# 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 $f\in L^1[0,1]$.

• moved from the Riemann Sum argument to a sum-integral bound to get a big-O estimate – robjohn May 8 '13 at 20:03
• I really appreciate the straightforwardness and clarity of this approach. (+1) – Antonio Vargas May 9 '13 at 19:49
• @AntonioVargas: I've added a Fourier Series approach, which is hopefully as straightforward as my earlier naive approach. – robjohn May 9 '13 at 22:33
• Dear robjohn, I think your "Fourier Series Approach" is essentially what the OP did in the description of his/her question, and the OP's question is simply why $(9)$ holds. – 23rd May 10 '13 at 5:51
• @Landscape: I have added a section covering the convergence for $f\in L^1[0,1]$. – robjohn May 10 '13 at 20:19

One way to look at this is as follows. Let $c = \int_0^1 f(x)\,dx$. Then $f(x) - c$ has integral $0$ over the period, and therefore is the derivative of a $C^1$ function $g(x)$ on ${\mathbb R}$ which also has period $1$. Then $$\int_{-\infty}^{\infty} f(x) e^{-Ax^2}\,dx = \int_{-\infty}^{\infty} c e^{-Ax^2}\,dx + \int_{-\infty}^{\infty} (f(x) - c) e^{-Ax^2}\,dx$$ The first term integrates to $c\sqrt{\frac{\pi}{A}} = \sqrt{\frac{\pi}{A}} \int_0^1 f(x)\,dx$, which is the main term. In the second term, if you integrate by parts you get $$\int_{-\infty}^{\infty} g(x)(2Ax e^{-Ax^2})\,dx$$ $g(x)$ is continuous and has period $1$, so $|g(x)| \leq M$ for some $M$. Thus we have $$\bigg|\int_{-\infty}^{\infty} g(x)(2Ax e^{-Ax^2})\,dx\bigg| \leq M \int_{-\infty}^{\infty} 2A|x| e^{-Ax^2}\,dx$$ $$=2M\int_{0}^{\infty} 2Ax e^{-Ax^2}\,dx$$ $$= 2M$$ This gives the error term.

Note that by the way $M$ is defined you have explicit bounds on $M$: For $0 \leq x \leq 1$ you have $$g(x) = \int_0^x f(y)\,dy - x\int_0^1f(y)\,dy = (1-x)\int_0^x f(y)\,dy - x\int_x^1 f(y)\,dy$$ So $$|g(x)| \leq (1-x)x\sup_{y \in \mathbb R}|f(y)| + x(1-x)\sup_{y \in \mathbb R}|f(y)|$$ $$=2x(1-x)\sup_{y \in \mathbb R}|f(y)|$$ Since the supremum of $x(1-x)$ on $[0,1]$ is ${1 \over 4}$, we get $$M = \sup_{x \in [0,1]}|g(x)| \leq {1 \over 2}\sup_{y \in \mathbb R}|f(y)|$$

• I think this my error is half yours, but that is good enough for a $O(1)$. – robjohn May 8 '13 at 20:31
• Since the integral gets multiplied by a constant $k$ when $f$ does, the $O(1)$ term will not have a single uniform bound like $2$. – Zarrax May 8 '13 at 20:34
• I should clarify, the inner $O(1)$ is bounded by $2$. The outer $O(1)$ is bounded by $2\int_0^1f(x)\,\mathrm{d}x$. See $(6)$ and $(7)$ in my answer. – robjohn May 8 '13 at 20:36
• Okay, this looks okay (+1) – robjohn May 8 '13 at 20:42
• I added the best error estimate I could think of. – Zarrax May 8 '13 at 20:52

I will also do it formally though maybe with a less number of steps to justify. It is clear that since $f$ is periodic, \begin{align} \int_{-\infty}^{\infty}f(x)e^{-Ax^2}dx&=\int_0^1 f(x)\sum_{n\in\mathbb{Z}}e^{-A(x+n)^2}dx=\\&= \int_0^1 f(x)e^{-A x^2}\vartheta_3\Bigl(iAx\Bigl|\Bigr.\frac{iA}{\pi}\Bigr)dx,\tag{1} \end{align} where $\vartheta_3(z|\tau)=\sum_{n\in\mathbb{Z}}e^{i\pi\tau n^2+2 i n z}$ denotes Jacobi theta function. Full asymptotic expansion of (1) in most easily obtained using

1. Jacobi imaginary transformation $$\vartheta_3(z|\tau)=(-i\tau)^{-\frac12}\exp\left\{-\frac{iz^2}{\pi\tau}\right\}\,\vartheta_3\Bigl(-\frac{z}{\tau}\Bigl|\Bigr.-\frac{1}{\tau}\Bigr)\tag{2}$$ after which (1) becomes $$\sqrt{\frac{\pi}{A}}\int_0^1f(x)\,\vartheta_3\Bigl(\pi x\Bigl|\Bigr.\frac{i\pi}{A}\Bigr)dx.\tag{3}$$ Here the prefactor comes from $(-i\tau)^{-\frac12}$, the exponential in the integral is cancelled by the second factor in (2) and we also used that $\vartheta_3(z|\tau)$ is an even function of $z$. Let me stress that so far no approximations were made, the only assumption is that the summation can be interchanged with the integration at the very first step.

