Let $\varphi:\mathbb R\to\mathbb C$, and suppose that the limit $$\lim_{\sigma\to\infty}\int_{-\infty}^\infty \varphi\left(x\right)\exp\left\{-\frac12\cdot\left(\frac x\sigma\right)^2\right\}\,dx$$ exists. Well, this limit converges intuitively to the integral of $\varphi$. What conditions do we need to impose on $\varphi$ so that the limit $$\lim_{\sigma\to\infty}\int_{-\sigma}^\sigma \varphi\left(x\right)\,dx$$ also exists? Could this limit be different from the other one above with the Gaussian? I know that if $\varphi\in L^1$, then it's easy, but I can't find non-$L^1$ counterexamples.
The context is: I'm trying to derive the Fourier inversion formula for a specific class of functions. The textbook I'm using defines the space $\mathcal M$ of "moderate decrease" functions as the class of functions $f:\mathbb R\to\mathbb C$ that are continuous and $O(1/x^2)$ at $\pm\infty$. I can show that, if $f\in\mathcal M$, then the Fourier transform of $f$ is continuous and bounded (actually, not only bounded but goes to zero at infinity). However, even knowing those properties, when I try to prove the Fourier inversion formula, all I can show is that $$f(0) = \lim_{\sigma\to\infty}\int_{-\infty}^\infty \varphi\left(\xi\right)\exp\left\{-\frac12\cdot\left(\frac \xi\sigma\right)^2\right\}\,d\xi$$ where $\varphi$ is the Fourier transform of $f$. Obviously, I actually wanted to show that: $$f(0) = \lim_{\sigma\to\infty}\int_{-\sigma}^\sigma \varphi\left(\xi\right)\,d\xi.$$
As mentioned in the OP, $\varphi\in L^1$ is enough (just do the Dominated Convergence Theorem thing), but I'm still not sure if this is too restrictive.
Again I'm still not sure, but this looks similar enough to Abel's theorem ("if a series is summable to $S$, then it is Abel-summable to $S$") that I'm willing to guess that if the "hard" limit exists, then the limit with the Gaussian kernel must the same. Maybe the proof itself of Abel's theorem can be adapted?
Well, now I can (which is why I'm posting this).
Let $$T\left(x\right)=\max\left\{1 - \left|x\right|, 0\right\}$$ be the "triangular window" or "triangular kernel". The counterexample is: $$\varphi\left(\xi\right) = \sum_{n\geqslant 1} \frac{\left(-1\right)^{n-1}}{n} T\left(\frac{\left|\xi\right|-n^2}{n}\right)\mbox{.}$$ (Take a moment to appreciate how those triangles fit next to each other.) This is easily seen to be continuous, zero at infinity, and the dull limit $\lim_{\sigma\to\infty}\int_{-\sigma}^\sigma\varphi\left(\xi\right)\,d\xi$ doesn't exist—that integral oscillates between $0$ and $2$ thanks to the $\left(-1\right)^{n-1}$ in the sum. There is nothing easy about that Gaussian integral, however; there might be less concocted ways, but what I did was the following.
$$\begin{align} \int_{-\infty}^{\infty}\varphi\left(\xi\right)e^{-\xi^2/2\sigma^2}\,d\xi &= 2\int_0^\infty\sum_{n\geqslant 1} \frac{\left(-1\right)^{n-1}}{n} T\left(\frac{\xi-n^2}{n}\right)e^{-\xi^2/2\sigma^2}\,d\xi \\ &= 2\sum_{n\geqslant 1}\int_{n^2-n}^{n^2+n}\frac{\left(-1\right)^{n-1}}{n} T\left(\frac{\xi-n^2}{n}\right)e^{-\xi^2/2\sigma^2}\,d\xi \end{align}$$
Write $g_\sigma\left(\xi\right)=e^{-\xi^2/2\sigma^2}$ so that $$g_\sigma'\left(\xi\right)=\frac{-\xi}{\sigma^2}\cdot e^{-\xi^2/2\sigma^2},\qquad g_\sigma''\left(\xi\right)=\frac{1}{\sigma^2}\left(\frac{\xi^2}{\sigma^2} - 1\right) e^{-\xi^2/2\sigma^2}$$ and then 1st-order Taylor with (2nd-order) Lagrange remainder in the interval $\xi\in\left[n^2-n,n^2+n\right]$: $$e^{-\xi^2/2\sigma^2} = e^{-n^4/2\sigma^2} + g_\sigma'\left(n^2\right)\cdot\left(\xi-n^2\right)+\frac{1}{2!}\left[\frac{1}{\sigma^2}\left(\frac{X_{n,\sigma}\left(\xi\right)^2}{\sigma^2} - 1\right) e^{-X_{n,\sigma}\left(\xi\right)^2/2\sigma^2}\right]\cdot\left(\xi-n^2\right)^2$$ where $X_{n,\sigma}\left(\xi\right)\in\left[n^2-n,n^2+n\right]$. Plugging this Taylor expansion back, the Gaussian integral becomes: $$\sum_{n\geqslant 1}2\left(-1\right)^{n-1} e^{-n^4/2\sigma^2} + \sum_{n\geqslant 1}\int_{n^2-n}^{n^2+n}\frac{\left(-1\right)^{n-1}}{n} T\left(\frac{\xi-n^2}{n}\right)\frac{1}{\sigma^2}\left(\frac{X_{n,\sigma}\left(\xi\right)^2}{\sigma^2} - 1\right) e^{-X_{n,\sigma}\left(\xi\right)^2/2\sigma^2}\left(\xi-n^2\right)^2\,d\xi\mbox{.}$$ (Note that the linear term, with the 1st derivative, vanished because $T\left(\left(\xi-n^2\right)/n\right)$ is even around $n^2$.) I will show that, when $\sigma\to\infty$, the first sum above converges to $1$, and the second converges to $0$.
