1
$\begingroup$

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?

$\endgroup$
1
2
$\begingroup$

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    *)
$\endgroup$
7
  • $\begingroup$ 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? $\endgroup$ Dec 4 '20 at 18:10
  • $\begingroup$ @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. $\endgroup$
    – Roman
    Dec 4 '20 at 18:18
  • $\begingroup$ How can I get imaginary part of this expression? $\endgroup$ Dec 4 '20 at 18:40
  • $\begingroup$ 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. $\endgroup$
    – Roman
    Dec 4 '20 at 19:14
  • $\begingroup$ Yes I am. But to calculate the value of a definite integral I am using the symbolic expression, i.e. antiderivative. $\endgroup$ Dec 4 '20 at 20:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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