Estimation of the eigenvalues and the integral of the eigenfunctions of the Newtonian potential operator

We split the proof into two steps, in the first one we justify the result in the case of a disc and in the second step we prove the result for a general shape satisfying Hypotheses 1.

2. (1)

   The case of a disc $D$ of radius $a$.
   We recall, from [9, Theorem 4.1], that the eigenvalues of the logarithmic potential operator for a disc are given by:

   (2.1)

   $$\lambda_{k,j}=a^{2}\,\left(\mu_{j}^{(k)}\right)^{-2},\qquad k=0,1,2,\cdots\quad\text{and}\quad j=1,2,\cdots$$

   and the corresponding eigenfunctions given by

   (2.2)

   $$u_{k,j}(r,\phi)=\LARGE\textbf{J}_{k}\left(\mu_{j}^{(k)}\,\frac{r}{a}\right)\,e^{ik\phi},\qquad r\in[0,a]\quad\text{and}\quad\phi\in[0,2\pi],$$

   where $\LARGE\textbf{J}_{k}$ is the Bessel function of the first kind of order $k$ and $\mu_{j}^{(k)}$ are the roots of the following transcendental equation

   (2.3)

   $$k\,\LARGE\textbf{J}_{k}\left(\mu_{j}^{(k)}\right)+\frac{\mu_{j}^{(k)}}{2}\,\left(\LARGE\textbf{J}_{k-1}\left(\mu_{j}^{(k)}\right)-\LARGE\textbf{J}_{k+1}\left(\mu_{j}^{(k)}\right)\right)=0,\qquad k=1,2,\cdots,$$

   and, for $k=0$,

   (2.4)

   $$\LARGE\textbf{J}_{0}\left(\mu_{j}^{(0)}\right)+2\,\log(a)\,\mu_{j}^{(0)}\,\LARGE\textbf{J}_{1}\left(\mu_{j}^{(0)}\right)=0.$$

   In [5, Appendix IV, Table III], for some specific values of the parameter $a$, the authors computed the first six roots of the transcendental equation $(\ref{EquaBessel})$. It is clear, from $(\ref{EquaBessel})$, that the solution $\left\{\mu_{j}^{(0)}\right\}_{j\geq 1}$ will be dependent on the parameter $a$. To see closely how the solutions depends on $a$, we start by plotting the graph⁴⁴4These graphs were produced with Mathematica. associated to the function, with parameter $a$, defined by

   (2.5)

   $$\Psi_{a}(x):=\LARGE\textbf{J}_{0}\left(x\right)+2\,\log(a)\,x\,\LARGE\textbf{J}_{1}\left(x\right).$$

   

   From the graphs, we can see clearly that the first root, i.e $\mu_{1}^{(0)}$, is small and the other roots $\mu_{j}^{(0)}$, for $j\geq 1$, are moderate. To determine the order of smallness of $\mu_{1}^{(0)}$, we use the asymptotic behavior of $\LARGE\textbf{J}_{0}$ and $\LARGE\textbf{J}_{1}$ for small argument. More precisely, for $0<x\ll\sqrt{n+1}$, see [11, Equation 25], we have

   (2.6)

   $$\LARGE\textbf{J}_{n}\left(x\right)\sim\frac{1}{n!}\,\left(\frac{x}{2}\right)^{n}.$$

   Now, using $(\ref{BesselNearZero})$ we approximate $(\ref{EquaBessel})$ as

   $$1+\log(a)\,\left(\mu_{1}^{(0)}\right)^{2}=0,$$

   and this implies that

   $$\mu_{1}^{(0)}\;\sim\;\frac{1}{\sqrt{\left|\log(a)\right|}}.$$

   Consequently,

   $$\lambda_{0,1}\;\sim\;a^{2}\,\left|\log(a)\right|.$$

   To study the other roots, we start by setting $\alpha_{k,j}$, for $k=0,1$ and $j\in\mathbb{N}$, to be the root of order $j$ associated to the Bessel function $\LARGE\textbf{J}_{k}$. Then, thanks to Dixon’s theorem, see [15], we know that $\mu_{j}^{(0)}$ will be interlaced between the roots of $\LARGE\textbf{J}_{0}$ and the roots of $\LARGE\textbf{J}_{1}$. More precisely,

