On Asymptotic Estimates for the Spectral Function of the Laplace Operator

F. G. Bulaevskaya--Maslova

Let \(D\) be the square in the \((x,y)\)-plane:

\[ 0 \leqslant x \leqslant 1,\quad 0 \leqslant y \leqslant 1, \tag{1} \]

and let \(B\) be the boundary of the square. Consider the boundary-value problem:

\[ \Delta u+\lambda u=0,\quad (x,y)\in D, \]

\[ u=0,\quad (x,y)\in B. \]

The eigenvalues of this problem are the numbers \(\lambda=\pi^{2}(n^{2}+m^{2})\), where \(n\geqslant 1,\ m\geqslant 1\) are integers. The multiplicity of the eigenvalue \(\lambda\) is equal to the number of representations of the number \(\lambda\) in the indicated form. The normalized eigenfunctions \(\omega(\lambda,x,y)\) corresponding to a given value \(\lambda\) are the functions

\[ \omega(\lambda,x,y)=2\sin \pi n x \sin \pi m y. \]

The spectral function for the Laplace operator in this case has the form

\[ \sum_{\lambda\leqslant T}\omega^{2}(\lambda,x,y) = 4 \sum_{\substack{n^{2}+m^{2}\leqslant \frac{T}{\pi^{2}}\\ n\geqslant 1,\ m\geqslant 1}} \sin^{2}\pi n x\,\sin^{2}\pi m y. \]

E. Titchmarsh [1] obtained an asymptotic formula for a fixed interior point \((x,y)\) of an arbitrary plane domain \(D\):

\[ \sum_{\lambda\leqslant T}\omega^{2}(\lambda,x,y) = \frac{T}{4\pi}+O(\sqrt{T}). \]

In the case where the domain \(D\) is the square (1), this estimate can be significantly improved if one applies I. M. Vinogradov’s method [2] for estimating trigonometric sums.

Theorem 1. As \(T\to\infty\), for an interior point \((x,y)\) of the square (1) we have the asymptotic formula

\[ \sum_{\lambda\leqslant T}\omega^{2}(\lambda,x,y) = \frac{T}{4\pi} + O\!\left(\frac{T^{1/3}\ln T}{l(x,y)}\right) + O\!\left(\frac{T^{1/4}}{l^{3/2}(x,y)}\right) + O\!\left(\frac{1}{l^{2}(x,y)}\right), \]

where \(l(x,y)\) is the distance from the point \((x,y)\) to the boundary of the square; here and below in the theorem the constants in the \(O\)-symbols are absolute.

Proof:

\[ \sum_{\lambda \leq T} \omega^2(\lambda,x,y) = 4 \sum_{\substack{n^2+m^2 \leq \frac{T}{\pi^2}\\ n\geq 1,\ m\geq 1}} \sin^2 \pi n x \sin^2 \pi m y. \]

Denote \(r=\dfrac{\sqrt{T}}{\pi}\). The summation is over all integer points lying inside the quadrant of the circle \(x^2+y^2\leq r^2\), \(x>0\), \(y>0\), with the points lying on the arc of the circumference included. We denote this region by \(U\). Introduce three regions into consideration:

\[ U_1:\quad x^2+y^2\leq r^2,\quad \frac{r}{\sqrt{2}}\geq x>0,\quad y>0; \]

\[ U_2:\quad x^2+y^2\leq r^2,\quad \frac{r}{\sqrt{2}}\geq y>0,\quad x>0; \]

\[ U_3:\quad 0<x\leq \frac{r}{\sqrt{2}},\quad 0<y\leq \frac{r}{\sqrt{2}}. \]

Obviously,

\[ \sum_{\lambda\leq T}\omega^2(\lambda,x,y) = 4\sum_{(n,m)\in U_1}\sin^2 \pi n x \sin^2 \pi m y + 4\sum_{(n,m)\in U_2}\sin^2 \pi n x \sin^2 \pi m y - 4\sum_{(n,m)\in U_3}\sin^2 \pi n x \sin^2 \pi m y = \]

\[ =4S_1+4S_2-4S_3. \]

We shall investigate successively the three quantities: \(4S_1, 4S_2, 4S_3\);

\[ 4S_1 = 4\sum_{(n,m)\in U_1}\sin^2 \pi n x \sin^2 \pi m y = \]

\[ = 4\sum_{1\leq n\leq \frac{r}{\sqrt{2}}}\sin^2 \pi n x \sum_{1\leq m\leq \sqrt{r^2-n^2}}\sin^2 \pi m y = \]

\[ = 2\sum_{1\leq n\leq \frac{r}{\sqrt{2}}}\sin^2 \pi n x \sum_{1\leq m\leq \sqrt{r^2-n^2}}(1-\cos 2\pi m y) = \]

\[ = 2\sum_{1\leq n\leq \frac{r}{\sqrt{2}}}[\sqrt{r^2-n^2}]\,\sin^2 \pi n x + \sum_{1\leq n\leq \frac{r}{\sqrt{2}}}\sin^2 \pi n x - \]

\[ - 2\sum_{1\leq n\leq \frac{r}{\sqrt{2}}}\sin^2 \pi n x \left( \frac{1}{2} + \sum_{1\leq m\leq \sqrt{r^2-n^2}}\cos 2\pi m y \right) = \]

\[ = \sum_{1\leq n\leq \frac{r}{\sqrt{2}}}[\sqrt{r^2-n^2}] - \sum_{1\leq n\leq \frac{r}{\sqrt{2}}}[\sqrt{r^2-n^2}]\cos 2\pi n x + \frac{1}{2}\left[\frac{r}{\sqrt{2}}\right] - \]

\[ -\frac{1}{2}\sum_{1\le n\le r/\sqrt{2}}\cos 2\pi n x -\sum_{1\le n\le r/\sqrt{2}}\sin^2 \pi n x\, \frac{\sin 2\pi\left(\left[\sqrt{r^2-n^2}\right]+\frac{1}{2}\right)y}{\sin \pi y} = \]

\[ =\sum_{1\le n\le r/\sqrt{2}}\sqrt{r^2-n^2} -\sum_{1\le n\le r/\sqrt{2}}\left\{\sqrt{r^2-n^2}\right\} -\sum_{1\le n\le r/\sqrt{2}}\sqrt{r^2-n^2}\cos 2\pi n x+ \]

