Path Integral For Free Field In the book "Quantum field Theory" by Mark Srednicki, in chapter 8, in order to evaluate the path integral, with a source, he introduces the change path integration variables to
$$ \tilde{\chi}(k) = \tilde{\psi}(k)-\frac{\tilde{J}(k)}{k^{2}+m^{2}}\tag{8.8}$$
and states that this is a merely shift by a constant. I do not understand why the second term is a constant. Any ideas?
 A: The path integral is a summation of values of functionals over contributions from values of functionals, which are defined over a small interval of field configurations (i.e. the interval $\mathcal{D}\tilde{\psi}=\prod_id\tilde\psi_i$), in the very same way the "normal" integral is a summation of values of functions $f(x)$ over an infinitesimal interval $dx$ for example. Shifting the field configuration $\tilde{\psi}$ corresponds to making the substitution $$\tilde{\chi}=\tilde{\psi}+\frac{\tilde{J}}{k^2+m^2}$$
The second term of the previously mentioned substitution does not depend on is a function that does not depend on the function $\tilde{\psi}$, in the very same way, as it was aptly noted in the comments, that a quantity $a$ satisfying the substitution $y=x+a$ is constant with respect to $x$... The correspondence is one to one here: $y\leftrightarrow\tilde{\chi},\ x\leftrightarrow\tilde{\psi}$ and $a\leftrightarrow\frac{\tilde{J}}{k^2+m^2}$. The only thing that you need to keep in mind is that, being already functions, instead of variables, the newly introduced function $\tilde{\chi}$ must be a substitution that shifts $\tilde{\psi}$ by a constant (i.e. not dependent on the FUNCTION $\tilde{\psi}$ and not some other variable, such as $k$ for example).
I do not wish to steal the glory of the comments so far, as they are correct, but I felt that I could elaborate a bit more. I hope this helps
