Wu Xiaobin said:
Thank you so much for all friends who replied.
I think I have worked it out, but maybe using another way to solve it.
However partial fractions really help when I solve this problem.
I needs great effort to type the integration process here, so I decide to attach the pdf
document here.
The attachment shows the whole process of solving the integral.
In the end, I want to give me special gratitude to haruspex and Ray Vickson.
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Best regards
Jacky Wu
Your final answer is incorrect, as I have already indicated; you made some errors in your computation.
However, it is possible to give a corrected answer, and it is not very difficult to get. Below, I use ##x_1, x_2, x_3## instead of ##\xi_1, \xi_2, \xi_3## and I use ##c_1, c_2, c_3## instead of ##\eta_1, \eta_2, \eta_3##. I will get the marginal density of ##x_3## instead of ##x_1##, but by a simple re-labelling you can get the other two marginals as well. The density function is
P(x_1,x_2,x_3) = k \frac{\delta(1 - \sqrt{x_1^2 + x_2^2 + x_3^2})}{( 1 - c_1 x_1 <br />
- c_2 x_2 - c_3 x_3)^2},
where k is a constant which you give in the pdf file. Let's change to polar coordinates:
x_1 = r \sin(\theta) \cos(\phi)\\<br />
x_2 = r \sin(\theta) \sin(\phi)\\<br />
x_3 = r \cos(\theta).
When we fix x_3 the area element is ##dA = dx_1 dx_2 = R\, dR\, d\phi##, where ##R = r \sin(\theta)##, so ##dR = dr \sin(\theta)## (since, ultimately, we are fixing θ). Thus, ##dA = r \sin^2(\theta) dr \, d\phi##, so the marginal density ##f(x_3)## is given as
f = k \sin^2(\theta) \int_0^{2 \pi} <br />
\frac{1}{(1 - c_1 \sin(\theta) \cos(\phi) - c_2 \sin(\theta) \sin(\phi) - c_3 \cos(\theta))^2} \, d \phi,
because the r-integration is simply ##\int \delta(r-1) q(r,\phi)\, dr##. Note that since r = 1 we have ##\cos(\theta) = x_3## and ##\sin(\theta) = \sqrt{1-x_3^2}.## We may write
c_1 \sin(\theta) \cos(\phi) + c_2 \sin(\theta) \sin(\phi) = C \cos(\omega - \phi), where ##C = \sin(\theta) \sqrt{c_1^2 + c_2^2}## and ##\cos(\omega) = c_1 \sin(\theta)/C, \: \sin(\omega) = \sin(\theta) c_2/C.## Since we are integrating ##\phi## over an entire cycle, we can omit ##\omega## and write
f = k (1-x_3^2) \int_0^{2 \pi} \frac{1}{(1 - C \cos(\phi) -c_3 x_3)^2}\, d\phi<br />
= k (1-x_3^2) \text{sign}(x_3 - 1 -C)\frac{2(c_3 x_3 - 1)}{( 1 - 2 c_3 x_3 + c_3^2 x_3^2 - c_1^2 - c_2^2 - C^2)^{3/2}}.
Since ##x_3 \in (-1,1)## we have sign = -1, so finally we have
f(x_3) = \text{const.} \frac{(1 - c_3 x_3)(1 - x_3^2)}{( (1 - c_3 x_3)^2 - (c_1^2 + c_2^2)(1 - x_3^2))^{3/2}}, \: -1 \leq x_3 \leq 1.
Here, we assume that ##c_1^2 + c_2^2 + c_3^2 < 1## and also that the denominator function ## (1 - c_3 x_3)^2 - (c_1^2 + c_2^2)(1 - x_3^2)## does not vanish for ##x_3 \in [-1,1].## A sufficient condition for this is that
1 - c_1^2 - c_2^2 - c_3^2 > 2 c_3. When the denominator function vanishes at some point in (-1,1) the integral is singular at some values of ##x_3##, and the computation breaks down.
Note that this formula gives ##f(x_3) \to 0## as ## x_3 \to \pm 1##, as it must be (since having x_3 = ± 1 gives an integration over a region of area = 0). The formula you gave in post #9 does not have this necessary property.