\[ +\sum_{1\le n\le r/\sqrt{2}}\left\{\sqrt{r^2-n^2}\right\}\cos 2\pi n x +\frac{r}{2\sqrt{2}}- \]

\[ -\frac{1}{2\sin \pi y}\sum_{1\le n\le r/\sqrt{2}} \sin 2\pi\left(\left[\sqrt{r^2-n^2}\right]+\frac{1}{2}\right)y+ \]

\[ +\frac{1}{4\sin \pi y}\sum_{1\le n\le r/\sqrt{2}} \sin 2\pi\left(\left(\left[\sqrt{r^2-n^2}\right]+\frac{1}{2}\right)y+nx\right)+ \]

\[ +\frac{1}{4\sin \pi y}\sum_{1\le n\le r/\sqrt{2}} \sin 2\pi\left(\left(\left[\sqrt{r^2-n^2}\right]+\frac{1}{2}\right)y-nx\right) +O\left(\frac{1}{(x)}\right), \]

where \((x)\) is the distance from the point \(x\) to the nearest integer. We have used the obvious estimate

\[ \sum_{1\le n\le r/\sqrt{2}}\cos 2\pi n x = O\left(\frac{1}{(x)}\right). \]

By N. Ya. Sonin’s formula ([2], p. 37),

\[ \sum_{0<t\le r/\sqrt{2}}\sqrt{r^2-t^2} = \int_{0}^{r/\sqrt{2}}\sqrt{r^2-t^2}\,dt +\rho\left(\frac{r}{\sqrt{2}}\right)\frac{r}{\sqrt{2}} -\frac{r}{2} +O(1), \tag{2} \]

where \(\rho(t)=\dfrac{1}{2}-\{t\}\), and from I. M. Vinogradov’s theorem on the sum of fractional parts ([3], p. 50),

\[ \sum_{1\le n\le r/\sqrt{2}}\left\{\sqrt{r^2-n^2}\right\} = \frac{r}{2\sqrt{2}}+O\left(r^{2/3}\ln r\right). \tag{3} \]

To the sum
\[ \sum_{1\le n\le r/\sqrt{2}}\sqrt{r^2-n^2}\cos 2\pi n x \]
we apply the formula given on p. 19 of [4]:

\[ \sum_{1\le n\le r/\sqrt{2}}\sqrt{r^2-n^2}\cos 2\pi n x = \int_{0}^{r/\sqrt{2}}\sqrt{r^2-u^2}\cos 2\pi u x\,du+ \]

\[ + \int_{0}^{r/\sqrt{2}}\left(u-[u]-\frac{1}{2}\right) \frac{d\bigl(\sqrt{r^{2}-u^{2}}\cos 2\pi xu\bigr)}{du}\,du -\frac{r}{2} +\rho\left(\frac{r}{\sqrt{2}}\right)\frac{r}{\sqrt{2}}\cos\frac{2\pi rx}{\sqrt{2}} . \]

Putting \(u=rt\), after integrating twice by parts we obtain

\[ \int_{0}^{r/\sqrt{2}}\sqrt{r^{2}-u^{2}}\cos 2\pi xu\,du = \frac{r\sqrt{2}}{4\pi x}\sin 2\pi\frac{r}{\sqrt{2}}x +O\left(\frac{1}{(x)^{2}}\right). \tag{4} \]

Expand \(u-[u]-\dfrac{1}{2}\) in a Fourier series and, changing the order of integration and summation, which is possible in view of the bounded convergence of the series, we shall have

\[ \int_{0}^{r/\sqrt{2}}\left(u-[u]-\frac{1}{2}\right) \frac{d\bigl(\sqrt{r^{2}-u^{2}}\cos 2\pi xu\bigr)}{du}\,du = \]

\[ = -\frac{1}{\pi}\sum_{\nu=1}^{\infty}\frac{1}{\nu} \int_{0}^{r/\sqrt{2}}\sin 2\pi\nu u\, \frac{d\bigl(\sqrt{r^{2}-u^{2}}\cos 2\pi xu\bigr)}{du}\,du = \]

\[ = x\sum_{\nu=1}^{\infty}\frac{1}{\nu} \int_{0}^{r/\sqrt{2}}\sqrt{r^{2}-u^{2}}\cos 2\pi u(\nu-x)\,du - \]

\[ - x\sum_{\nu=1}^{\infty}\frac{1}{\nu} \int_{0}^{r/\sqrt{2}}\sqrt{r^{2}-u^{2}}\cos 2\pi u(\nu+x)\,du + \]

\[ +\frac{1}{2\pi}\sum_{\nu=1}^{\infty}\frac{1}{\nu} \int_{0}^{r/\sqrt{2}}\frac{u}{\sqrt{r^{2}-u^{2}}}\sin 2\pi u(\nu+x)\,du - \]

\[ -\frac{1}{2\pi}\sum_{\nu=1}^{\infty}\frac{1}{\nu} \int_{0}^{r/\sqrt{2}}\frac{u}{\sqrt{r^{2}-u^{2}}}\sin 2\pi u(\nu-x)\,du . \]

Putting \(u=rt\) and integrating by parts, we find that

\[ \frac{1}{2\pi}\sum_{\nu=1}^{\infty}\frac{1}{\nu} \int_{0}^{r/\sqrt{2}}\frac{u}{\sqrt{r^{2}-u^{2}}}\sin 2\pi u(\nu-x)\,du = \]

\[ =O\left(\sum_{\nu=1}^{\infty}\frac{1}{\nu(\nu-x)}\right) =O\left(\frac{1}{(x)}\right), \]

and similarly

\[ \frac{1}{2\pi}\sum_{\nu=1}^{\infty}\frac{1}{\nu} \int_{0}^{r/\sqrt{2}} \frac{u}{\sqrt{r^{2}-u^{2}}}\sin 2\pi u(\nu+x)\,du=O(1). \]