   (2.7)

   $$\mu_{1}^{(0)}<\alpha_{0,1}<\alpha_{1,1}\quad\text{and}\quad\alpha_{1,j-1}<\mu_{j}^{(0)}<\alpha_{0,j},\quad j\geq 2.$$

   Using the previous relation and knowing that $\alpha_{1,1}=3.8317$, we deduce that $\mu_{j}^{(0)}>3.8317$, for $j\geq 2$. Therefore, for $j\geq 2$, we have $\mu_{j}^{(0)}\;\sim\;1$. Hence, from $(\ref{EigValDef})$, we obtain

   $$\lambda_{0,j}\;\sim\;a^{2},\quad\text{for}\quad j\geq 2.$$

   Next, contrary to $(\ref{EquaBessel})$, because the equation $(\ref{EigValk>1})$ is independent on the parameter $a$, we deduce that $\mu_{j}^{(k)}\sim 1$. Hence, the eigenvalues $\lambda_{k,j}$ defined by $(\ref{EigValDef})$ behave as

   $$\lambda_{k,j}\;\sim\;a^{2},\quad\text{for}\quad k\geq 1\quad\text{and}\quad j\geq 1.$$

   Arranging the obtained results we get:

   (2.8)

   $$\lambda_{0,1}\;\sim\;a^{2}\,\left|\log(a)\right|\quad\text{and}\quad\lambda_{k,j}\;\sim\;a^{2},\quad\text{for}\quad(k,j)\neq(0,1).$$

   
   

   In what follows, when we talk about the first eigenvalue of the Newtonian potential operator in the disc we refer to $\lambda_{0,1}$. This is because the eigenvalues of the Newtonian operator are decreasing and, as proved by $(\ref{BehaviourEigVal})$, $\lambda_{0,1}$ is the highest one.
   Similarly to $(\ref{BehaviourEigVal})$, we derive an analogous result for the integrals of the associated eigenfunctions $u_{k,j}(\cdot,\cdot)$, defined by $(\ref{Eigfcts})$. We start with the coming computations.

   	$\displaystyle\int_{D}u_{k,j}(x)\,dx=\int_{0}^{2\pi}\int_{0}^{a}u_{k,j}(r,\phi)\,r\ dr\,d\phi$	$\displaystyle=$	$\displaystyle\int_{0}^{2\pi}\int_{0}^{a}\LARGE\textbf{J}_{k}\left(\mu_{j}^{(k)}\,\frac{r}{a}\right)\,e^{ik\phi}\,r\ dr\,d\phi$
   		$\displaystyle=$	$\displaystyle 2\,\pi\,\int_{0}^{a}\LARGE\textbf{J}_{0}\left(\mu_{j}^{(0)}\,\frac{r}{a}\right)\,r\ dr\,\,\delta_{0,k}$
   		$\displaystyle=$	$\displaystyle 2\,\pi\,\left(\frac{a}{\mu_{j}^{(0)}}\right)^{2}\,\int_{0}^{\mu_{j}^{(0)}}\LARGE\textbf{J}_{0}\left(r\right)\,r\ dr\,\,\delta_{0,k},$

   where $\delta_{\cdot,\cdot}$ is the Kronecker-symbol. Thanks to [5, Formula (5), page 206], we know that

   $$x\,\LARGE\textbf{J}_{1}\left(x\right)=\int_{0}^{x}r\,\LARGE\textbf{J}_{0}\left(r\right)\,dr.$$

   Then,

   (2.9)

   $$\int_{D}u_{k,j}(x)\,dx=2\,\pi\,\frac{a^{2}}{\mu_{j}^{(0)}}\,\LARGE\textbf{J}_{1}\left(\mu_{j}^{(0)}\right)\,\delta_{0,k}.$$

   Later, to define the normalized eigenfunctions, we compute the $\left\|u_{0,j}\right\|_{\mathbb{L}^{2}(D)}$. We have,

   	$\displaystyle\left\|u_{0,j}\right\|^{2}_{\mathbb{L}^{2}(D)}:=\int_{D}\left|u_{0,j}\right|^{2}(x)\,dx=\int_{0}^{2\,\pi}\,\int_{0}^{a}\left|u_{0,j}\right|^{2}(r,\phi)\,r\,dr\,d\phi$	$\displaystyle=$	$\displaystyle 2\,\pi\,\int_{0}^{a}\left|\LARGE\textbf{J}_{0}\left(\mu_{j}^{(0)}\,\frac{r}{a}\right)\right|^{2}\,r\,dr$
   		$\displaystyle=$	$\displaystyle 2\,\pi\,\left(\frac{a}{\mu_{j}^{(0)}}\right)^{2}\,\int_{0}^{\mu_{j}^{(0)}}\left|\LARGE\textbf{J}_{0}\left(r\right)\right|^{2}\,r\,dr.$

