Let $f(x|\mu,\sigma^2)$ be the gaussian function (normal distribution):
$$f(x|\mu,\sigma^2)=\frac{1}{\sigma\sqrt{2\pi}}e^{ -\frac{(x-\mu)^2}{2\sigma^2} }$$
We know its integral over $\mathbb{R}$ is 1.
Now we divide the interval $[0,1]$ into $n$ subintervals, each with an equal length of $1/n$. For each subinterval $[\frac{i-1}{n},\frac{i}{n}], (i=1,2,\cdots,n)$, there is a function $g_i(x)$:
$$g_i(x)=\frac{1}{n}f(x|\frac{i-0.5}{n},\sigma^2)$$
Namely, a rescaled gaussian function whose mean is the center of the subinterval and integral over $\mathbb{R}$ is $1/n$.
Add them together:
$$G(x) = \sum_{i=1}^n g_i(x)$$
Problem: What's the expression of $G(x)$ when $n\rightarrow \infty$ ?
I simulated the result in Matlab by setting $n=10, \sigma=0.08$: ($g_i(x)$ in blue and $G(x)$ in red)
By the way, I plan to use $G(x)$ in image processing, so an analytical expression that can be computed directly may be more useful than a mathematical strict yet confusing solution (such as a series). Common special functions like Bessel function are also OK. And necessary approximation will also be acceptable.
Thank you in advance.
The function $G(x)$ has a simple expression in terms of erf function. $$ G(x) =\lim_{n\to \infty} \sum_{i=1}^n g_i(x)=\frac{1}{\sigma\sqrt{2\pi}}\lim_{n\to \infty} \frac{1}{n}\sum_{i=1}^n\exp \left(-\frac{(x-i/n)^2}{2 \sigma^2}\right)\\ =\frac{1}{\sigma\sqrt{2\pi}}\int_0^1 \exp \left(-\frac{(x-y)^2}{2 \sigma^2}\right) dy=\frac{1}{2} \left(\text{erf}\left(\frac{x}{\sqrt{2} \sigma }\right)-\text{erf}\left(\frac{x-1}{\sqrt{2} \sigma }\right)\right) $$
Below one can find the graph of this function.