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.

enter image description here

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]? $\endgroup$
    – Greg Hurst
    Commented Aug 16, 2016 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. $\endgroup$ Commented Aug 16, 2016 at 21:33
  • $\begingroup$ Not all integrals can be performed analytically Do you have reason to believe that this one can be? $\endgroup$
    – bbgodfrey
    Commented Aug 17, 2016 at 0:53
  • $\begingroup$ Yes, Mathematica calculates the indefinite integral. $\endgroup$ Commented Aug 17, 2016 at 6:51

1 Answer 1


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.

  • $\begingroup$ Thank you very much for the answer. When you apply the residue theorem, do the contour integrals along semi circle tend to 0? $\endgroup$ Commented Aug 18, 2016 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
    Commented Aug 18, 2016 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
    Commented Aug 18, 2016 at 11:40
  • $\begingroup$ You are definitely right, it needs a lot of work. Thanks for your help again. $\endgroup$ Commented Aug 18, 2016 at 15:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.