Integration with vector coefficients

Question

I asked this same question in Mathematics, and it was suggested I might try here. I'm more comfortable with Maple, but if I can get Mathematica to do what I'm after, so much the better.

Basically I'm trying to symbolically integrate something like this:

$\displaystyle\int \frac{a\mu-b}{||a\mu-b||^3} \mathrm{d}\mu$

where $a,b$ are vectors and $\mu$ is a scalar. The denominator is the cube of the 2-norm of the vector, and can be found by taking the dot product of a vector with itself, and raising it to the power of $\frac{3}{2}$.

Right now in Maple I'm explicitly multiplying out the denominator and making substitutions so that the denominator, at least, is only in terms of scalars ($a \cdot a = C$, etc. ), but I hate doing it this way, because it adds a lot of bookkeeping. Basically I'd like the computer to understand that $a * (b \cdot a)$ is not the same thing as $b * a^2$, but that $a \cdot b * c \cdot d = c \cdot d * a \cdot b$.

What's the most kosher way to do this integration in Mathematica?

UPDATE

This is the full integral I'm trying to do. I'm not sure it even has an answer, but the first integral is similar to what I have above. So I was hoping I could take any techniques that work on the simpler one above and apply them to the full problem below.

Let: $\vec{f} = (a - c) \mu_1 + (b - c) \upsilon_1 - (x - z) \mu_2 - (y-z) \upsilon_2 - (z - c) $

where $a, b, c, x, y, z$ are vectors representing positions, and $\mu_1, \nu_1, \mu_2, \nu_2$ are scalars.

I want to find:

$\vec{F_G} = \displaystyle\int_0^1 \int_0^{1-v_2} \int_0^{1} \int_0^{1-v_1} \! \frac{f}{||{f}||^3} \, \mathrm{d} \mu_1 \mathrm{d} \upsilon_1 \mathrm{d} \mu_2 \mathrm{d} \upsilon_2 $

Answer 1

It helps to do a little analysis to simplify the problem. This expression is integrating over a line through $\mathbf{b}$ in the direction of $\mathbf{a}$. By choosing a suitable coordinate system you can arrange for $\mathbf{a} = (x,0,0)$ where, to assure a unit Jacobian, $x = \|\mathbf{a}\|$ (and you can even make $\mathbf{b} = (0,b,0)$ if you like, but let's just stop here and generically take $\mathbf{b} = (a,b,c)$). Brute force now succeeds:

ClearAll[x, a, b, c];
Integrate[{x, 0, 0} μ / Norm[{x, 0, 0} μ - {a, b, c}]^3, 
  {μ, -Infinity, Infinity}, 
  Assumptions -> Im[a] == 0 && Im[b] == 0 && Im[c] == 0 && Im[x] == 0 && a b c != 0]

The output, after 5 seconds, is

{ConditionalExpression[(2 a Abs[x])/((b^2 + c^2) x^2), x != 0], 0, 0}

Change back to the original coordinates to obtain the general answer.

The key is to specify the assumptions implicit in the question: namely, that these are real vectors and that the line does not pass through the origin (where the integral diverges).

Answer 2

If I understood ...

av = Table[Subscript[a, i], {i, 3}];
bv = Table[Subscript[b, i], {i, 3}];
i = Integrate[(av mu - bv)/Dot[av mu - bv, av mu - bv]^(3/2), mu]
k = i /. {x_ __, x_ __, x_ __} -> x;

And then your integral is k * j

Where

$k =\frac{1}{\left(a_1^2 \left(b_2^2+b_3^2\right)-2 a_3 a_1 b_1 b_3+a_3^2 \left(b_1^2+b_2^2\right)-2 a_2 b_2 \left(a_1 b_1+a_3 b_3\right)+a_2^2 \left(b_1^2+b_3^2\right)\right) \sqrt{-2 a_1 b_1 \mu -2 a_2 b_2 \mu -2 a_3 b_3 \mu +a_1^2 \mu ^2+a_2^2 \mu ^2+a_3^2 \mu ^2+b_1^2+b_2^2+b_3^2}}$

and

$ j = \frac{i}{k}$