   Combining the previous formula with $(\ref{IntEigFcts})$, gives us a formula for the integral of the normalized eigenfunctions. More precisely,

   (2.10)

   $$\int_{D}v_{0,j}(x)\,dx:=\frac{\int_{D}u_{0,j}(x)\,dx}{\left\|u_{0,j}\right\|_{\mathbb{L}^{2}(D)}}=\dfrac{\sqrt{2\,\pi}\;a\;\LARGE\textbf{J}_{1}\left(\mu_{j}^{(0)}\right)}{\left[\int_{0}^{\mu_{j}^{(0)}}\left|\LARGE\textbf{J}_{0}\left(r\right)\right|^{2}\,r\,dr\right]^{\frac{1}{2}}}.$$

   As a result of the different behaviors of $\left\{\mu_{j}^{(0)}\right\}_{j\geq 1}$, with respect to the parameter $a$, to estimate $(\ref{IntNEigFcts})$ we split our computations into two steps.

   2. (a)

      For $j=1$, as $\mu_{1}^{(0)}$ is small, we have

      $$\int_{D}v_{0,1}(x)\,dx=\frac{\sqrt{2\,\pi}\;a\;\LARGE\textbf{J}_{1}\left(\mu_{1}^{(0)}\right)}{\left[\int_{0}^{\mu_{1}^{(0)}}\left|\LARGE\textbf{J}_{0}\left(r\right)\right|^{2}\,r\,dr\right]^{\frac{1}{2}}}\overset{(\ref{BesselNearZero})}{\sim}\frac{\sqrt{2\,\pi}\;a\;\dfrac{\mu_{1}^{(0)}}{2}}{\left[\int_{0}^{\mu_{1}^{(0)}}r\,dr\right]^{\frac{1}{2}}}=\sqrt{\pi}\,a\;\sim\;a.$$

   4. (b)

      For $j\geq 2$, as $\mu_{j}^{(0)}$ is moderate, we have

      $$\int_{D}v_{0,j}(x)\,dx=\dfrac{\sqrt{2\,\pi}\;a\;\LARGE\textbf{J}_{1}\left(\mu_{j}^{(0)}\right)}{\left[\int_{0}^{\mu_{j}^{(0)}}\left|\LARGE\textbf{J}_{0}\left(r\right)\right|^{2}\,r\,dr\right]^{\frac{1}{2}}}.$$

      By induction on the formula $(\ref{DixonResults})$, we obtain for $j\geq 2$ the following relation

      $$\alpha_{0,j-1}<\mu_{j}^{(0)}<\alpha_{0,j}.$$

      Then,

      $$\dfrac{\sqrt{2\,\pi}\;a\;\left|\LARGE\textbf{J}_{1}\left(\mu_{j}^{(0)}\right)\right|}{\left[\int_{0}^{\alpha_{0,j}}\left|\LARGE\textbf{J}_{0}\left(r\right)\right|^{2}\,r\,dr\right]^{\frac{1}{2}}}<\left|\int_{D}v_{0,j}(x)\,dx\right|<\dfrac{\sqrt{2\,\pi}\;a\;\left|\LARGE\textbf{J}_{1}\left(\mu_{j}^{(0)}\right)\right|}{\left[\int_{0}^{\alpha_{0,j-1}}\left|\LARGE\textbf{J}_{0}\left(r\right)\right|^{2}\,r\,dr\right]^{\frac{1}{2}}},$$

      or, equivalently,

      $$\dfrac{\sqrt{2\,\pi}\;a\;\left|\LARGE\textbf{J}_{1}\left(\mu_{j}^{(0)}\right)\right|}{\alpha_{0,j}\,\left[\int_{0}^{1}\left|\LARGE\textbf{J}_{0}\left(\alpha_{0,j}\,r\right)\right|^{2}\,r\,dr\right]^{\frac{1}{2}}}<\left|\int_{D}v_{0,j}(x)\,dx\right|<\dfrac{\sqrt{2\,\pi}\;a\;\left|\LARGE\textbf{J}_{1}\left(\mu_{j}^{(0)}\right)\right|}{\alpha_{0,j-1}\,\left[\int_{0}^{1}\left|\LARGE\textbf{J}_{0}\left(\alpha_{0,j-1}\,r\right)\right|^{2}\,r\,dr\right]^{\frac{1}{2}}}.$$