Replacing \(x\) in formula (4) by \(\nu-x\) or by \(\nu+x\), we obtain

\[ \begin{aligned} \sum_{1\le n\le r/\sqrt{2}}\sqrt{r^{2}-n^{2}}\cos 2\pi xn &=\frac{r\sqrt{2}}{4\pi x}\sin 2\pi\frac{r}{\sqrt{2}}x \\ &\quad+\frac{x^{2}r\sqrt{2}}{2\pi}\cos 2\pi\frac{r}{\sqrt{2}}x \sum_{\nu=1}^{\infty} \frac{\sin 2\pi \frac{r}{\sqrt{2}}\nu}{\nu(\nu^{2}-x^{2})} \\ &\quad-\frac{xr\sqrt{2}}{2\pi}\sin 2\pi\frac{r}{\sqrt{2}}x \sum_{\nu=1}^{\infty} \frac{\cos 2\pi \frac{r}{\sqrt{2}}\nu}{\nu^{2}-x^{2}} -\frac{r}{2} \\ &\quad+\rho\!\left(\frac{r}{\sqrt{2}}\right)\frac{r}{\sqrt{2}} \cos 2\pi\frac{r}{\sqrt{2}}x +O\!\left(\frac{1}{(x)^{2}}\right). \end{aligned} \tag{5} \]

Let us estimate the sum
\[ \sum_{1\le n\le r/\sqrt{2}}\{\sqrt{r^{2}-n^{2}}\}\cos 2\pi nx. \]
Take a positive number \(\delta<1\), depending on \(r\); this dependence will be specified later. Define on the half-interval \([-\delta,1-\delta)\) the function \(\Phi_{1}(\alpha)\) as follows:

\[ \Phi_{1}(\alpha)= \begin{cases} \dfrac{2\delta-1}{2\delta}\alpha+\dfrac{1}{2}, & -\delta\le \alpha\le \delta,\\[6pt] \alpha, & \delta<\alpha<1-\delta, \end{cases} \]

and extend this function periodically with period \(T=1\) to the whole real axis \(\alpha\). This will be a continuous function for all \(\alpha\). Denote by \(\Delta\) the union of two intervals: \((0,\delta)\) and \((1-\delta,1)\). It is obvious that inside them \(|\Phi_{1}(\alpha)-\{\alpha\}|<1\). The sum of the lengths of these intervals is \(2\delta\). Let \(N_{r}(\Delta)\) be the number of hits in the set \(\Delta\) of the fractional parts \(\{\sqrt{r^{2}-n^{2}}\}\), when \(n=1,2,\ldots,\left[\dfrac{r}{\sqrt{2}}\right]\). It is clear that

\[ \sum_{1\le n\le r/\sqrt{2}}\{\sqrt{r^{2}-n^{2}}\}e^{2\pi inx} = \sum_{1\le n\le r/\sqrt{2}}\Phi_{1}(\sqrt{r^{2}-n^{2}})e^{2\pi inx} +O\bigl(N_{r}(\Delta)\bigr). \]

By the theorem of I. M. Vinogradov ([3], p. 50),

\[ N_{r}(\Delta)=O(r\delta)+O(r^{2/3}\ln r). \]

Consequently,

\[ \sum_{1\le n\le r/\sqrt{2}}\{\sqrt{r^{2}-n^{2}}\}e^{2\pi inx} = \sum_{1\le n\le r/\sqrt{2}}e^{2\pi inx}\Phi_{1}(\sqrt{r^{2}-n^{2}}) +O(r\delta)+O(r^{2/3}\ln r). \]

The function \(\Phi_1(\alpha)\) is expanded in a Fourier series:

\[ \Phi_1(\alpha)=\frac{1}{2}+\sum_{s=-\infty}^{\infty}{}' a_s e^{2\pi i s\alpha} \]

(the prime on the summation sign means that the term with \(s=0\) is omitted), and

\[ |a_s|=O\left(\frac{1}{|s|}\right),\quad s\ne 0;\qquad |a_s|=O\left(\frac{1}{\delta s^2}\right),\quad s\ne 0. \]

We have

\[ \sum_{1\le n\le r/\sqrt{2}} e^{2\pi i n x}\left\{\sqrt{r^2-n^2}\right\} = O\left( \sum_{s=-\infty}^{\infty}{}' |a_s| \left| \sum_{1\le n\le r/\sqrt{2}} e^{2\pi i\left(nx+s\sqrt{r^2-n^2}\right)} \right| \right) + \]

\[ +O(r\delta)+O(r^{2/3}\ln r)+O\left(\frac{1}{(x)}\right). \]

Apply to the sum

\[ \sum_{1\le n\le r/\sqrt{2}} e^{2\pi i s\left(n\frac{x}{s}+\sqrt{r^2-n^2}\right)} \]

the van der Corput estimate ([4], p. 107):

\[ \left| \sum_{1\le n\le r/\sqrt{2}} e^{2\pi i\left(nx+s\sqrt{r^2-n^2}\right)} \right| = O\left(\sqrt{|s|\,r}\right). \]

Take some positive parameter \(P\):

\[ \sum_{s=-\infty}^{\infty}{}' |a_s| \left| \sum_{1\le n\le r/\sqrt{2}} e^{2\pi i\left(nx+s\sqrt{r^2-n^2}\right)} \right| = O\left( \sum_{|s|\le P}\sqrt{|s|\,r}\,\frac{1}{|s|} \right) + \]

\[ +O\left( \sum_{|s|>P} \frac{\sqrt{|s|\,r}}{\delta |s|^2} \right) = O(\sqrt{rP})+O\left(\frac{\sqrt r}{\delta\sqrt P}\right). \]

If \(P=\dfrac{1}{\delta}\), \(\delta=\dfrac{1}{r^{1/3}}\), then

\[ \sum_{1\le n\le r/\sqrt{2}} e^{2\pi i n x}\left\{\sqrt{r^2-n^2}\right\} = O(r^{2/3}\ln r)+O\left(\frac{1}{(x)}\right). \tag{6} \]

