4
$\begingroup$

In 14.0 under Windows 10 I try to find the improper integral $$\int\limits_{\sqrt{1-\sqrt{1-\psi }}}^{\sqrt{1+\sqrt{1-\psi }}} \frac{1}{r^2 \sqrt{-r^4+2 r^2-\psi }}\,dr$$ under the assumptions $\psi >0\land \psi <1$.

Unfortunately,

Integrate[1/r^2/Sqrt[2 r^2 - r^4 - \[Psi]], {r, Sqrt[1 - Sqrt[1 - \[Psi]]], 
Sqrt[1 + Sqrt[1 - \[Psi]]]},  Assumptions -> \[Psi] > 0 && \[Psi] < 1]

returns the input as well as

\[Psi] = 1/n; Integrate[1/r^2/Sqrt[2 r^2 - r^4 - \[Psi]], {r, Sqrt[1 - Sqrt[1 - \[Psi]]], 
Sqrt[1 + Sqrt[1 - \[Psi]]]}, Assumptions -> n \[Element] PositiveIntegers]

For many rational values of \[Psi](for example, 2/3,1/3,1/7,...) Mathematica calculates this integral. However, it is difficult to derive the general formula from those values.

The numerical value

f[\[Psi]_?NumericQ] := NIntegrate[ 1/r^2/Sqrt[2 r^2 - r^4 - \[Psi]], 
{r, Sqrt[1 - Sqrt[1 - \[Psi]]], Sqrt[1 + Sqrt[1 - \[Psi]]]}]
Plot[f[\[Psi]], {\[Psi], 0, 1}]

enter image description here

says nothing about the rate of the growth of f[\[Psi]] as \[Psi] tends to zero from above. Concerning \[Psi] tends to 1, the result of

\[Psi] = 0.99900000000000000000000000000; 
NIntegrate[1/r^2/Sqrt[2 r^2 - r^4 - \[Psi]], {r, Sqrt[1 - Sqrt[1 - \[Psi]]],  Sqrt[1 + Sqrt[1 - \[Psi]]]}, WorkingPrecision -> 20]

1.5722703994037859370

suggests this equals Pi/2.

I also try to find an antiderivative by

ad = Integrate[1/r^2/Sqrt[2 r^2 - r^4 - \[Psi]], r, 
Assumptions -> \[Psi] > 0 && \[Psi] < 1, GenerateConditions -> True]

(-((-2 r^2 + r^4 + \[Psi])/r) + (1/Sqrt[(1/(-1 + Sqrt[1 - \[Psi]]))]) I Sqrt[1 - r^2/( 1 + Sqrt[1 - \[Psi]])] (1 + Sqrt[1 - \[Psi]]) Sqrt[(-1 + r^2 + Sqrt[1 - \[Psi]])/(-1 + Sqrt[ 1 - \[Psi]])] (EllipticE[ I ArcSinh[r Sqrt[1/(-1 + Sqrt[1 - \[Psi]])]], ( 1 - Sqrt[1 - \[Psi]])/(1 + Sqrt[1 - \[Psi]])] - EllipticF[I ArcSinh[r Sqrt[1/(-1 + Sqrt[1 - \[Psi]])]], ( 1 - Sqrt[1 - \[Psi]])/(1 + Sqrt[1 - \[Psi]])]))/(Sqrt[ 2 r^2 - r^4 - \[Psi]] \[Psi])

Rubi produces the same result in a simpler form:

Get["Rubi`"]
adRubi = Int[1/r^2/Sqrt[2 r^2 - r^4 - \[Psi]], r]
Plot3D[Re[adRubi - ad], {r, 0, 1}, {\[Psi], 0, 1}]

Both results are doubtful in view of

u = ad /. r -> Sqrt[1 + Sqrt[1 - \[Psi]]];
l = ad /. r -> Sqrt[1 - Sqrt[1 - \[Psi]]];
FullSimplify[u - l, Assumptions -> \[Psi] > 0 && \[Psi] < 1]

Indeterminate

and

u - l /. \[Psi] -> 0.5

Indeterminate

The command

ClearAll[\[Psi]]; Limit[ad, r -> Sqrt[1 + Sqrt[1 - \[Psi]]], 
 Direction -> "FromBelow", Assumptions -> \[Psi] > 0 && \[Psi] < 1]

is running without any response for a long time.