      Now, thanks to [5, Formula (2), page 205], we know that

      $$\int_{0}^{1}\left(\LARGE\textbf{J}_{0}\left(\alpha_{0,k}\,r\right)\right)^{2}\,r\,dr=\frac{1}{2}\,\left(\LARGE\textbf{J}_{1}\left(\alpha_{0,k}\right)\right)^{2}\sim 1,\quad\text{for}\quad k\in\mathbb{N},$$

      hence,

      $$\dfrac{2\,\sqrt{\pi}\;a\;\left|\LARGE\textbf{J}_{1}\left(\mu_{j}^{(0)}\right)\right|}{\alpha_{0,j}\,\left|\LARGE\textbf{J}_{1}\left(\alpha_{0,j}\right)\right|}<\left|\int_{D}v_{0,j}(x)\,dx\right|<\dfrac{2\,\sqrt{\pi}\;a\;\left|\LARGE\textbf{J}_{1}\left(\mu_{j}^{(0)}\right)\right|}{\alpha_{0,j-1}\,\left|\LARGE\textbf{J}_{1}\left(\alpha_{0,j-1}\right)\right|}.$$

      As $\alpha_{0,k}\sim 1$ and $\left|\LARGE\textbf{J}_{1}\left(\alpha_{0,k}\right)\right|\sim 1$, for $k=j-1$ or $k=j$, with respect to the parameter $a$, we deduce that

      $$a\;\left|\LARGE\textbf{J}_{1}\left(\mu_{j}^{(0)}\right)\right|\lesssim\left|\int_{D}v_{0,j}(x)\,dx\right|\lesssim a\;\left|\LARGE\textbf{J}_{1}\left(\mu_{j}^{(0)}\right)\right|,$$

      or, equivalently,

      $$\left|\int_{D}v_{0,j}(x)\,dx\right|\;\sim\;a\;\left|\LARGE\textbf{J}_{1}\left(\mu_{j}^{(0)}\right)\right|\overset{(\ref{EquaBessel})}{=}\frac{a\;\left|\LARGE\textbf{J}_{0}\left(\mu_{j}^{(0)}\right)\right|}{2\,\mu_{j}^{(0)}\,\left|\log(a)\right|}.$$

      The final step consist in using the fact that⁵⁵5This can be proved using the integral representation of the Bessel function $$\LARGE\textbf{J}_{0}\left(x\right):=\frac{1}{\pi}\,\int_{0}^{\pi}\cos\left(x\,\sin(\tau)\right)\,d\tau,$$ see [14, Formula 9.19, page 230]. $\left|\LARGE\textbf{J}_{0}\left(x\right)\right|\leq 1,\,\forall\,x\in\mathbb{R}$, see [15, Formula (5), page 31]. Hence,

      $$\left|\int_{D}v_{0,j}(x)\,dx\right|\;\sim\;a\;\left|\log(a)\right|^{-1}.$$

   Analogously to $(\ref{BehaviourEigVal})$, after arranging the obtained results we deduce that:

   (2.11)

   $$\int_{D}v_{0,1}(x)\,dx\;\;\sim\;\;a\quad\text{and}\quad\int_{D}v_{0,j}(x)\,dx\;\;\sim\;\;a\,\left|\log(a)\right|^{-1},\quad j\geq 2.$$

   Obviously,

   $$\int_{D}v_{k,j}(x)\,dx=0,\quad k\geq 1\quad\text{and}\quad j\geq 1.$$

4. (2)

   The case of an arbitrary shape $\Omega$.
   To estimate the behavior of the eigenvalues of the Newtonian potential operator defined over an arbitrary domain $\Omega$, we proceed in two steps.