To estimate the modulus of the sum

\[ \sum_{1\le n\le r/\sqrt{2}} \sin 2\pi\left(\left[\sqrt{r^2-n^2}\right]+\frac{1}{2}\right)y \]

it is enough to estimate the modulus of the sum

\[ \left| \sum_{1\le n\le r/\sqrt{2}} e^{2\pi i\left(\left[\sqrt{r^2-n^2}\right]+\frac{1}{2}\right)y} \right| = \left| \sum_{1\le n\le r/\sqrt{2}} e^{2\pi i\sqrt{r^2-n^2}\,y}\cdot e^{-2\pi i\left\{\sqrt{r^2-n^2}\right\}y} \right|. \]

The function \(e^{-2\pi i\{\alpha\}y}\) is periodic in \(\alpha\) with period \(T=1\). Let \(\delta\)—a positive quantity depending on \(r\); we shall specify this dependence

Then. Define on the half-interval \([-\delta, 1-\delta)\) the function \(\Phi_2(\alpha)\) as follows:

\[ \Phi_2(\alpha)= \begin{cases} e^{-2\pi i(1-\delta)y}+\dfrac{\alpha+\delta}{2\delta} \left(e^{-2\pi i\delta y}-e^{-2\pi i(1-\delta)y}\right), & -\delta \leq \alpha \leq \delta,\\[6pt] e^{-2\pi i\alpha y}, & \delta<\alpha<1-\delta, \end{cases} \]

and extend this function periodically with period \(T=1\) to the entire real axis \(\alpha\). This will be a continuous function, bounded by a constant independent of \(y\) and \(\delta\). Arguing analogously to the investigation of the sum

\[ \sum_{1\leq n\leq \frac{r}{\sqrt{2}}} e^{2\pi i n x}\left\{\sqrt{r^2-n^2}\right\}, \]

we shall have

\[ \sum_{1\leq n\leq \frac{r}{\sqrt{2}}} e^{2\pi i\sqrt{r^2-n^2}\,y}\, e^{-2\pi i\{\sqrt{r^2-n^2}\}y} = \]

\[ = \sum_{1\leq n\leq \frac{r}{\sqrt{2}}} e^{2\pi i\sqrt{r^2-n^2}\,y}\, \Phi_2\!\left(\sqrt{r^2-n^2}\right) +O(r\delta)+O(r^{2/3}\ln r). \]

The Fourier series of the function \(\Phi_2(\alpha)\) will converge to it for every value of \(\alpha\):

\[ \Phi_2(\alpha)=\sum_{s=-\infty}^{\infty} b_s e^{2\pi i s\alpha}, \]

where

\[ |b_s|=O\!\left(\frac{1}{|s|}\right),\qquad s\ne 0, \]

\[ |b_s|=O\!\left(\frac{1}{|s|^2\delta}\right),\qquad s\ne 0, \]

since the variation of the function \(\Phi_2(\alpha)\) is estimated as \(O(1)\), while the variation of the function \(\Phi_2'(\alpha)\) has the estimate \(O\!\left(\dfrac{1}{\delta}\right)\).

Repeating almost word for word the same arguments as in estimating the sum

\[ \sum_{1\leq n\leq \frac{r}{\sqrt{2}}} e^{2\pi i n x}\left\{\sqrt{r^2-n^2}\right\}, \]

we obtain

\[ \sum_{1\leq n\leq \frac{r}{\sqrt{2}}} e^{2\pi i[\sqrt{r^2-n^2}]y} = O\!\left( \sum_{s=-\infty}^{\infty}|b_s| \sum_{1\leq n\leq \frac{r}{\sqrt{2}}} e^{2\pi i\sqrt{r^2-n^2}(y+s)} \right) +O(r\delta)+O(r^{2/3}\ln r). \]

By virtue of the above-mentioned estimate of van der Corput,

\[ \left| \sum_{1\leq n\leq \frac{r}{\sqrt{2}}} e^{2\pi i\sqrt{r^2-n^2}(y+s)} \right| = O\!\left(\sqrt{|s|\,r}\right), \qquad s\ne 0,\,-1. \]

\[ \left|\sum_{1\le n\le \frac{r}{\sqrt{2}}} e^{2\pi i\sqrt{r^2-n^2}\,(y+s)}\right| =O\left(\sqrt{\frac{r}{(y)}}\right),\qquad s=0,-1. \]

If \(P=\dfrac{1}{\delta}\), \(\delta=\dfrac{1}{r^{1/3}}\), then

\[ \left|\sum_{1\le n\le \frac{r}{\sqrt{2}}} e^{2\pi i\left([\sqrt{r^2-n^2}]+\frac12\right)y}\right| = O\left(\sqrt{\frac{r}{(y)}}\right) +O\left(\sum_{|s|<P}\frac{\sqrt{|s|\,r}}{|s|}\right)+ \]

\[ +O\left(\sum_{|s|>P}\frac{\sqrt{|s|\,r}}{\delta s^2}\right) +O(r\delta)+O(r^{2/3}\ln r) =O(r^{2/3}\ln r)+O\left(\sqrt{\frac{r}{(y)}}\right). \]

Taking into account that \(|\sin\pi y|\ge 2(y)\), we have

\[ \frac{1}{\sin\pi y} \sum_{1\le n\le \frac{r}{\sqrt{2}}} \sin 2\pi\left(\left[\sqrt{r^2-n^2}\right]+\frac12\right)y = O\left(\frac{r^{2/3}\ln r}{(y)}\right) +O\left(\frac{r^{1/2}}{(y)^{3/2}}\right). \tag{7} \]

Similarly,

\[ \frac{1}{\sin\pi y} \sum_{1\le n\le \frac{r}{\sqrt{2}}} \sin 2\pi\left(\left(\left[\sqrt{r^2-n^2}\right]+\frac12\right)y+nx\right) = \]

\[ =O\left(\frac{r^{2/3}\ln r}{(y)}\right) +O\left(\frac{r^{1/2}}{(y^{3/2})}\right). \tag{8} \]

