3
$\begingroup$

Working this problem on Mathematics Stack Exchange, I face the problem of computing $$I(t)=\int_{0}^{t}\frac{a \sin ^2(k)+b}{\sqrt{\left(a \sin ^2(k)+b\right)^2+\sin ^2(k)}}\,dk$$ where $(a,b)$ are positive constants $( 0 <b<\frac 1 4a)$ and $0<t\leq \frac \pi 2$.

Mathematica produces almost instantly the antiderivative in terms of elliptic integrals of the first and third kinds. However, no result for the definite integral (all above assumptions being provided).

$I(0)=0$ is immediately returned but asking for $I\left(\frac{\pi }{2}\right)$ Mathematica runs for even (I let my computer running $14$ hours for no result.

I suppose that the problem comes from the fact that the argument of the elliptic integrals is proportional to $\tan(k)$.

So, my question : is there any way to obtain the limit ?

    Integrate[(a Sin[t]^2 +b)/((a Sin[t]^2+b)^2+Sin[t]^2)^(1/2),{t,0,Pi/2}]
$\endgroup$
1
  • $\begingroup$ Please provide Mathematica code. This will increase chance to get helpful answers! $\endgroup$ Apr 26, 2023 at 8:26

2 Answers 2

1
$\begingroup$
$Version (*"12.2.0 for Microsoft Windows (64-bit) (December 12, 2020)"*)

The definite integral evaluates to

int = Values@DSolve[{f'[k] == (a Sin[k]^2 + b)/Sqrt[(a Sin[k]^2 + b)^2 + Sin[k]^2 ], f[0] == 0}, f, k][[1, 1]]

Unfortunately Asymptotic[int[k],k->Pi/2] doesn't evaluate.

With the help of substitution Tan[k]->tank

subst = Join[{Tan[k] -> tank},Table[Cos[i k] -> (Cos[i k] /. k -> ArcTan[tank] // TrigExpand), {i,1,4}]] // Simplify;

we get asymptotic of k->Pi/2

Asymptotic[int[k] /. subst, tank -> Infinity] //Simplify[#, Assumptions -> {tank > 0, 0 < b < a/4}] &

enter image description here

$\endgroup$
9
  • $\begingroup$ Thank you very much. I arrived to the point where I obtained (with your code) $-\frac{2 \sqrt{b^2}}{b \text{uk}}$ but I do not understand the next step. What do I do with this result to go to the final answer. I may look stupid but I am blind and my screen teller does not work properly (I have to get a new one). Would the result be independent of $a$ and $b$ ? Thanks agaib (please help me more !!). Cheers :-) $\endgroup$ Apr 26, 2023 at 10:36
  • $\begingroup$ Sorry, in the first version of my answer I evaluateted int[Pi]. See my modified answer! $\endgroup$ Apr 26, 2023 at 10:43
  • $\begingroup$ This is superb ! Thanks so much. $\endgroup$ Apr 26, 2023 at 11:11
  • $\begingroup$ @ClaudeLeibovici Hope it helps! $\endgroup$ Apr 26, 2023 at 11:19
  • $\begingroup$ Much more than that. I shall update my answer on MSE $\endgroup$ Apr 26, 2023 at 12:01
1
$\begingroup$

The limit is given by a rather complicated function of a and b containing complete Elliptic functions as:

(b*((b*(-1 + Sqrt[1 + 4*a*b]) + a*(1 + Sqrt[1 + 4*a*b]))*
EllipticK[(2*Sqrt[1 + 4*a*b])/(1 + 2*b*(a + b) + 
Sqrt[1 + 4*a*b])] + b*(1 + 2*a*(a + b) - Sqrt[1 + 4*a*b])*
EllipticPi[(1 + 2*a*(a + b)+Sqrt[1 + 4*a*b])/(2*(1 + (a + b)^2)), 
(2*Sqrt[1 + 4*a*b])/(1 + 2*b*(a + b) + Sqrt[1 + 4*a*b])]))/
(a*(1 + 2*b*(a + b) + Sqrt[1 + 4*a*b])*
Sqrt[b*(a + b) + (1/2)*(1 + Sqrt[1 + 4*a*b])])

I obtained it by transforming the imaginary argument of the incomplete elliptic functions returned by Mathematica into real form as well as applying a transformation for imaginary moduli (negative values in Mathematica speak)

$\endgroup$
10
  • 1
    $\begingroup$ Superb ! Thank you so much. $\endgroup$ Apr 26, 2023 at 12:07
  • $\begingroup$ When I think about my ridiculous asymptotics !! If you have time to waste, ask Mathematica for it : a pure monster !! $\endgroup$ Apr 26, 2023 at 12:49
  • $\begingroup$ @Andreas It would be very helpful if you show the solution steps which led to your result! $\endgroup$ Apr 26, 2023 at 16:56
  • $\begingroup$ They go like EllipticF[Izz, kk^2] -> IEllipticF[ArcTan[Sinh[zz]], 1 - kk^2] and EllipticF[zz, -kk] -> EllipticF[ArcTan[Sqrt[1 + kk]*Tan[zz]], kk/(1 + kk)]/Sqrt[1 + kk] $\endgroup$
    – Andreas
    Apr 26, 2023 at 17:03
  • $\begingroup$ @Andreas In addition case ` {b -> 1, a -> 0}` should evaluate to EllipticK[-1] but your formula gives Indeterminate! $\endgroup$ Apr 26, 2023 at 17:08

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.