2. Note that the first argument of the theta function in (3) does not depend on $A$ and, especially, that the half-period ratio $\tau$ now tends to $i\infty$ instead of $i0^+$. This allows to extract the asymptotic expansion from the triple product formula $$\vartheta_3(z|\tau)=\prod_{n=1}^{\infty}(1-q^{2n})\prod_{n=1}^{\infty}(1+e^{2iz}q^{n-\frac12})\prod_{n=1}^{\infty}(1+e^{-2iz}q^{n-\frac12}),\tag{4}$$ where $q=e^{i\pi\tau}$. The point is that as $A\rightarrow 0^+$, the nome $q=e^{-\frac{\pi^2}{A}}$ in (3) rapidly goes to $0$, so that (3) becomes $$\sqrt{\frac{\pi}{A}}\int_0^1f(x)\,\Bigl[1+2\cos2\pi x\,e^{-\frac{\pi^2}{A}}+O(e^{-\frac{2\pi^2}{A}})\Bigr]dx.$$ It is clear that, in principle, one can derive from (4) any desired number of terms in the asymptotic expansion.

Added: It may seem that the above procedure works only because of the specific Gaussian form of the kernel. This is not so: if we had some other function $g(x)$ instead of $e^{-Ax^2}$, Jacobi imaginary transformation used above would be replaced by Poisson summation formula.

• Wow, this is awesome. Could you point me toward a resource that illustrates how such an asymptotic could be derived from your equation $(4)$? (+1 by the way) – Antonio Vargas May 9 '13 at 20:06
• This is just expansion of $\theta_3(z|\tau)$ in powers of $q$. In fact we don't even need the triple product, one could use instead of it the initial representation $\vartheta_3(\pi x|\tau)=1+\sum_{n=1}^{\infty} 2q^{n^2}\cos2\pi nx$. – Start wearing purple May 9 '13 at 20:52
• Ah, I see it now. Thanks! – Antonio Vargas May 9 '13 at 20:58

Your estimate of error term is correct. The following are just some supplementary details to make your argument more rigorous.

Let $(f_n)_{n\ge 1}$ be the sequence of partial sums of the Fourier series of $f$, i.e. $$f_n(x) = \int_0^1 f(t)\,dt + \sum_{k=1}^n \big(a_k \cos(2\pi kx) + b_k \sin(2\pi kx)\big).$$ Note that $f_n$ converges to $f$ in $L^2([0,1])$, so by Cauchy-Schwarz's inequality, as $n\to\infty$, $$\int_0^1|f(t)-f_n(t)| dt\le \big(\int_0^1|f(t)-f_n(t)|^2dt\big)^{\frac{1}{2}}\to 0.\tag{1}$$ Also note that, for every $n\ge 1$,

$$|a_n|=2\cdot\big|\int_0^1 f(t)\cos(2\pi nt) dt\big|\le 2\int_0^1|f(t)|dt.\tag{2}$$

As you have shown, $$\int_{-\infty}^{+\infty}f_n(x)e^{-Ax^2}dx=\sqrt{\frac{\pi}{A}}(\int_0^1f(t)dt+\sum_{k=1}^na_ke^{\frac{-k^2\pi^2}{A}}).\tag{3}$$ From $(2)$ and $(3)$ we know, $$\big|\sqrt{\frac{A}{\pi}}\int_{-\infty}^{+\infty}f_n(x)e^{-Ax^2}dx-\int_0^1f(t)dt\big|\le 2\int_0^1|f(t)|dt\cdot\sum_{k=1}^\infty e^{\frac{-k^2\pi^2}{A}}\le M e^{\frac{-\pi^2}{A}},\tag{4}$$ where $M>0$ is independent of $n\ge 1$ and $0<A\le 1$. Since for every $m\in\mathbb{Z}$, $$\int_m^{m+1}|f_n(x)-f(x)|dx=\int_0^1|f_n(x)-f(x)|dx,$$

$$\int_{-\infty}^{+\infty}|f_n(x)-f(x)|e^{-Ax^2}dx\le2 \int_0^1|f_n(x)-f(x)|dx\cdot\sum_{m=0}^\infty e^{-Am^2}<\infty.\tag{5}$$ Letting $n\to\infty$ in $(5)$, from $(1)$ we know that

$$\lim_{n\to\infty}\int_{-\infty}^{+\infty}|f_n(x)-f(x)|e^{-Ax^2}dx=0.\tag{6}$$ Combining $(4)$ and $(6)$, it follows that

$$|\int_{-\infty}^{+\infty}f(x)e^{-Ax^2}dx-\sqrt{\frac{\pi}{A}}\int_0^1f(t)dt|\le M\sqrt{\frac{\pi}{A}} e^{\frac{-\pi^2}{A}}.\tag{7}$$

• Thank you, this is really helpful. Especially so since you've taught me some of the basic tools we can use to account for the "error" for the partial sums of a Fourier series. – Antonio Vargas May 9 '13 at 19:43
• @AntonioVargas: You are welcome! – 23rd May 9 '13 at 20:23
• @robjohn: After rereading my original answer and comparing it with yours, I realized that the $L^2$ estimates in my original answer seemed quite unnecessary and messy, especially when coupled with the rapidly decaying function $e^{-x^2}$. I kept the OP's continuity assumption on $f$ and simplified my answer with $L^1$ estimates, but now it looks just a weakened mimic of your Fourier approach. Therefore, I leave it to the community. :) – 23rd May 12 '13 at 8:26
• I've decided to award the bounty to this answer for the reasons outlined in my above comment. – Antonio Vargas May 14 '13 at 22:52
• @AntonioVargas: Thank you for your bounty award. Please note that I had edited my answer after your first comment and the reason was given in my comment to robjohn above. – 23rd May 15 '13 at 1:08