   2. (a)

      We start by proving the property of domain monotonicity for the Newtonian potential operator: eigenvalues of the Newtonian potential operator monotonously increase when the domain is enlarged, i.e., $\lambda_{n}\left(\Omega_{1}\right)\leq\lambda_{n}\left(\Omega_{2}\right)$ if $\Omega_{1}\subset\Omega_{2}$. To accomplish this, we define an extension operator $\bm{P}$, as follows:
      

      	$\displaystyle\bm{P}:=\mathbb{L}^{2}\left(\Omega_{1}\right)$	$\displaystyle\longrightarrow$	$\displaystyle\mathbb{L}^{2}\left(\Omega_{2}\right)$
      (2.12)		$\displaystyle f(\cdot)$	$\displaystyle\longrightarrow$	$\displaystyle\bm{P}\left(f\right)(\cdot):=\tilde{f}(\cdot):=f(\cdot)\;\underset{\Omega_{1}}{\chi}(\cdot)\;+\;0\;\underset{\Omega_{2}\setminus\Omega_{1}}{\chi}(\cdot),$

      where we have assumed that $\Omega_{1}\subset\Omega_{2}$. Manifestly, we have the following injection

      (2.13)

      $$\mathbb{L}^{2}\left(\Omega_{1}\right)\hookrightarrow\mathbb{L}^{2}\left(\Omega_{2}\right).$$

      In similar manner we define the extension of a function represented by the Newtonian operator

      $$u(x)=\int_{\Omega_{1}}\Phi_{0}(x,y)\,f(y)\,dy,\quad x\in\Omega_{1},$$

      to the domain $\Omega_{2}$, as follows

      $$\tilde{u}(x)=\int_{\Omega_{2}}\Phi_{0}(x,y)\,\tilde{f}(y)\,dy,\quad x\in\Omega_{2},$$

      where $\tilde{f}(\cdot)$ is the extension of $f(\cdot)$, with zero in $\Omega_{2}\setminus\Omega_{1}$, defined by $(\ref{Extension})$. Now, from the min-max principle applied to the Newtonian potential operator, see [12, Theorem 4, page 318-319], we have⁶⁶6In $(\ref{min-max})$, the positivity of the operator $N_{\Omega_{1}}(\cdot)$ ensures that all eigenvalues are non-negative.

      (2.14)

      $$\lambda_{k}\left(\Omega_{1}\right)=\underset{\Xi_{k}}{Sup}\quad\underset{u\in\left(\Xi_{k}\cap\mathbb{L}^{2}(\Omega_{1})\right)}{Inf}\quad\dfrac{\int_{\Omega_{1}}N_{\Omega_{1}}\left(u\right)(x)u(x)\;dx}{\int_{\Omega_{1}}\left|u\right|^{2}(x)\;dx},$$

      where $\Xi_{k}\subset\mathbb{L}^{2}(\mathbb{R}^{2})$ is $k$ dimensional subspace. Now, using the injection $(\ref{Injection})$ we deduce that:

      (2.15)

      $$\lambda_{k}\left(\Omega_{1}\right)\leq\underset{\Xi_{k}}{Sup}\quad\underset{u\in\left(\Xi_{k}\cap\mathbb{L}^{2}(\Omega_{2})\right)}{Inf}\quad\dfrac{\int_{\Omega_{1}}N_{\Omega_{1}}\left(u\right)(x)u(x)\;dx}{\int_{\Omega_{1}}\left|u\right|^{2}(x)\;dx}.$$

      Exploiting the properties of the extension function, defined by $(\ref{Extension})$, we obtain

      $$\int_{\Omega_{1}}N_{\Omega_{1}}\left(u\right)(x)u(x)\;dx=\int_{\Omega_{2}}N_{\Omega_{2}}\left(\tilde{u}\right)(x)\tilde{u}(x)\;dx\quad\text{and}\quad\int_{\Omega_{1}}\left|u\right|^{2}(x)\;dx=\int_{\Omega_{2}}\left|\tilde{u}\right|^{2}(x)\;dx.$$

      Hence, $(\ref{Black})$ becomes,

      $$\lambda_{k}\left(\Omega_{1}\right)\leq\underset{\Xi_{k}}{Sup}\quad\underset{\tilde{u}\in\left(\Xi_{k}\cap\mathbb{L}^{2}(\Omega_{2})\right)}{Inf}\quad\dfrac{\int_{\Omega_{2}}N_{\Omega_{2}}\left(\tilde{u}\right)(x)\tilde{u}(x)\;dx}{\int_{\Omega_{2}}\left|\tilde{u}\right|^{2}(x)\;dx}=\lambda_{k}\left(\Omega_{2}\right).$$

      Finally,

      (2.16)

      $$\text{if}\quad\Omega_{1}\subset\Omega_{2}\quad\text{we obtain}\quad\lambda_{k}\left(\Omega_{1}\right)\leq\lambda_{k}\left(\Omega_{2}\right),$$

      and this ends the proof of the monotonicity property for the eigenvalues of the Newtonian potential operator.

