Integrating over a symmetric-group function (elements being permutations)

285 Views Asked by At

I would like to integrate a permutation of a function. Namely I have the following:

$\sum_{\sigma, \sigma'\in S_{n+1}}\int_{-A}^A dz_1dz_2 ... dz_{n+1} \left(z_{\sigma(n+1)}e^{i\alpha\sum_{j=1}^{n+1}y_j(z_{\sigma(j)} - z_{\sigma'(j)})}\right)$

where the $S_{n+1}$ is the permutation groups and $\sigma(i)$ is the $i^\text{th}$ value of each element of the group. For example, $\sigma(2)$ has as elements the following: {{1,2},{2,1}}, and $\sigma(3)$ has as its elements the following: {{1,2,3},{1,3,2},{2,1,3},{2,3,1},{3,1,2},{3,2,1}}. So the integral requires a sum over all permutations and combinations of $\sigma(i)$ and $\sigma'(i)$.

Now for n=1, we would have:

$\int_{-A}^A dz_1dz_2 \left(z_{\sigma(2)}e^{(i\alpha y_1(z_1 - z_1) + y_2(z_2 - z_2))} + z_{\sigma(2)}e^{(i\alpha y_1(z_1 - z_2) + y_2(z_2 - z_1))} + z_{\sigma(1)}e^{(i\alpha y_1(z_2 - z_1) + y_2(z_1 - z_2))} + z_{\sigma(1)}e^{(i\alpha y_1(z_2 - z_2) + y_2(z_1 - z_1))}\right)$

i.e. for a particular value of $n$ we always have $((n+1)!)^2$ terms. I am required to evaluate the above integral generally (so in terms of $n$). This integral results from a combinatorics problem - the number of different ways of pairing bosons. In the hope that this integral can be analysed analytically, I would also like to evaluate the following second integral:

$\sum_{\sigma, \sigma'\in S_{n+1}}\int_{-A}^A dz_1dz_2 ... dz_{n+1} \left(z_{\sigma(n+1)} z_{\sigma'(n+1)}e^{i\alpha\sum_{j=1}^{n+1}y_j(z_{\sigma(j)} - z_{\sigma'(j)})}\right)$

Any help with this would be appreciated. Thanks!