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I am working on the following problem. However, the first task takes too long, although the answer 0 is correct. In the second task, to reduce the calculation time, I have used the Simplify function, but it gives wrong answer.

(*first task*)
Di = 3; (*Dimension of space *)
av = Table[a[i], {i, 1, Di}];
ξv = Table[ξ[i], {i, 1, Di}];
f[0] = a[0]/((π (4 a[Di + 1]/a[0] /Di))^(Di/2)) 
   Exp[-(ξv - av/a[0]).(ξv - av/a[0])/(4 a[Di + 1]/ a[0] /Di)];

Φ[0][0] = -Sum[Da[i][i], {i, 1, Di}];
Φ[0][k_] = -2/Di Da[k][4] - 
  a[k]/a[0] Sum[Da[i][i], {i, 1, Di}] - 
  Sum[a[i]/a[0] Da[i][k], {i, 1, Di}] + 
  Sum[a[i] a[k]/(a[0])^2 Da[i][0], {i, 1, Di}];
Φ[0][4] = -a[4]/a[0] Sum[Da[i][i], {i, 1, 3}] + 
  a[4]/(a[0])^2 Sum[a[i] Da[i][0], {i, 1, 3}] - 
  1/a[0] Sum[a[i] Da[i][4], {i, 1, 3}];

f[1] = -Sum[D[f[0], a[β]] Φ[0][β], {β, 0, 4}] -
  Sum[ξ[i] Sum[D[f[0], a[β]] Da[i][β], {β, 0, 4}], {i, 1, 3}];

Integrate[f[1], {ξ[1], -∞, ∞}, {ξ[2], -∞, ∞}, {ξ[3], -∞, ∞}, 
  Assumptions -> {a[Di + 1] ∈ Reals, a[Di + 1] > 0, a[0] ∈ Reals, a[0] > 0}]

(*second task*)
Integrate[Simplify[f[1]], {ξ[1], -∞, ∞}, {ξ[2], -∞, ∞}, {ξ[3], -∞, ∞}, 
  Assumptions -> {a[Di + 1] ∈ Reals, a[Di + 1] > 0, a[0] ∈ Reals, a[0] > 0}]
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What was the wrong answer that you got from "second task"? –  m_goldberg Feb 2 '13 at 15:19
    
-1/(32 [Pi]^(3/2) a[4]^4)3 Sqrt[3] Sqrt[a[4]/a[0]] Integrate[[ExponentialE]^(( 3 a[0] (-(-a[1]/a[0] + [Xi][1])^2 - (-a[2]/a[0] + [Xi][ 2])^2 - (-a[3]/a[0] + [Xi][3])^2))/( 4 a[4])) (-3 a[1]^3 (a[4] Da[1][0] + a[0] Da[1][4]) - 3 (a[2]^2 + a[3]^2) a[4] (a[2] Da[2][0] + a[3] Da[3][0])....... –  nes Feb 2 '13 at 15:26
    
The answer is not a calculated value but a kind of repeating my question. –  nes Feb 2 '13 at 15:27
    
For me, your code throws several Integrate::idiv messages on 9.0.1 - does it work for you? –  Yves Klett Feb 2 '13 at 19:36
    
f[1] does not appear to involve any of the variables of integration. –  Daniel Lichtblau Feb 4 '13 at 16:08
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