From formulas (2)—(8) we obtain

\[ 4S_1=\frac{\pi r^2}{8}+\frac{r^2}{4} +\rho\left(\frac{r}{\sqrt{2}}\right)\frac{r}{\sqrt{2}} -\frac{r\sqrt{2}}{4\pi x}\sin 2\pi\frac{r}{\sqrt{2}}x+ \]

\[ +\frac{xr\sqrt{2}}{2\pi}\sin 2\pi\frac{r}{\sqrt{2}}x \sum_{\nu=1}^{\infty} \frac{\cos 2\pi\frac{r}{\sqrt{2}}\nu}{\nu^2-x^2} - \]

\[ -\frac{x^2r\sqrt{2}}{2\pi}\cos 2\pi\frac{r}{\sqrt{2}}x \sum_{\nu=1}^{\infty} \frac{\sin 2\pi\frac{r}{\sqrt{2}}\nu}{\nu(\nu^2-x^2)} - \]

\[ -\rho\left(\frac{r}{\sqrt{2}}\right)\frac{r}{\sqrt{2}}\cos 2\pi\frac{r}{\sqrt{2}}x +O\left(\frac{1}{(x)^2}\right)+ \]

\[ +O\left(\frac{r^{2/3}\ln r}{(y)}\right) +O\left(\frac{r^{1/2}}{(y)^{3/2}}\right). \tag{9} \]

The sum \(4S_2\) is obtained from the sum \(4S_1\) by replacing \(x\) by \(y\).

We estimate the sum \(4S_3\). By elementary transformations, analogous to those used in estimating the sum \(4S_1\), we obtain

\[ 4S_3 = \left[\frac{r}{\sqrt{2}}\right]^2 + \left[\frac{r}{\sqrt{2}}\right] - \frac{1}{2}\frac{r}{\sqrt{2}} \left( \frac{\sin 2\pi\left(\left[\frac{r}{\sqrt{2}}\right]+\frac{1}{2}\right)x}{\sin \pi x} + \frac{\sin 2\pi\left(\left[\frac{r}{\sqrt{2}}\right]+\frac{1}{2}\right)y}{\sin \pi y} \right) + O\left(\frac{1}{(x)(y)}\right). \tag{10} \]

Let \(\dfrac{r}{\sqrt{2}}\) be an integer. From equalities (9), (10) we have

\[ \begin{aligned} \sum_{\lambda \le T} \omega^2(\lambda,x,y) &= \frac{\pi r^2}{4} - \frac{r\sqrt{2}}{4\pi} \left( \frac{\sin 2\pi \frac{r}{\sqrt{2}}x}{x} + \frac{\sin 2\pi \frac{r}{\sqrt{2}}y}{y} \right) \\ &\quad - \frac{r}{2\sqrt{2}} \left( \cos 2\pi \frac{r}{\sqrt{2}}x + \cos 2\pi \frac{r}{\sqrt{2}}y \right) + \\ &\quad + \frac{r\sqrt{2}}{2\pi} \left( x\sin 2\pi \frac{r}{\sqrt{2}}x \sum_{\nu=1}^{\infty}\frac{1}{\nu^2-x^2} + y\sin 2\pi \frac{r}{\sqrt{2}}y \sum_{\nu=1}^{\infty}\frac{1}{\nu^2-y^2} \right) + \\ &\quad + \frac{r\sqrt{2}}{4} \left( \sin 2\pi \frac{r}{\sqrt{2}}x \cdot \operatorname{ctg}\pi x + \cos 2\pi \frac{r}{\sqrt{2}}x + \right. \\ &\qquad\qquad\left. + \sin 2\pi \frac{r}{\sqrt{2}}y \cdot \operatorname{ctg}\pi y + \cos 2\pi \frac{r}{\sqrt{2}}y \right) + \\ &\quad + O\left(\frac{r^{2/3}\ln r}{l(x,y)}\right) + O\left(\frac{1}{l^2(x,y)}\right) + O\left(\frac{r^{1/2}}{l^{3/2}(x,y)}\right), \end{aligned} \]

or, taking into account the formula

\[ \operatorname{ctg}\pi x = \frac{1}{\pi x} - \frac{2x}{\pi} \sum_{\nu=1}^{\infty}\frac{1}{\nu^2-x^2} \]

for nonintegral \(x\), we also obtain the assertion of the theorem for the case when \(T\to\infty\), but in such a way that

\[ \frac{1}{\pi}\sqrt{\frac{T}{2}} \]

is an integer.

Let \(\dfrac{r}{\sqrt{2}}\) not be an integer. In equality (10), expanding the functions

\[ \cos 2\pi\left(\left\{\frac{r}{\sqrt{2}}\right\}-\frac{1}{2}\right)x \]

and

\[ \sin 2\pi\left(\left\{\frac{r}{\sqrt{2}}\right\}-\frac{1}{2}\right)x \]

in a Fourier series with respect to the variable \(\dfrac{r}{\sqrt{2}}\), we obtain

\[ \frac{\sin 2\pi\left(\left[\frac{r}{\sqrt{2}}\right]+\frac{1}{2}\right)x}{\sin \pi x} = \frac{\sin 2\pi \frac{r}{\sqrt{2}}x}{\pi x} + \]

\[ +\frac{2x}{\pi}\sin 2\pi\frac{r}{\sqrt{2}}x \sum_{\nu=1}^{\infty} \frac{\cos 2\pi\frac{r}{\sqrt{2}}\nu}{x^{2}-\nu^{2}} - \]

\[ -\frac{2}{\pi}\cos 2\pi\frac{r}{\sqrt{2}}x \sum_{\nu=1}^{\infty} \frac{\nu\sin 2\pi\frac{r}{\sqrt{2}}\nu}{x^{2}-\nu^{2}} . \tag{11} \]

