The free propagator for a massive $m\neq0$ real scalar field is the following: $$ G_{0}(x,y) \ = \ \int \frac{d^{4}p}{(2\pi)^4} \frac{e^{i p \cdot (x-y)}}{p^2 +m^2 - i \epsilon} $$
It is a well-known that integrating the above yields the following function involving the modified Bessel function of the second kind: $$ G_{0}(x,y) \ = \ - \frac{i}{4\pi^2} \frac{m}{\sqrt{ - t^2 + |\mathbf{r}|^2 }} K_{1}\left( m \sqrt{ - t^2 + |\mathbf{r}|^2} \right) - \frac{1}{4\pi} \delta\left( - t^2 + |\mathbf{r}|^2 \right) $$
Where $t := x^0 - y^0$ and $\mathbf{r} := \mathbf{x} - \mathbf{y}$.
Consider the following two-loop correction to the propagator in a $\phi^{4}$-interacting theory:
The above "double-tadpole" diagram results in 2 loop integrals and a Fourier transform over 3 propagators. Using the fact that: $$ \frac{1}{2} \frac{\partial^2}{\partial (m^2)^2} \left\{ \frac{1}{p^2 +m^2 - i \epsilon} \right\} \ = \ \frac{1}{\left(p^2 +m^2 - i \epsilon\right)^3} $$
we find that the above diagram is proportional (up to some $\mathbb{C}-$numbers) to the following function: $$ \propto \lambda^2 \mathcal{T}^2\frac{\partial^2 G_0(x,y)}{\partial (m^2)^2} \propto \lambda^2 \mathcal{T}^2 \frac{\sqrt{ - t^2 + |\mathbf{r}|^2}\ K_1\left( m \sqrt{ - t^2 + |\mathbf{r}|^2} \right)}{m} $$
Where $\mathcal{T}$ is the following tadpole loop (and $\Lambda \gg m$ is a UV cutoff): $$ \mathcal{T} \ = \ \int_{\Lambda} \frac{d^{4}k}{(2\pi)^4}\frac{1}{k^2+m^2-i\epsilon} \ = \ \frac{i}{16 \pi^2} \left[ \Lambda^2 - m^2 \log\left( \frac{\Lambda^2}{m^2} + 1 \right) \right] $$
This double tadpole function is alarming to me because of the following: when I set the spatial separation $\mathbf{r}=0$, and then consider the asymptotics as $t\to\infty$, the function looks like: $$ \sim \ \lambda^2 \mathcal{T}^2\ \frac{\sqrt{t}\ e^{-imt}}{m^{3/2}} $$
So this function grows as we take time $t\to \infty$. This seems to indicate a secular breakdown of our perturbative expansion in $\lambda$ - the idea being that no matter how tiny we make our $\lambda$-coupling, if we simply wait long enough, this term in the series will blow up ruining our series.
In general, for $N$-tadpole insertions:
I've found that the above treatment yields (for $\mathbf{r}=0$ and $t\to\infty$) the following asymptotics:
$$
\sim \lambda^N \mathcal{T}^N\ \frac{t^{N-\frac{3}{2}}}{m^{N-\frac{1}{2}}} e^{-im t}
$$
These asymptotics hold for all $N \geq 0$ (where $N=0$ is just the free propagator).
Why do these terms not result in a secularly-growing perturbative series? I have noticed that the terms that have this "time divergence" start at $N=2$ tadpoles, which of course, are the non-1PI graphs, so I have a feeling the answer has something to do with this and for some reason we need to ignore these diagrams perhaps.
