# Triple Improper Integral Involving Logarithm

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.

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]}]

• What is Subscript[y, 2]? Commented Aug 16, 2016 at 21:27
• 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. Commented Aug 16, 2016 at 21:33
• Not all integrals can be performed analytically Do you have reason to believe that this one can be? Commented Aug 17, 2016 at 0:53
• Yes, Mathematica calculates the indefinite integral. Commented Aug 17, 2016 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]} ]
(*0)


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

• Thank you very much for the answer. When you apply the residue theorem, do the contour integrals along semi circle tend to 0? Commented Aug 18, 2016 at 10:52
• @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. Commented Aug 18, 2016 at 11:36
• @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. Commented Aug 18, 2016 at 11:40
• You are definitely right, it needs a lot of work. Thanks for your help again. Commented Aug 18, 2016 at 15:18