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]])
Thanx
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 *)
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.
$\endgroup$
Commented
Jan 31, 2014 at 17:38
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]
(*
True
*)
Plot[s[x], {x, 1, 100}]
So
Limit[s[n], n -> Infinity]
(*
0
*)
n=2.1gives a revealing error. Do you mean fornto be integer? $\endgroup$