Let $d\ge 1$ be an integer. Let $\vec{B}_1 :=\left( B_{1,j}\right)_{j=1}^d\in {\mathbb R}_+^d$ and let $\vec{B} := \left( B_j \right)_{j=1}^d\in (0,1)^d$. We consider the following integral: \begin{eqnarray} {\mathfrak I}^{\vec{B}_1}_{\vec{B}} := \frac{1}{2\pi} \int\limits_0^{2 \pi} \frac{e^{\sum\limits_{j=1}^d B_{1,j} \sin(j \tau)}}{1+\sum\limits_{j=1}^d B_j \sin(j \tau)} d\tau \end{eqnarray} We obtained the result for $d=1$ . It reads: \begin{equation} {\mathfrak I}^{B_1}_{B} = \exp\left(-\frac{B_1}{B}\right) \frac{1}{\sqrt{1-B^2}}-\sum\limits_{p=0}^\infty \frac{(-B_1)^p}{p!} \sum\limits_{l=0}^{\lfloor \frac{p+1}{2} \rfloor -1}(-1)^l \binom{-1/2}{l} B^{2 l-p} \end{equation} The result was obtained by expanding the exponential in the numerator in a series and then integrating term by term in the complex plane by using the residue theorem .
My question is can we get the result for $d>1$ ?