So is it possible to find a closed-form expression for the integral under consideration with Mathematica? PS. Maple 2023 answers $$ \frac{\left(1-\sqrt{1-\psi}\right) \mathrm{EllipticE}\! \left(\sqrt{2}\, \sqrt{\frac{\sqrt{1-\psi}}{1+\sqrt{1-\psi}}}\right)}{\psi \sqrt{1+\sqrt{1-\psi}}\, \left(1-\frac{2 \sqrt{1-\psi}}{1+\sqrt{1-\psi}}\right)}$$

$\endgroup$
7
  • $\begingroup$ Elliptic functions in the possible answer are allowed. $\endgroup$
    – user64494
    Feb 21 at 16:27
  • 1
    $\begingroup$ Maple 2023 answers $$ \frac{\left(1-\sqrt{1-\psi}\right) \mathrm{EllipticE}\! \left(\sqrt{2}\, \sqrt{\frac{\sqrt{1-\psi}}{1+\sqrt{1-\psi}}}\right)}{\psi \sqrt{1+\sqrt{1-\psi}}\, \left(1-\frac{2 \sqrt{1-\psi}}{1+\sqrt{1-\psi}}\right)}$$ which is confirmed by numerics. $\endgroup$
    – user64494
    Feb 21 at 17:02
  • $\begingroup$ Since you now know what the correct answer is supposed to be, please include it in the question (it doesn't really belong in the comment) :) $\endgroup$
    – Domen
    Feb 21 at 17:09
  • $\begingroup$ @Domen: It is done. $\endgroup$
    – user64494
    Feb 21 at 17:12
  • 1
    $\begingroup$ this may be further simplified to (Sqrt[1 + Sqrt[1 - [Psi]]]/[Psi])* EllipticE[(2*Sqrt[1 - [Psi]])/(1 + Sqrt[1 - [Psi]])] where the argument of EllipticE is squared (Mathematica convention) $\endgroup$
    – Andreas
    Feb 21 at 17:51

1 Answer 1

6
$\begingroup$

A change of variables can help:

integrand = 1/r^2/Sqrt[2  r^2 - r^4 - ψ];

i2 = 
 IntegrateChangeVariables[
   Inactive[Integrate][
     integrand, 
     {r, Sqrt[1 - Sqrt[1 - ψ]], Sqrt[1 + Sqrt[1 - ψ]]}
    ], 
   s, 
   s == r^2 - 1, 
   Assumptions -> ψ > 0 && ψ < 1
  ]

(*Inactive[Integrate][1/(
 2 (1 + s)^(3/2) Sqrt[1 - s^2 - ψ]), {s, -Sqrt[1 - ψ], Sqrt[
  1 - ψ]}]*)

i3=Integrate[i2[[1]], i2[[2]], Assumptions -> ψ > 0 && ψ < 1]
(* (1/(2 ψ))(Sqrt[1 - Sqrt[1 - ψ]]
    EllipticE[(2 (-1 - Sqrt[1 - ψ] + ψ))/ψ] + 
  Sqrt[1 + Sqrt[1 - ψ]]
    EllipticE[(2 (-1 + Sqrt[1 - ψ] + ψ))/ψ]) *)

$$ \frac{\sqrt{1-\sqrt{1-\psi }} E\left(\frac{2 \left(\psi -\sqrt{1-\psi }-1\right)}{\psi }\right)+\sqrt{\sqrt{1-\psi }+1} E\left(\frac{2 \left(\psi +\sqrt{1-\psi }-1\right)}{\psi }\right)}{2 \psi } $$

A brief numeric test seems to support this result:

Table[i3, {ψ, 0.1, 0.9, 0.1}]
(* {14.4606, 7.33593, 4.94922, 3.75053, 3.02835, 2.54501, \
2.19849, 1.93769, 1.73415} *)

Table[NIntegrate[integrand,
  {r, Sqrt[1 - Sqrt[1 - ψ]], 
   Sqrt[1 + Sqrt[1 - ψ]]}], {ψ, 0.1, 0.9, 0.1}]
(* {14.4606, 7.33593, 4.94922, 3.75053, 3.02835, 2.54501, \
2.19849, 1.93769, 1.73415} *)

Also

Limit[i3, ψ -> 1]
(* π/2 *)
$\endgroup$
1
  • $\begingroup$ Also Series[(1/(2 \[Psi])) (Sqrt[ 1 - Sqrt[ 1 - \[Psi]]] EllipticE[(2 (-1 - Sqrt[1 - \[Psi]] + \[Psi]))/\[Psi]] + Sqrt[1 + Sqrt[1 - \[Psi]]] EllipticE[(2 (-1 + Sqrt[1 - \[Psi]] + \[Psi]))/\[Psi]]), {\[Psi], 0, 1}, Assumptions -> \[Psi] > 0 && \[Psi] < 1] results in Sqrt[2]/\[Psi]+(-3+4 Log[2]+Log[4]-Log[\[Psi]])/(8 Sqrt[2])+((-109+120 Log[2]+30 Log[4]-30 Log[\[Psi]]) \[Psi])/(512 Sqrt[2])+O[\[Psi]]^(3/2) $\endgroup$
    – user64494
    Feb 21 at 18:11

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.