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}]
Integrate::idivmessages on 9.0.1 - does it work for you? – Yves Klett Feb 2 at 19:36