   4. (b)

      At this stage, we use the proved monotonicity property, given by $(\ref{Monotonicity})$, by assuming the existence of two discs $B_{1}:=B(z_{1},\rho_{1})$ and $B_{2}:=B(z_{2},\rho_{2})$, where $z_{j}\in\Omega$ and $\rho_{j}>0$, for $j=1,2$, such that:

      $$B_{1}\subset\Omega\subset B_{2}\quad\text{and}\quad\left|B_{j}\right|\sim a^{2},\;\;j=1,2.$$

      Thanks to $(\ref{Monotonicity})$ we derive, from the previous formula, the following relation,

      (2.17)

      $$\lambda_{k}\left(B_{1}\right)\leq\lambda_{k}\left(\Omega\right)\leq\lambda_{k}\left(B_{2}\right).$$

      
      

      Now, by combining $(\ref{B1OmegaB2})$ with $(\ref{BehaviourEigVal})$, we derive the following behavior of the eigenvalues of the Newtonian potential operator, defined over an arbitrary shape $\Omega$.

      (2.18)

      $$\lambda_{0}\left(\Omega\right)\,\sim\,a^{2}\,\left|\log(a)\right|\quad\text{and}\quad\lambda_{k}\left(\Omega\right)\,\sim\,a^{2},\quad\text{for}\quad k\geq 1.$$

   Regarding the estimation of the integral of the eigenfunction $f_{n}$, of the Newtonian potential operator defined over $\Omega$, i.e.

   (2.19)

   $$N_{\Omega}\left(f_{n}\right)=\lambda_{n}(\Omega)\;f_{n},\,\quad\text{in}\quad\Omega,$$

   and as suggested by the behavior of the eigenvalues $\left\{\lambda_{n}\left(\Omega\right)\right\}_{n\geq 0}$, i.e. $(\ref{EstimationEigOmega})$, we split the study into two cases.

   2. i)

      For $n=0$, we know from $(\ref{EstimationEigOmega})$, that $\lambda_{0}(\Omega)=\beta_{0}\,a^{2}\,\left|\log(a)\right|$, where $\beta_{0}$ is a positive constant independent on the parameter $a$. Moreover, after scaling the equation $(\ref{AddedEqua})$, from $\Omega$ to $\Omega^{\star}$, and taking the inner product with respect to $\tilde{f}_{n}$, we end up with

      $$\left\langle\tilde{f}_{n};N_{\Omega^{\star}}\left(\tilde{f}_{n}\right)\right\rangle_{\mathbb{L}^{2}(\Omega^{\star})}=\frac{\lambda_{n}(\Omega)}{a^{2}}\;\left\|\tilde{f}_{n}\right\|_{\mathbb{L}^{2}(\Omega^{\star})}+\frac{1}{2\,\pi}\,\log(a)\,\left(\int_{\Omega^{\star}}\tilde{f}_{n}\right)^{2}.$$

      Next, we set $\overline{f}_{n}:=\dfrac{\tilde{f}_{n}}{\left\|\tilde{f}_{n}\right\|_{\mathbb{L}^{2}\left(\Omega^{\star}\right)}}$ and $\tilde{\lambda}_{n}:=\left\langle\tilde{f}_{n};N_{\Omega^{\star}}\left(\tilde{f}_{n}\right)\right\rangle_{\mathbb{L}^{2}(\Omega^{\star})}$ we obtain from the previous equation

      (2.20)

      $$\tilde{\lambda}_{n}=\frac{\lambda_{n}(\Omega)}{a^{2}}+\frac{1}{2\,\pi}\,\log(a)\,\left(\int_{\Omega^{\star}}\overline{f}_{n}(x)\,dx\right)^{2},$$

      where $\Omega^{\star}=\dfrac{-z+\Omega}{a},\,\left|\Omega^{\star}\right|\,\sim\,1$ and $\left\{\tilde{\lambda}_{n}\right\}_{n\geq 0}$ is a positive and uniformly bounded sequence, i.e. $0<\tilde{\lambda}_{n}\leq\underset{n}{Sup}\;\tilde{\lambda}_{n}<C^{te}$, where $C^{te}$ is a uniformly constant. Using the scale of $\lambda_{0}(\Omega)$ and the previous formula give us

