How to Simplify This Complex Integral Calculation?

[kassem84]

Hello,
I have some difficulties of calculating the following integral:
$I=\int _{D}\:\:\:d^{3}q\: d^{3}k\: d^{3}p\:\:F(q^{2}, q.k, q.p, k^{2}, p^{2})$
where:
D=|k|>1, |k+q|<1 and |p-q|<1

Thanks in advance.

 

[muppet]

What is the function F?

 

[kassem84]

What is the function F?

Hello,
F=e$^{-(q^{2}+q.k+q.p)}$
The most important thing is how to obtain the boundaries of the integrals. i.e. q,p,k go from where to where?
Thanks.

 

[jackmell]

I'm not sure I'm interpreting your notation correctly. Is that dot products in there? Also, the d^3 notation. Is that vectors? Tell you what, if it was just:

$$\int\int\int f(q,k,p) dqdkdp$$

then I think we can use Mathematica to obtain the boundaries.

 

[kassem84]

I'm not sure I'm interpreting your notation correctly. Is that dot products in there? Also, the d^3 notation. Is that vectors? Tell you what, if it was just:

$$\int\int\int f(q,k,p) dqdkdp$$

then I think we can use Mathematica to obtain the boundaries.

Yes, it is the correct notation of the integral I. All the vectors are 3-dimensional in the definition of the function and in the boundary D.
Thanks.

 

[jackmell]

The scalar version is quite interesting. There are two rhomboid regions to integrate over since |k|>1. I believe this is the integral for the region k>1:


$$\mathop\iiint\limits_{D} f(p,q,k)dpdqdk=\int_{1}^{\infty}\int_{-1-k}^{1-k}\int_{q-k}^{q+k} f(p,q,k)dpdqdk$$


Perhaps the vector version is similar and you can adapt it to this one.

 

[kassem84]

I'm not sure I'm interpreting your notation correctly. Is that dot products in there? Also, the d^3 notation. Is that vectors? Tell you what, if it was just:

$$\int\int\int f(q,k,p) dqdkdp$$

then I think we can use Mathematica to obtain the boundaries.

How can we use mathematica to determine the boundaries- the intersection of the three spheres?

 

[kassem84]

The scalar version is quite interesting. There are two rhomboid regions to integrate over since |k|>1. I believe this is the integral for the region k>1:


$$\mathop\iiint\limits_{D} f(p,q,k)dpdqdk=\int_{1}^{\infty}\int_{-1-k}^{1-k}\int_{q-k}^{q+k} f(p,q,k)dpdqdk$$


Perhaps the vector version is similar and you can adapt it to this one.

Thanks. For the vector version, it difficult for me to determine the boundaries on the angles θ and $\phi$.

 

[jackmell]

Could you or someone else tell me if I'm interpreting this correctly since I've never worked on one like this before. But first, let's just restrict it to a double integral for now:

$$\mathop\iint\limits_{D} f(k,q)d^3q\, d^3 k$$

where each integral is a triple integral in spherical coordinates and:
D={|k|>1, |k+q|<1}

We can compute the outer one easily. Since |k|>1, then for spherical coordinate r, we can write:

$$\mathop\int_{r>1}\left( \mathop\int\limits_{S} f(k,q) d^3 q\right)\,d^3 k$$

So what is S? Since |k+q|<1, then that means we need:

$$\sqrt{(k_x+q_z)^2+(k_y+q_y)^2+(k_z+q_z)^2}<1$$

for every point in k-space (k_x, k_y, k_z). Now suppose we have for a particular point:

$$k=(3,4,7)$$

Then for |k+q|<1, we would have to integrate in q-space over a sphere centered at q=(-3,-4,-7) with radius one. The boundary for that one k-point would be:

$$(q_x+3)^2+(q_y+4)^2+(q_z+7)^2=1$$

So for just that one k-point, the integral would be:

$$\mathop\iiint\limits_{(q_x+3)^2+(q_y+4)^2+(q_z+7)^2\leq 1} f(k,q)d^3q$$

and therefore for all of the k-space, we could then write:

$$\mathop\iiint\limits_{k_x^2+k_y^2+k_z^2>1}\iiint_{(q_x-k_x)^2+(q_y-k_y)^2+(q_z-k_z)^2\leq 1} f(k,q)d^3q d^3k$$

Ok, so just for now, can we let p be what ever it has to be to work, say p=(1,1,1) or whatever, can we now compute:

$$\mathop\iiint\limits_{k_x^2+k_y^2+k_z^2>1}\iiint_{(q_x-k_x)^2+(q_y-k_y)^2+(q_z-k_z)^2\leq 1} \exp\{-(q^2+q\cdot k+q\cdot(1,1,1))\}d^3q d^3k$$