I'm looking to prove \begin{align} I(\mathbf{A},\mathbf{j}) &= \int d\mathbf{v}\exp\left(-\frac12\mathbf{v}^\top \mathbf{A} \mathbf{v} + \mathbf{j}^\top\mathbf{v}\right) \\ &=(2\pi)^{N/2}(\det\mathbf{A})^{-1/2}\exp\left(-\frac12\mathbf{j}^\top\bf{A}^{-1}\bf{j}\right) \end{align} where $\bf{A}$ is a real, symmetric, positive definite $N\times N$ matrix, $\bf{v}$ and $\bf{j}$ are $N$-dimensional real vectors, and $d\vec{v}=dv_1\dots dv_N$.
I already solved the problem, first handling the $\bf{j}=0$ case by diagonalizing $\bf{A}$, and then solving the general case via $\bf{u}$-substitution with $\bf{v}=\bf{u}+\bf{A}^{-1}\bf{j}$. But when I look at the original problem, my mind goes straight to $N=1$, and I think "complete the square."
That being said, I am having difficulty approaching this via index notation (Einstein summation convention)... How can I complete the square using index notation? Is it better to diagonalize before completing the square? Can I even complete the square!? I'm having a case of indexicitis here! (that last part was a joke)
\begin{align} -\frac12v_lA_{kl}v_k + j_mv_m &= -\frac12A_{kl}\left( v_lv_k + 2A_{kl}^{-1}j_mv_m\right) \end{align}
You do already know the answer, but the noncommutativity and the inverse components are probably what's tripping you up. If we write $B_{ij}=(A^{-1})_{ij}$, then $ A_{ij} B_{jk} = \delta_{ik} $ and $ B_{ij}A_{jk} = \delta_{ik} $, so $$ A_{ij}(v_i-B_{ik}j_k)(v_j - B_{jl}j_l) = A_{ij}v_iv_j - A_{ij} B_{ik}j_k v_j - A_{ij}v_iB_{jl}j_l + A_{ij} B_{ik}j_k B_{jl}j_l \\ = A_{ij}v_iv_j - \delta_{jk} j_k v_j - \delta_{il} v_i j_l + B_{ik} \delta_{il}j_kj_l \\ = v^T A v - 2 j^Tv + j^T A^{-1}j. $$ Remember that $(A^{-1})_{ij} \neq (A_{ij})^{-1}$, which is perhaps one of your problems: your notation should distinguish these. Also, $A_{kl}(A^{-1})_{kl} = \delta_{kk}=N$, so whatever you think your attempt means, it doesn't work. The key in the calculation is really that $A^{-1}AA^{-1}=A^{-1}$.
So $$ -\frac{1}{2}A_{ij}v_iv_j + j_iv_i = -\frac{1}{2}A_{ij}(v_i-B_{ik}j_k)(v_j - B_{jl}j_l) + \frac{1}{2} B_{ij}j_ij_j, $$ or $$ -\frac{1}{2}v^T A v + j^Tv = -\frac{1}{2}(v-A^{-1}j)^T A (v - A^{-1}j) + \frac{1}{2} j^T A^{-1} j $$