hypergeometric2F1[-(1/2), (1 + n)/2, (3 + n)/2, Sin[phi]^2] Sec[ phi] Sin[phi]^(1 + n)

Plot3D[%, {phi, 0, Pi}, {n, 0, 3}]

enter image description here

$\endgroup$

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.