Can't find the mass scale; calculation using the modified minimal subtraction scheme and dimensional regularisation I am taking a course on quantum field theory where there is some confusion regarding the renormalisation scheme we are using (and a corresponding one in my mind). Apparently the lecturer meant MS-bar when he said MS. Anyway, on with the question.
Consider the* 1-loop contribution to the 1PI 4-point correlation function of phi-4 theory in $d=4-\epsilon$ dimensions, i.e the Lagrangian is $$L=-\frac{1}{2}(\partial \phi)^2 -\frac{1}{2}m^2 \phi^2-\frac{1}{4!} \mu^\epsilon \lambda \phi^4$$ where $\mu$ is a mass scale required to make $\lambda$ dimensionless as we vary $\epsilon$.
The Feynman rules yield an expression of the form $$\frac{(\mu^{ \epsilon}\lambda)^2}{2} \int \frac{d^d k}{(2 \pi)^d}\frac{1}{(k^2 +m^2)((k+p)^2+m^2)}$$where $p$ is a certain sum of external momenta, which eventually reduces to $$\frac{\lambda^2}{2 (4 \pi)^2} \int_0^1 d\alpha \ (m^2 + p^2 \alpha (1-\alpha))^{-\frac{\epsilon}{2}} \Gamma\left(\frac{\epsilon}{2}\right) \mu^{2 \epsilon} [4\pi]^{\frac{\epsilon}{2}}$$I can clearly pull one factor of $\mu^\epsilon$ into $ (m^2 + p^2 \alpha (1-\alpha))^{-\frac{\epsilon}{2}}$ to get $\left(\frac{m^2 + p^2 \alpha (1-\alpha)}{\mu^2}\right)^{-\frac{\epsilon}{2}}$ so that the final, finite result at 1 loop looks like $$ -\frac{\lambda^2}{2 (4 \pi)^2} \int_0^1 d\alpha \log\left(\frac{m^2 + p^2 \alpha (1-\alpha)}{\mu^2}\right) +\text{2 other terms*}$$
But where did the other factor of $\mu^\epsilon$ disappear to? Was it implicitly 'countertermed' away along with the factor $ [4\pi e^{-\gamma}]^{\frac{\epsilon}{2}}$? It looks as if this can be done at this order, but does this generalise to higher loop orders, i.e, is there a prescription telling us how many factors $\mu^\epsilon$ to counterterm away? 
Note that in phi-3 theory we get an identical loop integral by considering the 1-loop contribution to the 1PI 2-point correlation function, the difference being that $(\mu^{ \epsilon}\lambda)^2 \to (\mu^{ \frac{\epsilon}{2}} g)^2$ (where $g$ is the phi-3 coupling constant) in front of the integral. So that diagram does not have this problem somehow.
*There are two other contributions obtained by permuting the external momenta. These have identical integral expressions.
EDIT: I'll write this down fully. The term inside the integral can be written as $$(m^2 + p^2 \alpha (1-\alpha))^{-\frac{\epsilon}{2}} \Gamma\left(\frac{\epsilon}{2}\right) \mu^{2 \epsilon} [4\pi]^{\frac{\epsilon}{2}} =\left(\frac{m^2 + p^2 \alpha (1-\alpha)}{\mu^4}\right)^{-\frac{\epsilon}{2}}\Gamma\left(\frac{\epsilon}{2}\right) [4\pi]^{\frac{\epsilon}{2}}=\left(1-\frac{\epsilon}{2}\log\left(\frac{m^2 + p^2 \alpha (1-\alpha)}{\mu^4}\right)\right) \frac{2}{\epsilon} [4\pi e^{-\gamma}]^{\frac{\epsilon}{2}}$$
OR as $$=\left(1-\frac{\epsilon}{2}\log\left(\frac{m^2 + p^2 \alpha (1-\alpha)}{\mu^2}\right)\right) \frac{2}{\epsilon} [4\pi e^{-\gamma}\mu^2]^{\frac{\epsilon}{2}}$$ where the last two equalities are both asymptotic expansions in $\epsilon$ as $\epsilon \to 0$. I can absorb that last square bracket into the counterterm if I want to, at this order, and effectively ignore it.
 A: Your $\mu^\epsilon$ is still there, it's just that you have expanded in small $\epsilon$ so you got 
$$ \mu^\epsilon \approx 1 + \epsilon \log \mu  = 1 - \epsilon \log \frac{1}{\mu} $$
The $\log \frac{1}{\mu}$ is there in your expression
$$ -\frac{\lambda^2}{2 (4 \pi)^2} \int_0^1 d\alpha \log\left(\frac{m^2 + p^2 \alpha (1-\alpha)}{\mu^2}\right)$$
Now if we had instead 
$\mu^\epsilon \mu^\epsilon = \mu^{2\epsilon}$
we'd just get
$$ \mu^{2\epsilon} \approx 1 + 2\epsilon \log \mu  = 1 - 2 \epsilon \log \frac{1}{\mu} $$
I think what is confusing you is your are writing down a $d = 4+\epsilon$ dimension lagrangian with a $\mu^\epsilon \lambda$ interaction term, and then you want to get a loop correction that is for arbitrary dimensions too. That is fine, I suppose you can calullate that, (although I don't know how far you can take it without specifying the number of dimensions, or equivalently $\epsilon$). At any rate, the calculation you are performing isn't for arbitrary number of dimensions, it is for $d = 4$ since you are expanding in small $\epsilon$ so your corrections are coming out $\sim\lambda \mu^0 = \lambda$.
A: Finally got this cleared up: after a discussion that was somewhat longer than it really should be it was agreed that you pack the remaining $\mu^\epsilon$ into the counterterms like I describe above.
