Let $G(t,x)$ be the heat kernel
$$
G(t,x)=\frac{1}{\sqrt{2\pi t}}e^{-\frac{x^2}{2t}}, \quad t>0, \:x\in\mathbb{R}.
$$


Here is one approximation to $G(t,x)$:


$$
G\_\epsilon(t,x)=e^{-t/\epsilon} \sum\_{k=1}^\infty \left(\frac{t}{\epsilon}\right)^k \frac{1}{k!} G(k\epsilon,x).
$$


The question is: can one show that for some $a>0$ and $C>0$


$$
\int\_{\mathbb{R}}\left|G(t,x)-G\_\epsilon(t,x)\right| d x \le e^{-t/\epsilon}+C \left(\frac{\epsilon}{t}\right)^{1/3},\quad \text{for $0<\epsilon/t\le a$}?
$$


Thanks a lot for any hints!


-----EDIT------


Thanks Professor Lucia for his nice solution. It turns out that it is not sufficient for what we actually need. Here is the revised question:


Fix $\epsilon>0$. Is there a constant $C>0$ and $0<\beta<1/2$, such that


$$
\int\_{\mathbb{R}}\left|G(t,x)-G\_\epsilon(t,x)\right| d x \le e^{-t/\epsilon}+C \left(\frac{\epsilon}{t}\right)^{\beta},\quad \text{for all $t>0$}?
$$


The exponent $1/3$ is mysterious. But any order $\beta<1/2$ will be fine. Or probably this will never happen?



About the convergence rate for an approximation to the heat kernel
One can prove a stronger estimate in fact. Suppose $u$ and $v$ are positive with $v>u$ say. Note that 
$$ 
|G(u,x)-G(v,x)| \le \int\_{u}^v \Big| \frac{d}{dt} G(t,x)\Big| dt =\int\_{u}^{v} 
\frac{e^{-x^2/2t}}{\sqrt{2\pi t}} \Big|\frac{x^2}{2t^2}-\frac{1}{2t}\Big| dt.
$$
Integrating this over $x\in {\Bbb R}$ we obtain 
$$ 
\int\_{-\infty}^{\infty} |G(u,x)-G(v,x)| dx \le \int\_u^v \frac{1}{2t} \Big(\frac{1}{\sqrt{2\pi t}} \int\_{-\infty}^{\infty} e^{-x^2/(2t)} \Big(\frac{x^2}{t}+1\Big) dx \Big) dt \le C \int\_{u}^{v} \frac{dt}{t} = C \log \Big(\frac{v}{u}\Big),
$$ 
for some constant $C$. 


Now we use this in the problem; assume throughout that $t/\epsilon$ is bounded away from zero, say it is at least $1$. We have (the first term accounts for the missing $k=0$ term)
$$ 
|G(t,x) - G\_{\epsilon}(t,x)| \le e^{-t/\epsilon} G(t,x) + e^{-t/\epsilon}\sum\_{k=1}^{\infty} \Big(\frac{t}{\epsilon}\Big)^k \frac{1}{k!} |G(t,x)-G(k\epsilon, x)|.
$$
Integrating both sides over $x \in {\Bbb R}$ and using our first estimate we get 
$$ 
\int\_{{\Bbb R}} |G(t,x)-G\_{\epsilon}(t,x)| dx \le e^{-t/\epsilon} + Ce^{-t/\epsilon}\sum\_{k=1}^{\infty} 
\Big(\frac{t}{\epsilon}\Big)^k \frac{1}{k!} |\log (k\epsilon/t)|. \tag{1}
$$


It remains lastly to estimate the sum over $k$ above. Note that $|\log (v/u)| 
\le |u-v|/(\min(u,v)) \le |u-v|(1/u+1/v)$. So the sum over $k$ is 
$$
\le \sum\_{k=1}^{\infty} \Big(\frac{t}{\epsilon}\Big)^k \frac{1}{k!} |k - t/\epsilon|\Big( \frac{1}{k} +\frac{\epsilon}{t}\Big). 
$$
Using Cauchy-Schwarz 
$$
\sum\_{k=1}^{\infty} \Big(\frac{t}{\epsilon}\Big)^k \frac{1}{k!} \frac{1}{k}|k-t/\epsilon|
\le \Big(\sum\_{k=1}^{\infty} \Big(\frac{t}{\epsilon}\Big)^{k} \frac{1}{k!} \frac{1}{k^2}\Big)^{\frac 12} \Big( \sum\_{k=1}^{\infty} \Big(\frac{t}{\epsilon}\Big)^k \frac{1}{k!} (k-t/\epsilon)^2\Big)^{\frac 12},
$$  

and it is easy to see that the first factor above is $O((\epsilon/t)e^{t/(2\epsilon)})$ and the second factor is $O((t/\epsilon)^{\frac 12} e^{t/(2\epsilon)})$ so that our quantity above is $O((\epsilon/t)^{\frac 12} e^{t/\epsilon})$. Similarly 
$$ 
\sum\_{k=1}^{\infty} \Big(\frac{t}{\epsilon}\Big)^k \frac{1}{k!} \frac{\epsilon}{t} |k-t/\epsilon| = O\Big( \Big(\frac{\epsilon}{t}\Big)^{\frac 12} e^{t/\epsilon}\Big).
$$
Using these estimates in (1) we get 
$$ 
|G(t,x)-G\_{\epsilon}(t,x)| \le e^{-t/\epsilon} + C\_1 \Big(\frac{\epsilon}{t}\Big)^{\frac 12},
$$
which is stronger than the bound you wanted. 


