Singular asymptotics of Gaussian integrals with periodic perturbations At the bottom of page 5 of this paper by Giedrius Alkauskas it is claimed that, for a $1$-periodic continuous function $f$,
$$
\int_{-\infty}^{\infty} f(x) e^{-Ax^2}\,dx = \sqrt{\frac{\pi}{A}} \int_0^1 f(x)\,dx + O(1)
\tag{1}
$$
as $A \to 0^+$.

How can I prove $(1)$?

I'm having a hard time doing it rigorously since I'm unfamiliar with Fourier series.
If I ignore convergence and just work formally then I can get something that resembles statement $(1)$.  Indeed, since $f$ is $1$-periodic we write
$$
f(x) = \int_0^1 f(t)\,dt + \sum_{n=1}^{\infty} \Bigl[a_n \cos(2\pi nx) + b_n \sin(2\pi nx)\Bigr].
$$
Multiply this by $e^{-Ax^2}$ and integrate term by term, remember that $\sin(2\pi nx)$ is odd, and get
$$
\begin{align}
\int_{-\infty}^{\infty} f(x) e^{-Ax^2}\,dx &= \int_{-\infty}^{\infty} e^{-Ax^2}\,dx \int_0^1 f(x)\,dx + \sum_{n=1}^{\infty} a_n \int_{-\infty}^{\infty} cos(2\pi nx) e^{-Ax^2}\,dx \\
&= \sqrt{\frac{\pi}{A}} \int_0^1 f(x)\,dx + \sqrt{\frac{\pi}{A}} \sum_{n=1}^{\infty} a_n e^{-\pi^2 n^2/A} \\
&= \sqrt{\frac{\pi}{A}} \int_0^1 f(x)\,dx + O\left(A^{-1/2} e^{-\pi^2/A}\right)
\end{align}
$$
as $A \to 0^+$.
I suppose the discrepancy in the error stems from the fact that $f$ need not be smooth (I think in the paper it's actually nowhere differentiable).  Obviously there are some issues, namely the convergence of the series and the interchange of summation and integration.  Resources concerning the analytic properties of Fourier series which are relevant would be much appreciated.
Answers which do not use Fourier series are also very welcome.

Naive Approach:
$$
\begin{align}
\int_{-\infty}^\infty f(x)\,e^{-Ax^2}\,\mathrm{d}x
&=\sum_{k=-\infty}^\infty\int_0^1f(x)\,e^{-A(x+k)^2}\,\mathrm{d}x\\
&=\int_0^1f(x)\sum_{k=-\infty}^\infty e^{-A(x+k)^2}\,\mathrm{d}x\tag{1}
\end{align}
$$
For $x\in[0,1]$ and $k\ge0$,
$$
e^{-A(x+k+1)^2}
\le\int_{x+k}^{x+k+1} e^{-At^2}\,\mathrm{d}t
\le e^{-A(x+k)^2}\tag{2}
$$
and
$$
e^{-A(x-k-2)^2}
\le\int_{x-k-2}^{x-k-1} e^{-At^2}\,\mathrm{d}t
\le e^{-A(x-k-1)^2}\tag{3}
$$
Summing $(2)$ and $(3)$ yields
$$
-1+\int_{-\infty}^\infty e^{-At^2}\,\mathrm{d}t
\le\sum_{k=-\infty}^\infty e^{-A(x+k)^2}
\le2+\int_{-\infty}^\infty e^{-At^2}\,\mathrm{d}t\tag{4}
$$
Since $\int_{-\infty}^\infty e^{-At^2}\,\mathrm{d}t=\sqrt{\frac\pi{A}}$, $(4)$ says
$$
\sum_{k=-\infty}^\infty e^{-A(x+k)^2}=\sqrt{\frac\pi{A}}+O(1)\tag{5}
$$
Combining $(1)$ and $(5)$ yields
$$
\int_{-\infty}^\infty f(x)\,e^{-Ax^2}\,\mathrm{d}x=\left(\sqrt{\frac\pi{A}}+O(1)\right)\int_0^1f(x)\,\mathrm{d}x\tag{6}
$$
For a given $f$, we can move the $O(1)$ outside to get
$$
\int_{-\infty}^\infty f(x)\,e^{-Ax^2}\,\mathrm{d}x=\sqrt{\frac\pi{A}}\int_0^1f(x)\,\mathrm{d}x+O(1)\tag{7}
$$

Fourier Series Approach:
Contour integration yields
$$
\int_{-\infty}^\infty e^{2\pi inx}e^{-Ax^2}\,\mathrm{d}x
=e^{-\pi^2n^2/A}\sqrt{\frac\pi A}\tag{8}
$$
If $f=\sum\limits_na_ne^{2\pi inx}$, then $a_0=\int_0^1f(x)\,\mathrm{d}x$
$$
\int_{-\infty}^\infty f(x)e^{-Ax^2}\,\mathrm{d}x=\sqrt{\frac\pi{A}}\left(\int_0^1f(x)\,\mathrm{d}x+\sum_{n\ne0}a_ne^{-\pi^2n^2/A}\right)\tag{9}
$$
and for $A\le1$,
$$
\begin{align}
\left|\sum_{n\ne0}a_ne^{-\pi^2n^2/A}\right|
&\le\sup_{n\ne0}|a_n|\sum_{n\ne0}e^{-\pi^2n^2/A}\\
&\le\|f\|_{L^1[0,1]}2\sum_{n=1}^\infty e^{-\pi^2n/A}\\
&\le\|f\|_{L^1[0,1]}\frac{2e^{-\pi^2/A}}{1-e^{-\pi^2}}\tag{10}
\end{align}
$$
Combining $(9)$ and $(10)$ yields
$$
\int_{-\infty}^\infty f(x)e^{-Ax^2}\,\mathrm{d}x
=\sqrt{\frac{\pi}{A}}\int_0^1f(x)\,\mathrm{d}x+O\left(\frac1{\sqrt{A}}e^{-\pi^2/A}\right)\|f\|_{L^1[0,1]}\tag{11}
$$

Convergence in the Fourier Series Approach:
Using the Fejér Kernel, we can show that trigonometric polynomials are dense in $L^1[0,1]$.
For trigonometric polynomials, there is no convergence problem in $(9)$ since all sums are finite. Likewise, $(11)$ also holds for trigonometric polynomials.
For any $\epsilon\gt0$, we can find a trigonometric polynomial $p$ so that
$$
\|f-p\|_{L^1[0,1]}\le\epsilon\tag{12}
$$
For $f-p$, both terms on the right side of $(11)$ are controlled by $(12)$. The only thing we need to control is the left hand side of $(11)$:
$$
\begin{align}
\int_{-\infty}^\infty|f(x)-p(x)|e^{-Ax^2}\,\mathrm{d}x
&\le\|f-p\|_{L^1[0,1]}2\sum_{k=0}^\infty e^{-Ak^2}\\
&\le\|f-p\|_{L^1[0,1]}2\sum_{k=0}^\infty e^{-Ak}\\
&=\|f-p\|_{L^1[0,1]}\frac2{1-e^{-A}}\\
&\le\|f-p\|_{L^1[0,1]}\frac{2e}{e-1}\tag{13}
\end{align}
$$
Using $(12)$ and $(13)$ and the fact that $(11)$ holds for any trigonometric polynomial $p$, $(11)$ also holds for any $f\in L^1[0,1]$.
