limit of a sequence of integral functions

Question

I have to study punctual and uniform convergence of the following succession $$f_n(x)=n\int_{x-\frac{1}{n}}^{x+\frac{1}{n}}\dfrac{1}{e^{t^2}}dt.$$ I showed that the functions $f_n$ are uniformly convergent in $\mathbb R$ but I'm not able to calculate explicitly the limit function $f(x):=\underset{n\to +\infty}{\lim}f_n(x)$.
My idea was to split the integral in half and use the characteristic function of the domain of integration but this seems to be inconclusive.

Thank you for your help and your time.

Answer 1

For the updated question, the pointwise limit is $2e^{-x^2}$. Its easy to see that we just need to check $$ 2n\int_{x-1/n}^{x+1/n}|e^{-t^2} - e^{-x^2}| dt \to 0$$ Since $e^{-t^2}$ is continuous at $t=x$, for each $\epsilon$ there is $\delta$ such that for all $|t-x| < \delta$, $$ |e^{-t^2} - e^{-x^2}|\le \epsilon$$ Hence for all $n$ such that $1/n<\delta$, $$2n\int_{x-1/n}^{x+1/n}|e^{-t^2} - e^{-x^2}| dt\le \epsilon $$ this shows exactly what we need. This basically proves a special case of the Lebesgue Differentiation Theorem $$ f\in L^1_{loc}\implies f(x) = \lim_{r\to 0} \frac1{|B_r(x)|}\int_{B_r(x)} f(y)dy \quad \text{a.e.}$$ which is not trivial to prove in full generality, but much easier when $f$ is continuous.

Answer 2

$$f_n(x)=\int_{x-\frac{1}{n}}^{x+\frac{1}{n}}\dfrac{1}{e^{t^2}}dt=\frac{1}{2} \sqrt{\pi } \left(\text{erf}\left(\frac{1}{n}-x\right)+\text{erf}\left(\frac{1}{n}+x\right)\right)$$ Using the infinite series representation of $\text{erf}(t)$ and continuing with Taylor series, when $n$ is large $$f_n(x)\sim g_n(x)=\frac{2 e^{-x^2}}{n}\Bigg[1+\frac{2 x^2-1}{3 n^2}+\frac{4 x^4-12 x^2+3}{30 n^4}+O\left(\frac{1}{n^6}\right) \Bigg]$$

For example, using $x=\pi$ and $n=2^k$ $$\left( \begin{array}{cc} k & \frac {f_n(x)} {g_n(x)} \\ 0 & 1.28419669254952350166175610011 \\ 1 & 1.02205560798133497330954918511 \\ 2 & 1.00074377793925422180746341333 \\ 3 & 1.00001499797854822520136495734 \\ 4 & 1.00000025126261683720744981856 \\ 5 & 1.00000000399669102804908420208 \\ 6 & 1.00000000006272933330307433397 \\ 7 & 1.00000000000098124840982741539 \\ 8 & 1.00000000000001533631803233576 \\ 9 & 1.00000000000000023964681613758 \\ 10 & 1.00000000000000000374454731476 \\ 11 & 1.00000000000000000005850880888 \\ 12 & 1.00000000000000000000091420114 \\ 13 & 1.00000000000000000000001428440 \\ 14 & 1.00000000000000000000000022319 \\ 15 & 1.00000000000000000000000000349 \\ 16 & 1.00000000000000000000000000005 \\ 17 & 1.00000000000000000000000000000 \end{array} \right)$$

Answer 3

$$0\le f_n(x)=\int_{x-1/n}^{x+1/n}e^{-t^2}dt\le\int_{x-1/n}^{x+1/n}dt=\left(x+\frac 1n\right) - \left(x-\frac 1n\right)=\frac2n. $$ So $f_n(x)$ congerves uniformly to the null function over $\Bbb R$.