Constructing Ward identity associated with conserved currents Consider constructing the Ward identity associated with Lorentz invariance.  It is possible to find a 3rd rank tensor $B^{\rho \mu \nu}$ antisymmetric in the first two indices, then the stress-energy tensor can be made symmetric.  Once done, the conserved current coming from the classical analysis is of the form 
$$j^{\mu \nu \rho} = T_B^{\mu \nu}x^{\rho} - T_B^{\mu \rho}x^{\nu}$$  
This ensures the symmetry of the conserved current which can be seen most easily be invoking the conservation law $$\partial_{\mu}j^{\mu \nu \rho} = 0 $$ and $$\partial_{\mu}T_B^{\mu \nu} =\partial_{\mu} (T^{\mu \nu}_C + \partial_{\rho}B^{\rho \mu \nu}) =  0.$$  
Let $X$ denote a set of $n$ fields.  The Ward identity associated with Lorentz invariance is then 
$$\partial_{\mu} \langle (T^{\mu}x^{\rho} - T^{\mu \rho}x^{\nu})X\rangle = \sum_i \delta(x-x_i)\left[ x^{\nu}_i \partial^{\rho}_i - x^{\rho}_i\partial^{\nu}_i\langle X \rangle - iS^{\nu \rho}_i \langle X \rangle\right].\tag{1}$$ 
This is then equal to 
$$\langle (T^{\rho \nu} - T^{\nu \rho})X \rangle = -i\sum_i \delta (x-x_i)S^{\nu \rho}_i\langle X \rangle,$$ 
which states that the stress tensor is symmetric within correlation functions, except at the position of the other fields of the correlator.
My question is: how is this last equation and statement derived?
I think the Ward identity associated with translation invariance is used after perhaps splitting (1) up like so:
$$\sum_i^n x^{\nu}_i \sum_i^n \delta(x-x_i)\partial^{\rho}_i \langle X \rangle - \sum_i^n x^{\rho}_i \sum_i^n \delta(x-x_i)\partial^{\nu}_i \langle X \rangle - i\sum_i^n\delta(x-x_i)S^{\nu \rho}_i\langle X \rangle $$ and then replacing $$\partial_{\mu}\langle T^{\mu}_{\,\,\,\rho}X \rangle = -\sum_i \delta (x-x_i)\frac{\partial}{\partial x^{\rho}_i} \langle X \rangle$$ 
for example.  The result I am getting is that $$\langle ((\partial_{\mu}T^{\mu \nu})x^{\rho} - (\partial_{\mu}T^{\mu \rho})x^{\nu} + T^{\rho \nu} - T^{\nu \rho})X \rangle = \sum_i x^{\nu}_i \partial_{\mu}\langle T^{\mu \rho}X \rangle + \sum_i x^{\rho}_i \partial_{\mu} \langle T^{\mu \nu} X \rangle - i\sum_i\delta(x-x_i)S^{\nu \rho}_i\langle X \rangle$$  To obtain the required result, this means that  e.g$$ \sum_i x^{\nu}_i \partial_{\mu} \langle T^{\mu \rho}X \rangle = \langle(\partial_{\mu}T^{\mu \rho})x^{\nu} X \rangle,$$ but why is this the case? Regarding the statement at the end, do they mean that when the position in space $x$ happens to coincide with one of the points where the field $\Phi_i \in X$ takes on the value $x_i$ (so $x = x_i$) then the r.h.s tends to infinity and the equation is then nonsensical?
 A: This step is wrong:
$$
\langle ((\partial_{\mu}T^{\mu \nu})x^{\rho} - (\partial_{\mu}T^{\mu \rho})x^{\nu} + T^{\rho \nu} - T^{\nu \rho})X \rangle = \sum_i x^{\nu}_i \partial_{\mu}\langle T^{\mu \rho}X \rangle + \sum_i x^{\rho}_i \partial_{\mu} \langle T^{\mu \nu} X \rangle - i\sum_i\delta(x-x_i)S^{\nu \rho}_i\langle X \rangle
$$
because
$$
\sum_i x^{\nu}_i \partial_{\mu}\langle T^{\mu \rho}X \rangle\neq \sum_i \delta(x-x_i) x^{\nu}_i \partial_i^{\rho}\langle X \rangle ,
$$
and the spliting are wrong too
$$
\sum_i \delta(x-x_i) x^{\nu}_i \partial_i^{\rho}\langle X \rangle 
\neq\sum_i x^{\nu}_i\sum_i \delta(x-x_i) \partial_i^{\rho}\langle X \rangle .
$$
The right way to proceed here is to note that the distribution
$$
\partial_{\mu}\langle T^{\mu}_{\,\,\,\rho}X \rangle = -\sum_i \delta (x-x_i)\frac{\partial}{\partial x^{\rho}_i} \langle X \rangle
$$
acting on a test function $g(x)$, when composed with another function $f (x)$, gives:
$$
\int g (x)f (x)\partial_{\mu}\langle T^{\mu}_{\,\,\,\rho}X\rangle= - \int g (x)\sum_i f (x_i)\delta (x-x_i)\frac{\partial}{\partial x^{\rho}_i} \langle X \rangle
$$
in the sense of distribution, we have the equality:
$$
f (x)\partial_{\mu}\langle T^{\mu}_{\,\,\,\rho}X\rangle=-\sum_i f (x_i)\delta (x-x_i)\frac{\partial}{\partial x^{\rho}_i} \langle X \rangle
$$
In your case, the function $f(x)$ that composes with you distribution is $x^{\rho}$ and $x^{\nu}$. Note that $x^{\mu}$ is a c-number, and can jump out of the expectation value. Now, in eq. 1, the terms on the l.h.s. originated by the derivative acting only on $\langle T^{\mu\nu}X\rangle $ cancels the first two terms on the r.h.s. because they are the same distribution.
Now you have that $\langle T^{\mu\nu}(x) X\rangle$ is symmetric if $X$ don't have fields at $x$. This implies that as operator, $T^{\mu\nu}$ is symmetric. 
