I have the following integral involving a matrix $\alpha$:
$$\int^{\infty}_{-\infty}\left[\prod_{k=1}^Ndx_k\right]\frac{\partial}{\partial x_n}\left[x_me^{-\frac{1}{2}\sum^N_{k,l=1}x_k\alpha_{kl}x_l}\right]$$
Now, I used the product rule. Apparently, the correct answer from the document I got this integral is the following:
$$\delta_{nm}\int^{\infty}_{-\infty}\left[\prod_{k=1}^Ndx_k\right]e^{-\frac{1}{2}\sum^N_{k,l=1}x_k\alpha_{kl}x_l}-\sum_{s=1}^N\int^{\infty}_{-\infty}\left[\prod_{k=1}^Ndx_k\right]x_m\alpha_{ns}x_se^{-\frac{1}{2}\sum^N_{k,l=1}x_k\alpha_{kl}x_l}$$
For the first term, I'm okay with; I calculated it correctly. For the second term, I don't understand why there's $x_m\alpha_{ns}x_s$ instead of something else. It appears that the derivative in that second term is only taken for when $k=l=n$, as it appears that it's finding the derivative of a constant multiplied by $x_k^2$.
So I would like to know how the derivative of the second term is done.
Let's use an uppercase $A$, instead of a greek $\alpha$ to represent the matrix. And for convenience, let's define the scalar variable $$\beta = -\frac{1}{2}x^TAx \,\,\implies\, \nabla\beta=-Ax$$ Also note that $$\eqalign{ \nabla e^\beta &= e^\beta\,\nabla\beta = -e^\beta\,Ax \cr\cr }$$ Working with the differential version of your problem $$\eqalign{ \nabla(e^\beta x^T) &= e^\beta\,\nabla x^T + (\nabla e^\beta)\,x^T \cr &= e^\beta\,I - e^\beta Axx^T \cr }$$ Or in component form $$\eqalign{ \delta_{nm}\,e^\beta - \sum_s A_{ns}x_sx_m \,e^\beta \cr }$$ In integral form, you can pull the $\delta_{nm}$ and the $\sum_s$ out of the integral to obtain the result in the document.
In fact, you can pull both matrices out of the integral, if you'd like: $$\eqalign{ I\,\Bigg[\int d\Omega\,e^\beta\Bigg] - A\,\Bigg[\int d\Omega\,e^\beta xx^T\Bigg] \cr\cr }$$