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&gt;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}&lt;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&gt;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&gt;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