[Community Question] Calculus: integral of differences of vector

One of our user asked:

I have a vector function $f: \mathbb{R}^n \to \mathbb{R}^n$ defined with components $$ f_i(a) = \sum_{j=1}^n \sin(a_i - a_j) $$ which I want to integrate from ${\bf{\alpha}}^0$ to ${\bf{\alpha}}^1$ where ${\bf{\alpha}}^k = [\alpha^k_1, \ldots, \alpha^k_n]$ for $k \in \{1,2\}$. So the problem looks like $$ \int_^0}^^1} f(a)^{\top} {\rm d}\, a. $$ I thought that I could integrate as below $$ \int_{\alpha^0}^{\alpha^1} \sum_{i=1}^n \left\{ \sum_{j=1}^n \sin(a_i - a_j)\right\} {\rm d}a_i $$ by expanding the inner product in the integrand. I think that I can then write the integral as $$ \int_{(\alpha_1^0, \ldots, \alpha_n^0)}^{(\alpha_1^1, \ldots, \alpha_n^1)} \sum_{i=1}^n \left\{ \sum_{j=1}^n \sin(a_i - a_j)\right\} {\rm d}a_i = \sum_{i=1}^n \int_{\hat{\alpha}_i^0}^{\hat{\alpha}_i^1} \left\{ \sum_{j=1}^n \sin(a_i - a_j)\right\} {\rm d}a_i $$ where $\hat{\alpha}_i^0$ treats every component of $a$ as fixed $\alpha_j^0$ for $j \neq i$, which I think would give $$ \sum_{i=1}^n \sum_{j=1}^n \left[ \cos(\alpha_i^1 - \alpha_j^1) - \cos(\alpha_i^0 - \alpha_j^0) \right]. $$ Is this correct?