I'm working in the $\varphi^4$ QFT where
$$S(\varphi)=\frac{1}{2} \mu \varphi^{2}+\frac{1}{4 !} \lambda_{4} \varphi^{4}$$
and the text says that we can expand (assuming small $\lambda_4$) as
$$\exp (-S(\varphi))=\exp \left(-\frac{1}{2} \mu \varphi^{2}\right) \sum_{k \geq 0} \frac{1}{k !}\left(-\frac{\lambda_{4}}{24}\right)^{k} \varphi^{4 k}$$
which makes sense, but then it says that we interchange the series expansion in $\lambda_4$ with an integration over $\varphi$ and arrive at the following Green's functions
$$G_{2 n}=H_{2 n} / H_{0}$$
where
$$H_{2 n}=\frac{1}{\mu^{n}} \sum_{k \geq 0} \frac{(4 k+2 n) !}{2^{2 k+n}(2 k+n) ! k !}\left(-\frac{\lambda_{4}}{24 \mu^{2}}\right)^{k}$$
and for the life of me I can't figure out how they are getting to this.
I understand how it worked for the free theory, where
$$\begin{aligned} Z(J) &=N \int \exp \left(-\frac{1}{2} \mu \varphi^{2}+J \varphi\right) d \varphi \\ &=N \int \exp \left(-\frac{1}{2} \mu\left(\varphi-\frac{J}{\mu}\right)^{2}+\frac{J^{2}}{2 \mu}\right) d \varphi=\exp \left(\frac{J^{2}}{2 \mu}\right) \end{aligned}$$ and we can use the taylor expansion of $\exp \left(\frac{J^{2}}{2 \mu}\right)$ to get $$G_{2 n}=\frac{(2 n) !}{2^{n} n !} \frac{1}{\mu^{n}}$$
But I'm miffed how I can't get to the result stated for the $\varphi^4$ theory. If I were to follow the logic directly from the free theory case, then I wouldn't even taylor expand to get $$\exp (-S(\varphi))=\exp \left(-\frac{1}{2} \mu \varphi^{2}\right) \sum_{k \geq 0} \frac{1}{k !}\left(-\frac{\lambda_{4}}{24}\right)^{k} \varphi^{4 k}$$ but instead complete the square on this expression.
Any insights would be greatly appreciated!