$$x \in \mathbb{R^n},$$ M is a positive symmetrical nonsingular nxn Matrix and j is an arbitrary vector in $$\mathbb{R}^n.$$ The following has to be calculated:
$$Z(j) = \int_\mathbb{R^n} e^{-x^TMx+2j^Tx} d^{n}x.$$
Further, calculate
$\dfrac{\partial {\log(Z)}}{\partial {M}} $ at $j=0.$
I m not really sure how I should approach the problem, however, it should somehow be solved with the help of completing the square, but am not sure, how exactly. Any help would be appreciated.
To answer your question about completing the full square. First, note that you can write $j = M y$, where $y := M^{-1} j$. Then, you get: $$ \exp\left\{-x^\top M x + 2 j^\top x\right\} = \exp\left\{-\left(x^\top M x - 2 y^\top M x + y^\top M y\right) + y^\top M y\right\} = \exp\left\{-(x - y)^\top M (x - y) + y^\top M y\right\} $$
The expression for $Z(j)$ will turn into $$ Z(j) = e^{y^\top M y} \int_{\mathbb{R}^n} e^{-\frac{1}{2}(x - y)^\top (2M) (x - y)} d^n x $$ Note that now the integral resembles PDF of a multivariate normal distribution with precision matrix $2M$. The only difference is that there is no normalizing constant. Therefore, integration will give you precisely the following: $$ \int_{\mathbb{R}^n} e^{-\frac{1}{2}(x - y)^\top (2M) (x - y)} d^n x = \frac{(2\pi)^{n/2}}{\det (2M)^{1/2}} = \sqrt{\frac{\pi^n}{\det M}} $$ We can plug in $y = M^{-1}j$ and get the final expression for $Z(j)$: $$ Z(j) = \sqrt{\frac{\pi^n}{\det M}} \exp\left\{j^\top M^{-1} j\right\} $$
To answer the second part of the question, you need to compute the gradient of the scalar function $Z$ with a matrix argument. Here is how you can do that. $$ \frac{\partial}{\partial M}\ln Z = -\frac{1}{2} \frac{\partial}{\partial M}\ln \det M + \frac{\partial}{\partial M} j^T M^{-1} j. $$ For these two derivatives, you can look the standard expressions in the Matrix Cookbook or consult the first section of a nice tutorial on matrix algebra. You get the following: $$ \frac{\partial}{\partial M} \ln \det M = M^{-1} $$ and $$ \frac{\partial}{\partial M} j^T M^{-1} j = \frac{\partial}{\partial M} \mathrm{Tr}\left(j^T M^{-1} j\right) = \frac{\partial}{\partial M} \mathrm{Tr}\left(j j^T M^{-1}\right) = - M^{-1} j j^T M^{-1} $$ Finally: $$ \frac{\partial}{\partial M}\ln Z = -\frac{1}{2} M^{-1} - M^{-1} j j^T M^{-1} $$ and you have $\partial\ln Z / \partial M = -M^{-1}/2$ when $j = 0$.