From equalities (9), (10), (11) we have

\[ \sum_{\lambda\leq T}\omega^{2}(\lambda,x,y) = \frac{\pi r^{2}}{4} + \frac{r^{2}}{2} + 2\frac{r}{\sqrt{2}}\rho\left(\frac{r}{\sqrt{2}}\right) - \left[\frac{r}{\sqrt{2}}\right]^{2} - \]

\[ -\frac{r}{\sqrt{2}} - \rho\left(\frac{r}{\sqrt{2}}\right)\frac{r}{\sqrt{2}} \left(\cos 2\pi\frac{r}{\sqrt{2}}x+\cos 2\pi\frac{r}{\sqrt{2}}y\right) + \]

\[ +\frac{r\sqrt{2}}{2\pi}\cos 2\pi\frac{r}{\sqrt{2}}x \sum_{\nu=1}^{\infty} \left( \frac{\nu}{\nu^{2}-x^{2}} - \frac{x^{2}}{\nu(\nu^{2}-x^{2})} \right) \sin 2\pi\frac{r}{\sqrt{2}}\nu + \]

\[ +\frac{r\sqrt{2}}{2\pi}\cos 2\pi\frac{r}{\sqrt{2}}y \sum_{\nu=1}^{\infty} \left( \frac{\nu}{\nu^{2}-y^{2}} - \frac{y^{2}}{\nu(\nu^{2}-y^{2})} \right) \sin 2\pi\frac{r}{\sqrt{2}}\nu + \]

\[ + O\left(\frac{1}{l^{2}(x,y)}\right) + O\left(\frac{r^{2/3}\ln r}{l(x,y)}\right) + O\left(\frac{r^{1/2}}{l^{3/2}(x,y)}\right). \]

But

\[ \frac{r^{2}}{2} + 2\frac{r}{\sqrt{2}}\rho\left(\frac{r}{\sqrt{2}}\right) - \left[\frac{r}{\sqrt{2}}\right]^{2} - \frac{r}{\sqrt{2}} = O(1). \]

We obtain

\[ \sum_{\lambda\leq T}\omega^{2}(\lambda,x,y) = \frac{\pi r^{2}}{4} - \frac{r}{\sqrt{2}}\rho\left(\frac{r}{\sqrt{2}}\right) \cos 2\pi\frac{r}{\sqrt{2}}x - \]

\[ - \frac{r}{\sqrt{2}}\rho\left(\frac{r}{\sqrt{2}}\right) \cos 2\pi\frac{r}{\sqrt{2}}y + \frac{r\sqrt{2}}{2\pi}\cos 2\pi\frac{r}{\sqrt{2}}x \sum_{\nu=1}^{\infty} \frac{\sin 2\pi\frac{r}{\sqrt{2}}\nu}{\nu} + \]

\[ + \frac{r\sqrt{2}}{2\pi}\cos 2\pi\frac{r}{\sqrt{2}}y \sum_{\nu=1}^{\infty} \frac{\sin 2\pi\frac{r}{\sqrt{2}}\nu}{\nu} + \]

\[ + O\left(\frac{r^{2/3}\ln r}{l(x,y)}\right) + O\left(\frac{r^{1/2}}{l^{3/2}(x,y)}\right) + O\left(\frac{1}{l^{2}(x,y)}\right). \]

But since \(\dfrac{r}{\sqrt{2}}\) is not an integer, we have

\[ \frac{1}{\pi}\sum_{\nu=1}^{\infty} \frac{\sin 2\pi \dfrac{r}{\sqrt{2}}\,\nu}{\nu} = \rho\left(\frac{r}{\sqrt{2}}\right). \]

Hence

\[ \sum_{\lambda\leq T}\omega^{2}(\lambda,x,y) = \frac{T}{4\pi} + O\left(\frac{T^{1/3}\ln T}{l(x,y)}\right) + O\left(\frac{T^{1/4}}{l^{3/2}(x,y)}\right) + O\left(\frac{1}{l^{2}(x,y)}\right). \]

Thus the theorem is proved in this case as well.

For the point \((x,y)=\left(\dfrac12,\dfrac12\right)\), the sum
\(\sum_{\lambda\leq T}\omega^{2}(\lambda,x,y)\) is equal to the number of integer points with both coordinates odd lying in the disk

\[ x^{2}+y^{2}\leq \frac{T}{\pi^{2}}. \]

We may study this sum by analogy with the study of integer points inside a disk. Our considerations will be analogous to the exposition of the question of the number of integer points inside a disk given in the textbook of E. Landau [7].

First of all, we shall prove a refinement of the general theorem for the case

\[ (x,y)=\left(\frac12,\frac12\right). \]

Theorem 2. As \(T\to\infty\),

\[ \sum_{\lambda\leq T}\omega^{2}\left(\lambda,\frac12,\frac12\right) = \frac{T}{4\pi} + O(T^{1/3}). \]

The constant in the symbol “\(O\)” is absolute.

Proof. Let \(J_k(z)\) be the Bessel function of order \(k\). Further, let \(r(n)\) be the number of representations of the natural number \(n\) in the form

\[ n=x^{2}+y^{2} \]

with integral \(x\) and \(y\). From number theory it is known ([5], p. 52):

\[ r(n)=4\sum_{d\mid n}(-1)^{\frac{d-1}{2}}, \]

where \(d\) is odd. Hence it is clear that

\[ r(2n)=r(n). \tag{12} \]

Further,

\[ r(n)\leq 4\tau(n), \]

where \(\tau(n)\) is the number of divisors of \(n\). We shall use the well-known estimate from number theory (see, for example, [6], p. 162):

\[ \tau(n)=O(n^{\varepsilon}),\qquad \varepsilon>0. \]

We shall use this assertion in the weakened form

\[ \tau(n)=O(n^{1/8}). \]

Denote by \(A(y)\) the number of integer points with both coordinates odd, lying in the circle \(\xi^2+\eta^2\leqslant y\). We have the following

Lemma. The equality holds
\[ \int_0^x A(y)\,dy=\frac{\pi x^2}{8}+\frac{x}{\pi}\sum_{n=1}^{\infty}(-1)^n\frac{r(n)}{n}J_2(\pi\sqrt{nx}). \]

The proof of the lemma is based on a well-known summation formula from the theory of Fourier series ([7], p. 204): let \(\alpha<\beta\), let \(f(y)\) be real-valued for \(\alpha\leqslant y\leqslant\beta\), continuous, piecewise monotone, and let \(f(\alpha)=0,\ f(\beta)=0\). Then
\[ \sum_{\alpha<n<\beta} f(n) = \sum_{k=-\infty}^{\infty}\int_{\alpha}^{\beta} f(y)\cos 2\pi ky\,dy. \tag{13} \]

