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Define for all $n\in\mathbb{N}$, a definite double integral, $$I_n= \int_{0}^{1}\int_{0}^{1}\left(\frac{x^2(1-x)y^2(1-y)}{1+x^2 y^2}\right)^n\frac{\cos((2n+1)\tan^{-1}(xy))}{\sqrt{1+x^2y^2}}\ dxdy$$

I need asymptotic integral expansion of $I_n$ as $n\to\infty$

I tried the following code on Wolfram Cloud:

ee1[z_]:=(x^16*(1-x^4)y^16*(1-y^4)/(1+x^2*y^2)^8)^z*Cos[(16*n+1)ArcTan[x*y]]/Sqrt[1+x^2*y^2]
i1 =Integrate[ee1[n],{x,0,1},{y,0,1}]
s1 = Expand[Normal[Series[i1, {n, Infinity, 2}]]]

I am getting the output as: The computation has exceeded the time limit for your plan or it is showing "aborted".

Any help would be highly appreciated. Thank you.

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  • $\begingroup$ Careful, you've got ArcTan[xy] not ArcTan[x y]. It's treating xy as a unique symbol instead of the product x*y $\endgroup$
    – flinty
    Commented Mar 9 at 12:03
  • $\begingroup$ @flinty Thanks a lot. Then what is the solution code? Please post it. I will happily accept your answer. Thanks again $\endgroup$
    – Max
    Commented Mar 9 at 12:20
  • $\begingroup$ I don't know, I was just pointing out an error. It may not have a nice closed form solution. $\endgroup$
    – flinty
    Commented Mar 9 at 12:27
  • $\begingroup$ @flinty Okay. Thanks. Please tell me if anything comes to your mind. $\endgroup$
    – Max
    Commented Mar 9 at 12:48
  • $\begingroup$ The code does not correspond to the expression above it. Which is correct? $\endgroup$
    – bbgodfrey
    Commented Mar 10 at 1:36

1 Answer 1

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Solving this problem analytically is difficult. However, it is possible to get exact values of the integral In in symbolic form for each n.

 Integrate[(x^2*(1-x)*y^2*(1-y)/(1 + x^2*y^2))^n * Cos[(2*n+1)*ArcTan[x*y]]/Sqrt[1 + x^2*y^2]/.n->1, {x, 0, 1}, {y, 0, 1}]
 (* -Catalan + (7 \[Pi])/8 - \[Pi]^2/16 - (7 Log[2])/4 *)

 Integrate[(x^2*(1-x)*y^2*(1-y)/(1 + x^2*y^2))^n * Cos[(2*n+1)*ArcTan[x*y]]/Sqrt[1 + x^2*y^2]/.n->2, {x, 0, 1}, {y, 0, 1}] // Expand
 (* 183/4 - 14 Catalan - (215 \[Pi])/24 + (5 \[Pi]^2)/12 - (77 Log[2])/6 *)

 Integrate[(x^2*(1-x)*y^2*(1-y)/(1 + x^2*y^2))^n * Cos[(2*n+1)*ArcTan[x*y]]/Sqrt[1 + x^2*y^2]/.n->3, {x, 0, 1}, {y, 0, 1}] // Expand
 (* -(19817/60) + 251 Catalan - (2927 \[Pi])/240 + (7 \[Pi]^2)/16 + (23263 Log[2])/120 *)

Such expressions can be obtained for any n, although for larger n it is time consuming.

 val = Table[PrintTemporary[n]; Integrate[(x^2*(1-x)*y^2*(1-y)/(1 + x^2*y^2))^n * Cos[(2*n+1) * ArcTan[x*y]]/Sqrt[1 + x^2*y^2], {x, 0, 1}, {y, 0, 1}], {n, 1, 30}] // Simplify

The numerical values are as follows. To add to this, I want to note that using NIntegrate directly leads to error messages for larger n and I didn't have the patience to experiment with different parameters.

 N[val,80]
 (*
 {0.0030701366658608634428975453931124728091424020885493502245540460203846958760120243,
 -6.1569550293516657371596971272056926107472850125261746835292024037224512690511116*10^-6,
 -4.9573022175486282585575903013967374515098952666994856787282182478234213973647599*10^-7,
 -8.5871437907848650515391861076818180834313731950437975833725264035944670144903503*10^-9,
 -8.2142613428795714287988511790242728239919040649430859172613586937930470981678207*10^-11,
 1.3847044937167217659296663716706036001611975173651185822939781255019231276009811*10^-14,
 1.7892711527483957052283014626334519856293879722336760989258099932899062104608207*10^-14,
 3.8791957450524600173559499447778556429519581856113982851227359366936649427375936*10^-16,
 4.4688571870968641320765463778878683437005599116429526971429151154358579120972046*10^-18,
 8.3914481409477945438635344664629732959305998561577003644293574597060624394066275*10^-21,
 -8.8842543487659752620985127585283776322638503832978387121964432909803043947819284*10^-22,
 -2.2290195466141809766398229643702338390952818432313565865626406605990221043109963*10^-23,
 -2.8983596441944885187441110949288862623074934486867401780613159046323596287604000*10^-25,
 -1.0673137173487623000052380055834484778466784813621684416239454595648813068896001*10^-27,
 4.9006181350152860567285197769086010465930052551654212673223247677394737465405394*10^-29,
 1.4132103240916322425538872528472675573780649372052167023402247594655330128862634*10^-30,
 2.0299922968246319562700270872831819565809052640092620068950499565550618199971013*10^-32,
 1.0714203393584658262026710589507429970849323052551850498820947807378893482387450*10^-34,
 -2.8044769253431543286543633334807384981760786222564720235651555740854098503653913*10^-36,
 -9.4249586240745017219079387181752435641221544903175439338869668772753997907515923*10^-38,
 -1.4810457081865403694515877019984012183433706845338026319683004142022100468049472*10^-39,
 -9.9447329589238538069328666737557393046528488632778909386241092885628838297836445*10^-42,
 1.6006426807866253855268101926575800966972719359475211857554988048656892969265927*10^-43,
 6.4660883377643630555372355872487779640521658948433098252689782981852945262890840*10^-45,
 1.1065363131572522311341783258591031212125083232318385492269815642245960609566333*10^-46,
 8.8910007141678446811617375861730741584245887596260006038714285361964366761591463*10^-49,
 -8.7618481707892914117384800850822389947781885829734616101664622703838618438789519*10^-51,
 -4.5060965601973851295504650297384592444283858749128591326562530173333338764852701*10^-52,
 -8.3869959324995378766606753663225941713348751399167746330285724111916441670602139*10^-54,
 -7.7770904821425894968307520711548913135922351319128695952114560085955580373479135*10^-56}
 *)

It is clear from these values that In -> 0, as n -> oo

But the following graphs are interesting:

enter image description here

From here it seems that Abs[In]^(1/n) converges somewhere towards 0.016..., which is supported by other graphs

enter image description here

enter image description here

enter image description here

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