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I have a function of six variables in total - x1, x2, x3, x11, x22, x33 and I need to perform symbolic integration to get a symbolic expression.

Before showing my function, I want to show from where does the function originates.

The 3D function f[x1_, x2_, x3_] := -(1/(8 \[Pi])) Sqrt[x1^2 + x2^2 + x3^2] is a fundamental solution of biharmonic equation: https://en.wikipedia.org/wiki/Biharmonic_equation

∂^4f/∂x1^4 + ∂^4f/∂x2^4 + \
∂^4f/∂x3^4 + 
2 ∂^4f/(∂x1^2 ∂x2^2) + 
2 ∂^4f/(∂x1^2 ∂x3^2) + 
2 ∂^4f/(∂x2^4 ∂x3^2) == 0

I have written a partial differential equation in Wolfram Mathematica just for clarification. My three variables are x1, x2, x3.

Now, if I differentiate my function f four times with respect to variable x1 I will get

difFunction = FullSimplify@D[f[x1, x2, x3], {x1, 4}]

which is a little bit more complicated expression.

Finally, the function I need to integrate symbolically is obtained by a change of variables in difFunction from x1 to x1 - x11, from x2 to x2 - x22 and from x3 to x3 - x33.

I end up with the function g of six variables in total x1, x2, x3, x11, x22, x33.

g[x1_, x2_, x3_, x11_, x22_, x33_] := -((
3 (4 (x1 - x11)^2 - (x2 - x22)^2 - (x3 - x33)^2) ((x2 - 
x22)^2 + (x3 - x33)^2))/(
8 \[Pi] ((x1 - x11)^2 + (x2 - x22)^2 + (x3 - x33)^2)^3.5))

I tried integrating this function with Wolfram Mathematica 7

Integrate[g[x1, x2, x3, x11, x22, x33], x1, x2, x3, x11, x22, x33]

but I Wolfram Mathematica wasn't able to evaluate this integral.

Is there a way to simplify function g[x1_, x2_, x3_, x11_, x22_, x33_] so that I can evaluate this integral?

Or is there some other method which I can use to get the symbolic expression?

Avrana
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Cro Simpson2.0
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1 Answers1

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It integrates if (1) we replace your exponent $3.5$ by the exact $7/2$, and (2) we change the order of integration:

g[x1_, x2_, x3_, x11_, x22_, x33_] =
   -((3(4(x1-x11)^2-(x2-x22)^2-(x3-x33)^2)((x2-x22)^2+(x3-x33)^2))/
    (8π((x1-x11)^2+(x2-x22)^2+(x3-x33)^2)^(7/2)));

Integrate[g[x1, x2, x3, x11, x22, x33], x3, x33, x2, x22, x1, x11]
(*    lengthy output    *)
Roman
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  • Dear Roman! Thank you for your answer! Is there a way to get rid of a complex part of a expression? Or to get complex and real part of this expression? – Cro Simpson2.0 Dec 04 '20 at 18:10
  • @CroSimpson2.0 indefinite integrals (antiderivatives) include an arbitrary offset constant, which may be complex-valued. If you don't like the offset constant, you can modify it as you wish. For example, evaluate the imaginary part and subtract it. Alternatively, if you compute a definite integral you won't get any imaginary parts. – Roman Dec 04 '20 at 18:18
  • How can I get imaginary part of this expression? – Cro Simpson2.0 Dec 04 '20 at 18:40
  • Aren't you rather looking for a definite integral at the end? If you are, then that's going to be much easier than figuring out the story of the six integration constants. – Roman Dec 04 '20 at 19:14
  • Yes I am. But to calculate the value of a definite integral I am using the symbolic expression, i.e. antiderivative. – Cro Simpson2.0 Dec 04 '20 at 20:45
  • Have you tried simply specifying the integration limits in Integrate instead of using the antiderivative? Multi-dimensional antiderivatives aren't quite as simple as one-dimensional ones. – Roman Dec 05 '20 at 10:19
  • I will try that as well. Thx for the advice. – Cro Simpson2.0 Dec 05 '20 at 11:12