Consider the sum
\[ \sum_{-\frac{\sqrt{x}-1}{2}\leqslant m\leqslant \frac{\sqrt{x}-1}{2}} \ \sum_{-\frac{\sqrt{x-(2m+1)^2}-1}{2}\leqslant n\leqslant \frac{\sqrt{x-(2m+1)^2}-1}{2}} \bigl(x-(2m+1)^2-(2n+1)^2\bigr) = \int_0^x \sum_{(2m+1)^2+(2n+1)^2\leqslant y} 1\,dy = \int_0^x A(y)\,dy . \]

Applying the summation formula (13) to the outer, and then to the inner sum, we obtain
\[ \int_0^x A(y)\,dy = \sum_{a=-\infty}^{\infty} \int_{-\frac{\sqrt{x}-1}{2}}^{\frac{\sqrt{x}-1}{2}} \cos 2\pi au \left( \sum_{b=-\infty}^{\infty} \int_{-\frac{\sqrt{x-(2u+1)^2}-1}{2}}^{\frac{\sqrt{x-(2u+1)^2}-1}{2}} \cos 2\pi bv\,(x- \right. \]
\[ \left. -(2u+1)^2-(2v+1)^2)\,dv \right)\,du, \]
or, after changing the order of summation with respect to \(b\) and integration with respect to \(u\), which is possible by virtue of the uniform convergence of the series, and after the change of variables \(2u+1=t,\ 2v+1=w\),
\[ \int_0^x A(y)\,dy = \frac14 \sum_{a=-\infty}^{\infty} \sum_{b=-\infty}^{\infty} \cos\pi a\cos\pi b\,Q(x,a,b), \]
where
\[ Q(x,a,b)= \iint_{t^2+w^2\leqslant x} \cos\pi at\cos\pi bw\,(x-t^2-w^2)\,dt\,dw . \]

Obviously,
\[ Q(x,a,b)= \int_0^x \left( \iint_{t^2+w^2\leqslant y} \cos\pi at\cos\pi bw\,dt\,dw \right)\,dy = \]

\[ = \int_0^x dy \iint_{t^2+w^2\le y} \cos \pi(at+bw)\,dt\,dw. \]

It is clear that \(Q(x,0,0)=\dfrac{\pi x^2}{2}\). If, however, \(a^2+b^2>0\), then after the change of variables

\[ \frac{at+bw}{\sqrt{a^2+b^2}}=T,\qquad \frac{-bt+aw}{\sqrt{a^2+b^2}}=W \]

we have

\[ Q(x,a,b)=\int_0^x dy \iint_{T^2+W^2\le y} \cos\bigl(\pi T\sqrt{a^2+b^2}\bigr)\,dT\,dW = \]

\[ =4\int_0^x y\left(\int_0^1 \sqrt{1-u^2}\cos \pi u\sqrt{y(a^2+b^2)}\,du\right)dy = \]

\[ =\frac{2}{\sqrt{a^2+b^2}}\int_0^x \sqrt{y}\, J_1\bigl(\pi\sqrt{(a^2+b^2)y}\bigr)\,dy = \]

\[ =\frac{4}{\pi^3(a^2+b^2)^2} \int_0^{\pi\sqrt{x(a^2+b^2)}} z^2J_1(z)\,dz =\frac{4x}{\pi(a^2+b^2)}J_2\bigl(\pi\sqrt{x(a^2+b^2)}\bigr). \]

Thus,

\[ \int_0^x A(y)\,dy =\frac{\pi x^2}{8} +\frac{x}{\pi} \sum_{a=-\infty}^{\infty}\sum_{b=-\infty}^{\infty}{}' \frac{\cos \pi a\cos \pi b}{a^2+b^2} J_2\bigl(\pi\sqrt{x(a^2+b^2)}\bigr), \]

where the prime indicates that the term with \(a=0,\ b=0\) is omitted. The series on the right-hand side is absolutely convergent. Further, if the natural number \(n\equiv 3\pmod 4\), then it cannot be represented as a sum of two squares. If \(n\equiv 1\pmod 4\), then in every representation of \(n\) as a sum of two squares one of the summands is even and the other odd. If \(n\equiv 2\pmod 4\), then in every representation of \(n\) both summands are odd. If \(n\equiv 0\pmod 4\), then in every representation of \(n\) both summands are even. Therefore

\[ \int_0^x A(y)\,dy =\frac{\pi x^2}{8} +\frac{x}{\pi}\sum_{n=1}^{\infty}\sum_{a^2+b^2=n} \frac{\cos \pi a\cos \pi b}{a^2+b^2} J_2\bigl(\pi\sqrt{x(a^2+b^2)}\bigr) = \]

\[ =\frac{\pi x^2}{8} +\frac{x}{\pi}\sum_{n=1}^{\infty} \frac{(-1)^n r(n)}{n}J_2(\pi\sqrt{xn}). \]

The lemma is proved.

Let us now prove Theorem 2. Put

\[ R(x)=A(x)-\frac{\pi x}{4}. \]

Let \(x>1\). We have

\[ \int_x^{x\pm x^{1/3}} R(y)\,dy = \frac1\pi \sum_{n=1}^{\infty}(-1)^n\frac{r(n)}{n} \left. yJ_2(\pi\sqrt{yn})\right|_x^{x\pm x^{1/3}} . \]

For \(n>x^{1/3}\), by virtue of the estimate \(|J_2(z)|<C/\sqrt z,\ z>0\),

\[ \left. yJ_2(\pi\sqrt{yn})\right|_x^{x\pm x^{1/3}} = O\!\left(\frac{x^{3/4}}{n^{1/4}}\right). \]

For \(n\leq x^{1/3}\), by virtue of the estimate \(|J_1(z)|<C/\sqrt z,\ z>0\),

\[ \begin{aligned} \left. yJ_2(\pi\sqrt{yn})\right|_x^{x\pm x^{1/3}} &= \int_x^{x\pm x^{1/3}} \frac{d\bigl(yJ_2(\pi\sqrt{ny})\bigr)}{dy}\,dy \\ &= \frac{1}{\pi^2 n} \int_x^{x\pm x^{1/3}} (\pi\sqrt{nz})^2 J_1(\pi\sqrt{nz})\, \frac{\pi\sqrt n}{2\sqrt z}\,dz = O(n^{1/4}x^{7/12}). \end{aligned} \]

On the basis of the estimates in [7], p. 202, we shall have

\[ \int_x^{x\pm x^{1/3}} R(y)\,dy = O\!\left(x^{7/12}\sum_{n\leq x^{1/3}}\frac{r(n)}{n^{3/4}}\right) + O\!\left(x^{3/4}\sum_{n>x^{1/3}}\frac{r(n)}{n^{5/4}}\right) = O(x^{2/3}) \]

and, carrying out further arguments analogous to [7], p. 208, we obtain the assertion of the theorem.

