# Multiple integral involving trigonometric functions over a hypercube.

Let $d\ge 1$ be an integer and $\vec{A} := \left(A_j\right)_{j=1}^d \in {\mathbb R}^d$ subject to $\sum\limits_{j=1}^d A_j^2 \le 1$. Define a following integral: $${\mathfrak I}^{(d)}(\vec{A}) := \frac{1}{(2 \pi)^d} \cdot \int_{[0,2\pi]^d} \frac{\prod\limits_{j=1}^d d\phi_j}{\left(1+\sum\limits_{j=1}^d A_j \cos(\phi_j)\right)} = \int\limits_0^\infty \exp(-s) \left(\prod\limits_{j=1}^d I_0(s A_j)\right) ds$$ Now using the Cauchy' residue theorem and then the definition of elliptic integrals we found the following: \begin{eqnarray} {\mathfrak J}^{(1)}(\vec{A})&=& \frac{1}{\sqrt{1-A_1^2}}\\ {\mathfrak J}^{(2)}(\vec{A})&=& \frac{2}{\pi} \frac{1}{\sqrt{1-(A_1-A_2)^2}} \cdot K\left( \frac{4 A_1 A_2}{1-(A_1-A_2)^2}\right)\\ {\mathfrak J}^{(3)}(\vec{A})&=& \frac{2}{\pi^2} \cdot \\ &&\int\limits_{-\infty}^\infty \frac{K(\frac{4 A_2 A_3 (1+t^2)^2}{(1+A_1+A_2-A_3+(1-A_1+A_2-A_3) t^2)((1+A_1-A_2+A_3+(1-A_1-A_2+A_3) t^2)})} {\sqrt{(1+A_1+A_2-A_3+(1-A_1+A_2-A_3) t^2)((1+A_1-A_2+A_3+(1-A_1-A_2+A_3) t^2)}} dt \end{eqnarray} Here $K()$ is the complete elliptic integral of the first kind. Now my question is is it possible to find a closed form expression for the quantities in question when $d\ge 3$?

• Are you sure the listed condition $\sum_{j=1}^{d}A_{j}^{2}\le1$ shouldn't actually be $\sum_{j=1}^{d}|A_{j}|<1$? I'm fairly sure this alternative condition is a necessary and sufficient one for your integral to converge. – David H Jul 28 '18 at 9:38
• Alright, thank you for this comment.I'd be grateful if you could tell me where your condition comes from. This would give me more insight into this integral. – Przemo Jul 30 '18 at 11:38