      (2.21)

      $$\tilde{\lambda}_{0}=\frac{\beta_{0}\,a^{2}\,\left|\log(a)\right|}{a^{2}}+\frac{1}{2\,\pi}\,\log(a)\,\left(\int_{\Omega^{\star}}\overline{f}_{0}(x)\,dx\right)^{2},$$

      and this implies

      (2.22)

      $$\left|\int_{\Omega^{\star}}\overline{f}_{0}(x)\,dx\right|=\sqrt{2\,\pi\,\left(\beta_{0}-\frac{\tilde{\lambda}_{0}}{\left|\log(a)\right|}\right)}\;\;\sim\;\;1.$$

   4. ii)

      For $n\geq 1$, we know from $(\ref{EstimationEigOmega})$, that $\lambda_{n}(\Omega)=\beta_{n}\,a^{2}\,$, where $\beta_{n}$ is a positive constant independent on the parameter $a$. Hence, using $(\ref{Formulafrom2DWork})$, we obtain

      $$\tilde{\lambda}_{n}=\frac{\beta_{n}\,a^{2}}{a^{2}}+\frac{1}{2\,\pi}\,\log(a)\,\left(\int_{\Omega^{\star}}\overline{f}_{n}(x)\,dx\right)^{2},$$

      and, knowing that $\tilde{\lambda}_{n}>0$,

      $$\left(\int_{\Omega^{\star}}\overline{f}_{n}(x)\,dx\right)^{2}\leq\beta_{n}\;2\,\pi\;\left|\log(a)\right|^{-1}.$$

      This implies,

      $$\left|\int_{\Omega^{\star}}\overline{f}_{n}(x)\,dx\right|\;\;\lesssim\;\;\left|\log(a)\right|^{-\frac{1}{2}}.$$

      Next, we show that the previous obtained estimation can be improved to be of order $\left|\log(a)\right|^{-1}$, instead of $\left|\log(a)\right|^{-\frac{1}{2}}$. Now, after scaling the equation $(\ref{AddedEqua})$ and integrating again the obtained equation we obtain

      (2.23)

      $$\int_{\Omega^{\star}}\overline{f}_{n}(x)\,dx=\frac{\int_{\Omega^{\star}}N_{\Omega^{\star}}\left(1\right)(x)\overline{f}_{n}(x)\,dx}{\left[\frac{\lambda_{n}\left(\Omega\right)}{a^{2}}-\frac{1}{2\,\pi}\,\left|\log(a)\right|\,\left|\Omega^{\star}\right|\right]},$$

      and knowing that $\lambda_{n}\left(\Omega\right)=\beta_{n}\,a^{2}$ we get

      $$\int_{\Omega^{\star}}\overline{f}_{n}(x)\,dx=\frac{\int_{\Omega^{\star}}N_{\Omega^{\star}}\left(1\right)(x)\overline{f}_{n}(x)\,dx}{\left[\beta_{n}-\frac{1}{2\,\pi}\,\left|\log(a)\right|\,\left|\Omega^{\star}\right|\right]}.$$

      By taking the modulus, in both sides, of the previous relation, using the fact that $\left\|N_{\Omega^{\star}}\right\|_{\mathcal{L}\left(\mathbb{L}^{2}(\Omega^{\star});\mathbb{L}^{2}(\Omega^{\star})\right)}\;\sim\;1$ and recalling that $\overline{f}_{n}(\cdot)$ are orthonormalized eigenfunctions in $\mathbb{L}^{2}(\Omega^{\star})$, we deduce that

      $$\left|\int_{\Omega^{\star}}\overline{f}_{n}(x)\,dx\right|\;\lesssim\;\left|\log(a)\right|^{-1}.$$

   Finally, correspondingly to $(\ref{IranProtests})$ and for an arbitrary shape domain $\Omega$, we obtain after rescaling back to $\Omega$ the following behaviour of the integral of the eigenfunctions of the Newtonian potential operator with respect to the parameter $a$.

   (2.24)

   $$\int_{\Omega}f_{0}(x)\,dx\;\;\sim\;\;a\quad\text{and}\quad|\int_{\Omega}f_{n}(x)\,dx|\;\;\lesssim\;\;a\,\left|\log(a)\right|^{-1},\quad\text{for}\;\;n\geq 1.$$