To reason about the first sum, write it as a power series: $$\sum2\left(-1\right)^{n-1} e^{-n^4/2\sigma^2} = \sum a_m r^m$$ where $a_m=0$ except when $m=n^4$ is a perfect fourth power, in which case $a_m=2\left(-1\right)^{n-1}$, and $r=e^{-1/2\sigma^2}$. Saying that sum converges to $1$ amounts to saying that $\sum a_m$ is Abel-summable to $1$; which is a consequence of the fact that it is Cesàro-summable to $1$. To see that the Cesàro means indeed converge to $1$, write down the sequence of partial cummulative sums: $$0, 0,\quad \underbrace{2, \ldots, 2}_{\mbox{length $2^4-1^4$}},\quad \underbrace{0, \ldots, 0}_{\mbox{length $3^4-2^4$}},\quad \underbrace{2, \ldots, 2}_{\mbox{length $4^4-3^4$}},\quad 0, \ldots$$ Whenever a block of consecutive zeros or twos ends at index $n^4$, its length is $\sim 4n^3$. Consider a pair of consecutive blocks: one with "zeros" and the following with "twos", each having length $\sim 4n^3$. Because the lengths are similar, the higher $n$ is, the closer the mean of the coefficients in this whole pair of blocks is to $\left(0+2\right)/2=1$. Each time we add another pair of blocks, we bring the whole Cesàro mean closer to $1$, q.e.d.
Almost there! Now we only have to prove that that second sum: $$S_2\left(\sigma\right)=\sum_{n\geqslant 1}\int_{n^2-n}^{n^2+n}\frac{\left(-1\right)^{n-1}}{n} T\left(\frac{\xi-n^2}{n}\right)\frac{1}{\sigma^2}\left(\frac{X_{n,\sigma}\left(\xi\right)^2}{\sigma^2} - 1\right) e^{-X_{n,\sigma}\left(\xi\right)^2/2\sigma^2}\cdot\left(\xi-n^2\right)^2\,d\xi$$ converges to zero when $\sigma\to\infty$. Not too trivial, but not too hard either: $$\begin{align} \left|S_2\left(\sigma\right)\right| & \leqslant \sum_{n\geqslant 1}\int_{n^2-n}^{n^2+n}\frac{1}{n\sigma^2}\left(\frac{\left(n^2+n\right)^2}{\sigma^2} + 1\right) e^{-\left(n^2-n\right)^2/2\sigma^2}\cdot n^2\,d\xi \\ &= \sum_{n\geqslant 1}\frac{2n^2}{\sigma^2}\left(\frac{n^2\left(n+1\right)^2}{\sigma^2} + 1\right) e^{-\left(n^2-n\right)^2/2\sigma^2} \end{align}$$ When $\sigma\to\infty$, each term in this sum converges to zero. By the Dominated Convergence Theorem, it only remains to show that those terms are dominated by a summable sequence which is independent of $\sigma$. To walk this last step, note that $\alpha e^{-\alpha}\leqslant e^{-1}$, and $\alpha^2 e^{-\alpha} \leqslant 4e^{-2}$, and if $n$ is big then $n^2-n\geqslant n^2/2$: $$\begin{align} \frac{2n^2}{\sigma^2}\left(\frac{n^2\left(n+1\right)^2}{\sigma^2} + 1\right) e^{-\left(n^2-n\right)^2/2\sigma^2} & \leqslant \frac{2n^2}{\sigma^2}\left(\frac{n^2\left(n+1\right)^2}{\sigma^2} + 1\right) e^{-n^4/8\sigma^2} \\ & = \frac{2n^4\left(n+1\right)^2}{\sigma^4} e^{-n^4/8\sigma^2} + \frac{2n^2}{\sigma^2} e^{-n^4/8\sigma^2} \\ & = \frac{128\left(n+1\right)^2}{n^4}\left[\frac{n^8}{64\sigma^4}e^{-n^4/8\sigma^2}\right] + \frac{16}{n^2} \left[\frac{n^4}{8\sigma^2} e^{-n^4/8\sigma^2}\right] \\ & \leqslant \frac{128\left(n+1\right)^2}{n^4} \cdot 4e^{-2} + \frac{16}{n^2} \cdot e^{-1} \\ & = O\left(n^{-2}\right)\mbox{,} \end{align}$$ q.e.d.
Hmmm I didn't do all the calculations, but I guess this works: $$\varphi\left(\xi\right) = \sum_{n\geqslant 1} \frac{\left(-1\right)^{n-1}}{n} T\left(\left|\xi\right|-\left(2n-1\right)\right)\mbox{.}$$
This function: $$\varphi\left(\xi\right) = \sum_{n\geqslant 1} \frac{\left(-1\right)^{n-1}}{n} T\left(\frac{\left|\xi\right|-n^2}{n}\right)$$ may be $C_0\left(\mathbb R\right)$, but it doesn't look like the transform of an $\mathcal M$-function; the (inverse) Fourier transform of $\varphi$ is apparently discontinuous at zero (again, I didn't do the calculations). Anyway, maybe this could be fixed with a $\left(\log n\right)$-ish factor?
Still looking for an example of an $\mathcal M$-function $f$ such that its transform $\hat f$ can't be integrated with $\lim_{\sigma\to\infty}\int_{-\sigma}^{\sigma}\hat f$.