As a part of a conditional differential entropy calculation, I need to calculate the following improper integral and I need a closed form expression, not a numerical result. However, Mathematica 10.4 does not give any result although I have waited more than 12 hours. Is there anything I can do in order to accelerate the computation? Here gamma is real and positive scale parameter of the probability distribution.

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Integrate[(-((8*γ^3*(x^4*(4*γ^2 + y^2 + z^2) + 
   (4*γ^2 + y^2)*(4*γ^2 + z^2)*(20*γ^2 + y^2 + 
     z^2) + x^2*(96*γ^4 + y^4 + 28*γ^2*z^2 + z^4 + 
     y^2*(28*γ^2 - 6*z^2))))/(Pi^3*(x^4 + x^2*(8*γ^2 - 2*y^2) + 
   (4*γ^2 + y^2)^2)*(x^4 + x^2*(8*γ^2 - 2*z^2) + 
   (4*γ^2 + z^2)^2)*(y^4 + y^2*(8*γ^2 - 2*z^2) + 
   (4*γ^2 + z^2)^2))))*
    Log[x^4*(4*γ^2 + y^2 + z^2) + 
 (4*γ^2 + y^2)*(4*γ^2 + z^2)*(20*γ^2 + y^2 + z^2) + 
 x^2*(96*γ^4 + y^4 + 28*γ^2*z^2 + z^4 + y^2*(28*γ^2 - 6*z^2))], 
 {z, -Infinity, Infinity}, {y, -Infinity, Infinity}, 
 {x, -Infinity, Infinity}, Assumptions -> 
 {γ > 0 && Element[γ, Reals]}]
  • $\begingroup$ What is Subscript[y, 2]? $\endgroup$ – Chip Hurst Aug 16 '16 at 21:27
  • $\begingroup$ I am sorry, it must be z. I have changed the notation from y0,y1,y2 to x,y,z in order to be simple. $\endgroup$ – Arda Atalık Aug 16 '16 at 21:33
  • $\begingroup$ Not all integrals can be performed analytically Do you have reason to believe that this one can be? $\endgroup$ – bbgodfrey Aug 17 '16 at 0:53
  • $\begingroup$ Yes, Mathematica calculates the indefinite integral. $\endgroup$ – Arda Atalık Aug 17 '16 at 6:51

You can use the method of residues. Let $e$ be the integrand. A sketch of the derivation is given below.

1) Find $x$-residues

rx = Solve[Denominator[e] == 0, x]

$\{\{x\to -y-2 i \gamma \},\{x\to y-2 i \gamma \},\{x\to -z-2 i \gamma \},\{x\to z-2 i \gamma \},\{x\to -y+2 i \gamma \},\{x\to y+2 i \gamma \},\{x\to -z+2 i \gamma \},\{x\to z+2 i \gamma \}\}$

Close the contour in the upper complex half-plane, select residues with positive imaginary part.

ex = 2 I \[Pi] Table[Residue[e, {x, x /. res}], {res, rx[[5 ;; 8]]}] //
     Total // FullSimplify // Together

2) Find $y$-residues:

ry = Solve[Denominator[ex] == 0, y]

$\{\{y\to -z\},\{y\to z\},\{y\to -z-2 i \gamma \},\{y\to z-2 i \gamma \},\{y\to -z+2 i \gamma \},\{y\to z+2 i \gamma \}\}$

First two poles cancel each other. Only last two are needed:

ey = 2 I \[Pi] Table[
      Residue[ex, {y, y /. res}], {res, ry[[5 ;; 6]]}] // Total // 
   FullSimplify // Together

3) Find $z$-residues

rz = Solve[Denominator[ey] == 0, z]

$\{\{z\to 0\},\{z\to -i \gamma \},\{z\to i \gamma \},\{z\to -2 i \gamma \},\{z\to 2 i \gamma \}\}$

Last one is the second-order pole (please, check), therefore can be omitted. What remains is 1 and 3 poles. Pole 3 gives 0 contribution

FullSimplify[2 I \[Pi] Residue[ey, {z, z /. rz[[3]]}], 
 Assumptions -> {\[Gamma] > 0 && Element[\[Gamma], Reals]} ]

However, method of residues is not directly applicable to $z=0$ because it is located exactly on the contour. Needs to be treated manually.

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  • $\begingroup$ Thank you very much for the answer. When you apply the residue theorem, do the contour integrals along semi circle tend to 0? $\endgroup$ – Arda Atalık Aug 18 '16 at 10:52
  • $\begingroup$ @ArdaAtalık Not sure, one needs to carefully investigate. You can add a convergence parameter and after you got the final result set it to 0. In any case, your integral needs a lot of work. I do not think MA can just compute it. $\endgroup$ – yarchik Aug 18 '16 at 11:36
  • $\begingroup$ @ArdaAtalık The aim of posting this answer was to show the capabilities of MA to deal with residue calculations and to propose a way to tackle your problem. You probably have some extra knowledge like symmetry of the integrand function. This should facilitate your calculation. $\endgroup$ – yarchik Aug 18 '16 at 11:40
  • $\begingroup$ You are definitely right, it needs a lot of work. Thanks for your help again. $\endgroup$ – Arda Atalık Aug 18 '16 at 15:18

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