Trasformig a sum in an integral
I posted this in the mathematical forums. Maybe you will help me. I found an hard article http://prola.aps.org/abstract/PR/v105/i3/p776_1 of <PERSON> and <PERSON>. 346
The authors begins with the sum: $$\frac{{E_2 }}{{E_0 }} = \frac{{16\pi ^2 a^2 \lambda ^2 }}{{V^2 }}\sum\limits_{} {'\frac{{\langle n_\alpha \rangle \langle n_\gamma \rangle \langle n_\lambda \rangle }}{{\frac{1}{2}\left( {k_\alpha ^2 + k_\beta ^2 - k_\gamma ^2 - k_\lambda ^2 } \right)}}\delta \left( {\vec k{}\alpha + \vec k{}\beta - \vec k{}\gamma - \vec k{}\lambda } \right)} $$ which represents the second order perturbation term of the energy of a bose gas. ${\langle n_\alpha \rangle }$ is: $$\langle n_\alpha \rangle = \sum\limits_{n = 0}^\infty {n\left( {ze^{ - \beta \varepsilon \alpha } } \right)} ^n /\sum\limits{n = 0}^\infty {\left( {ze^{ - \beta \varepsilon \alpha } } \right)} ^n = \frac{{ze^{ - \beta \varepsilon \alpha } }}{{1 - ze^{ - \beta \varepsilon \alpha } }} $$ In the sum the terms with a vanishing denominator are omitted. There is also the restriction ${\vec k{}\alpha \ne \vec k{}\beta }\ \ $ and ${\vec k{}\gamma \ne \vec k{}\lambda }\ \ $. Now the passage that is unclear to me is the passage to the integral: $$\frac{{E_2 }}{{E_0 }} = \frac{{16\pi a^2 \lambda ^2 }}{{V^2 }}\left( {\frac{{4V^3 }}{{\pi ^3 \lambda ^5 }}} \right)\sum\limits{i,j,k}^\infty {\frac{{z^{j + k + l} }}{{\left( {j + k + l} \right)^{1/2} \left( {j - k} \right)l}}} \frac{{\partial J}}{{\partial u}} $$ where: $$J = \int\limits_0^\infty {\int\limits_0^\infty {dqdq\frac{{\cosh \left( {upq} \right)}}{{q^2 - p^2 }}} } e^{ - vq^2 - wp^2 } $$ and: $$u = \frac{{\hbar ^2 \beta }}{{2m}}\left( {\frac{{2\left( {j - k} \right)l}}{{j + k + l}}} \right) $$ $$v = \frac{{\hbar ^2 \beta }}{{2m}}\left( {\frac{{\left( {j + k} \right)l}}{{j + k + l}}} \right) $$ $$w = \frac{{\hbar ^2 \beta }}{{2m}}\left( {\frac{{\left( {j + k} \right)l + 4jk}}{{j + k + l}}} \right) $$ $\frac{{\partial J}}{{\partial u}}$ is calculated here: http://math.stackexchange.com/questions/63534/difficult-integral Maybe we don't have to do thermal average of n and we need : $$n = \frac{1}{{z^{ - 1} e^{\varepsilon /kT} - 1}} = \sum\limits_{n = 1}^\infty {\left( {ze^{ - \varepsilon /kT} } \right)^l } $$ so: $$\sum\limits_{} {'... 592