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Could someone explain why Mathematica can't finish computing this limit (this is a limit of an array, when n -> Infinity (n is an integer), whose elements are area-integrals as shown):

 Limit[Integrate[(x^(2n) + y^(2n)) Boole[x^(2n) + y^(2n) < 1], {x, -1, 1}, {y, -1, 1}], n -> Infinity]

In the case when n=2, the integral is Pi/2, for n=4 it is (2 Gamma[1/4] Gamma[5/4])/(3 Sqrt[\[Pi]])


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Your integrand does give values for some n, but not for others. For example, n=2.1 gives a revealing error. Do you mean for n to be integer? – bill s Jan 31 '14 at 14:27
Yes, n is an integer, sorry I forgot to specify. – Dragisa Jan 31 '14 at 17:27

I can get the integral by using the first quadrant (for positivity), placing an assumption on n, and recasting without Boole (I don't know why that was needed).

in = 4 Integrate[(x^(2 n) + y^(2 n)), {x, 0, 1}, {y, 
    0, (1 - x^(2 n))^(1/(2*n))}, Assumptions -> n >= 1]

(* Out[112]= (2^(2 - 1/n) Sqrt[\[Pi]]
  Gamma[1 + 1/(2 n)])/((1 + n) Gamma[(1 + n)/(2 n)]) *)

As @Belisarius notes, the limit is now straightforward.

Limit[in, n -> Infinity]

(* t[113]= 0 *)
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The Boole is there to specify the area over which an integral is computed (i think this is one way to do that). For example when n=2 the area is a circle with unit area, centered at the origin. – Dragisa Jan 31 '14 at 17:38
I realize why you had the Boole there. This is a perfectly appropriate usage but, for whatever reason, Integrate did not seem able to handle it. That's why I recast with the more awkward endpoint of integration. – Daniel Lichtblau Jan 31 '14 at 17:57

By manual inspection I got a closed formula for your integral:

s[n_] := 2/(n + 1) Gamma[1/(2 n)] Gamma[(2  n + 1)/(2 n)]/Gamma[1/n]
f[n_] := Integrate[(x^(2 n) + y^(2 n)) Boole[x^(2 n) + y^(2 n) < 1], {x, -1, 1}, {y, -1, 1}]

For example

s[16] == f[16]

Plot[s[x], {x, 1, 100}]

Mathematica graphics


Limit[s[n], n -> Infinity]
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Thanks, seems this really goes to zero, which is not what I had expected. – Dragisa Jan 31 '14 at 17:47

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