Stress tensor trace anomaly in two dimensions

Question

I'm trying to calculate the expectation value of the stress tensor in 2D following the book "Quantum fields in curved space" (Birrell and Davies). In 2D the divergent contribution to the one-loop effective action is \begin{equation} W_{div}=\frac{1}{2(4\pi)^{n/2}}\left(\frac{m}{\mu}\right)^{n-2}\Gamma\left(1-\frac{n}{2}\right)\int d^{n}x[-g(x)]^{1/2}\frac{R(x)}{6}. \end{equation} So, the contribution to the trace of the stress tensor should be \begin{align} \langle T_{\mu}^{\mu}\rangle_{div}&=-\frac{2}{\sqrt{-g}}g^{\mu\nu}\frac{\delta W_{div}}{\delta g^{\mu\nu}}=\notag\\ &=-\frac{2}{\sqrt{-g}}g^{\mu\nu}\frac{1}{2(4\pi)^{n/2}}\left(\frac{m}{\mu}\right)^{n-2}\Gamma\left(1-\frac{n}{2}\right)\frac{1}{6}\frac{\delta}{\delta g^{\mu\nu}}\int d^{n}x\sqrt{-g}R(x)=\notag\\ &=\frac{1}{2(4\pi)^{n/2}}\left(\frac{m}{\mu}\right)^{n-2}\Gamma\left(1-\frac{n}{2}\right)\frac{1}{6}g^{\mu\nu}\left(R_{\mu\nu}-\frac{1}{2}g_{\mu\nu}R\right)=\notag\\ &=\frac{1}{2(4\pi)^{n/2}}\left(\frac{m}{\mu}\right)^{n-2}\Gamma\left(1-\frac{n}{2}\right)\frac{1}{6}\left(R-\frac{1}{2}nR\right)=\notag\\ &=\frac{1}{2(4\pi)^{n/2}}\left(\frac{m}{\mu}\right)^{n-2}\frac{2}{(2-n)}\frac{(2-n)}{2}\frac{R}{6}. \end{align} Taking $n=2$ one has \begin{equation} \langle T_{\mu}^{\mu}\rangle_{div}=\frac{R}{48\pi}. \end{equation} But the correct result is \begin{equation} \langle T_{\mu}^{\mu}\rangle_{div}=\frac{R}{24\pi}. \end{equation} Can someone help me understand where the mistake is?