For the sum \(\sum_{\lambda<T}\omega^2\!\left(\lambda,\frac12,\frac12\right)\) a theorem is valid analogous to Hardy’s theorem on the number of integral points in a circle.

Theorem 3. There exists a constant \(C>0\) and a sequence of positive values \(T_i\to\infty\) for which

\[ \left| \sum_{\lambda\leq T_i}\omega^2\!\left(\lambda,\frac12,\frac12\right) - \frac{T_i}{4\pi} \right| > CT_i^{1/4}. \]

Proof. Denote

\[ Q(x)=\int_0^x R(y)\,dy = \frac{x}{\pi}\sum_{n=1}^{\infty}\frac{(-1)^n}{n}\,r(n)J_2(\pi\sqrt{nx}). \]

The series on the right-hand side converges uniformly on any interval \(0\leq x\leq z\). Therefore, taking into account the equality

\[ \frac{d\bigl(z^{3/2}J_3(l\sqrt z)\bigr)}{dz} = \frac{l}{2}zJ_2(l\sqrt z) \]

and the estimate \(|J_3(z)|<C/\sqrt z\), we shall have

\[ \int_0^z Q(x)\,dx = -\frac{z^{3/2}}{2\pi^2} \sum_{n=1}^{\infty}(-1)^n\frac{r(n)}{n^{3/2}} J_3(\pi\sqrt{nz}) = O(z^{5/4}). \]

We shall show that the assertion

\[ Q(x)=O(x^{3/4}) \tag{14} \]

is false. Using the inequality

\[ \left|J_2(z)+\sqrt{\frac{2}{\pi}}\, \frac{\cos\left(z-\frac{\pi}{4}\right)}{\sqrt z}\right| \leq \frac{C}{z^{3/2}}, \]

we obtain

\[ \frac{Q(x)}{x^{3/4}} = -\frac{\sqrt2}{\pi^2} \sum_{n=1}^{\infty}(-1)^n\frac{r(n)}{n^{5/4}} \cos\left(2\pi\sqrt{\frac{nx}{4}}-\frac{\pi}{4}\right) + O\left(\frac{1}{x^{1/2}}\right). \]

Let us apply a lemma from the theory of linear Diophantine approximations ([7], p. 133). Given a natural number \(k_i\), a real number \(\eta_i\), and a natural number \(q_i\), there exists a real number \(v_i\) \((\eta_i\leq v_i\leq q_i^{k_i}\eta_i)\) and integers \(a_1^i,a_2^i,\ldots,a_{k_i}^i\) such that

\[ |v_i\sqrt n-a_n|<\frac{1}{q_i},\qquad n=1,2,\ldots,k_i. \]

Putting \(x_i=4v_i^2\), we shall have

\[ \frac{Q(x_i)}{-x_i^{3/4}} = -\frac{1}{\pi^2} \sum_{n=1}^{k_i}(-1)^n\frac{r(n)}{n^{5/4}} + O\left(\frac{1}{q_i}\sum_{n=1}^{k_i}\frac{r(n)}{n^{5/4}}\right) + \]

\[ + O\left(\sum_{k_i+1}^{\infty}\frac{r(n)}{n^{5/4}}\right) + O\left(\frac{1}{x_i^{1/2}}\right) \]

or

\[ \frac{Q(x_i)}{x_i^{3/4}}-A = O\left(\frac{1}{q_i}\right) + O\left(\frac{1}{k_i^{1/8}}\right) + O\left(\frac{1}{x_i^{1/2}}\right), \]

where

\[ A=-\frac{1}{\pi^2} \sum_{n=1}^{\infty}\frac{(-1)^n r(n)}{n^{5/4}}. \]

From (12) it follows that

\[ A= \frac{1}{\pi^2} \sum_{n=1}^{\infty}\frac{r(n)}{n^{5/4}} \left(1-\frac{1}{2^{1/4}}\right) >0. \]

Letting \(k_i,\eta_i,q_i\) tend to infinity, we find a sequence \(x_i\to\infty\) such that

\[ \lim_{i\to\infty}\frac{Q(x_i)}{x_i^{3/4}}=A, \]

which proves that (14) is false.

Now we shall prove Theorem 3 with the aid of an elementary lemma ([7], p. 233).

Lemma. Let \(f(x)\) be a real-valued function for \(x \geq 0\) and integrable on the interval \([0,\omega]\) for every \(\omega>0\). Denote

\[ f_1(x)=\int_0^x f(y)\,dy,\qquad f_2(x)=\int_0^x f_1(y)\,dy. \]

If

\[ \varlimsup_{x\to\infty}\frac{f(x)}{x^{1/4}}>0 \quad \text{and} \quad f_2(x)=O(x^{5/4}), \quad \text{then} \]

\[ f_1(x)=o(x^{3/4}). \]

Putting \(f(x)=R(x)\), we obtain that

\[ \varliminf_{x\to\infty}\frac{R(x)}{x^{1/4}}<0. \]

The theorem is proved.

References

1. Titchmarsh E. Expansions in eigenfunctions associated with second-order differential equations; 2. Moscow, 1961.
2. Vinogradov I. M. Selected works. Moscow, 1952.
3. Vinogradov I. M. Elements of number theory. Moscow, 1949.
4. Titchmarsh E. The theory of the Riemann zeta-function. Moscow, 1953.
5. Venkov B. A. Elementary number theory. Moscow, 1937.
6. Segal B. I. Uspekhi Mat. Nauk, 1, issue 3—4, 1946.
7. Landau E. Vorlesungen über Zahlentheorie, Bd II, Leipzig, 1927.

Received by the editors
March 22, 1965

Moscow Institute of Physics and Technology