积分-微分算子$L:=L\left(q, M, h, H\right)$定义如下

(1.1)$\begin{equation}\label{E1.1} Ly:=-y''+q(x)y+\int_{0}^{x}M(x, t)y(t) {\rm d}t=\lambda y, \quad x\in(0, \pi), \\ \end{equation} $

满足边界条件

(1.2)$y'\left(0, \lambda\right)-hy\left(0, \lambda\right)=0, $

(1.3)$y'\left(\pi, \lambda\right)+Hy\left(\pi, \lambda\right)=0, $

其中$h, H\in \mathbb{R} \cup\{\infty\}$, $q(x)$和$M(x, t)$是实值函数, $q\in L^2\left[0, \pi\right]$, $M\in L^2(D_0)$, $D_0:=\{(x, t):0\leq t\leq x\leq \pi, x, t\in \mathbb{R} \}$.当$h=\infty, H\in \mathbb{R} $时,称问题(1.1)-(1.3)为Dirichlet-Robin边界条件下积分-微分算子；当$h=H=\infty$时,称问题(1.1)-(1.3)为Dirichlet边界条件下积分-微分算子.

在$M(x, t)$已知的条件下, Kuryshova^[6]讨论了积分-微分算子$L$的逆谱问题及其稳定性; Kuryshova和谢忠村^[7]研究了积分-微分算子$L$的逆结点问题.当$M(x, t)=M(x-t)$时, Yurko^[1]研究了Neumann边界条件$y'(0, \lambda)=y'(\pi, \lambda)=0$下的积分-微分算子$L$的逆谱问题,在$q(x)$已知的条件下,用一组谱重构了$M(x)$；Buterin^[2-3]研究了积分-微分算子$L$的逆谱问题,并建立了$M(x)$的唯一性定理及逆谱问题的可解性、稳定性理论.王玉华和魏广生^[4]用混合谱数据研究了积分-微分算子的逆谱问题.在$M\in W_2^{1}(D_0)$条件下,王於平^[5]证明了积分-微分算子$L$稠定的结点集能够唯一确定$[0, \pi]$上的势函数$q(x)$和积分扰动核$M(x)$,并且给出了这个逆结点问题的解的重构算法.

经典的Sturm-Liouville算子的谱理论也得到深入的研究^[8-14].特别地, 1953年, Gelfand和Levitan^[14]研究了下面的Sturm-Liouville问题

$\begin{eqnarray*} \left\{\begin{array}{ll} -y''+q(x)y=\lambda y, \quad x\in(0, \pi), \\ y'(0)=y'(\pi)=0, \end{array} \right. \end{eqnarray*} $

其中$q\in C\left[0, \pi\right]$.他们得到了Neumann边界条件下Sturm-Liouville算子的迹公式

$\begin{eqnarray*} \sum\limits_{n=0}^{\infty}\left(\lambda_{n}-n^{2}-\frac{1}{\pi}\int_{0}^{\pi}q(x){\rm d}x\right)= \frac{1}{4}\left(q(0)+q(\pi)\right)-\frac{1}{2\pi}\int_{0}^{\pi}q(x){\rm d}x. \end{eqnarray*} $

他们还得到Dirichlet边界条件下Sturm-Liouville算子的迹公式如下

(1.4)$ \begin{equation}\label{E1.4} \sum\limits_{n=1}^{\infty}\left(\lambda_{n}-n^{2}-\frac{1}{\pi}\int_{0}^{\pi}q(x){\rm d}x\right)= -\frac{1}{4}\left(q(0)+q(\pi)\right)+\frac{1}{\pi}\int_{0}^{\pi}q(x){\rm d}x. \end{equation} $

之后,不少学者对迹公式产生了浓厚的兴趣,得到一些有意义的结果^[15-18].当$h, H\in \mathbb{R} $时,王於平等^[18]得到了积分-微分算子(1.1)-(1.3)的迹公式如下

$ \begin{eqnarray*} \sum\limits_{n=0}^{\infty}\left(\lambda_n-n^2-\frac{2\omega}{\pi}\right)=\frac{1}{4}\left(q(0)+q(\pi)\right)-\frac{\omega}{\pi} +\frac{1}{2}\int_{0}^{\pi}M(t, t){\rm d}t-\frac{h^2+H^2}{2}, \end{eqnarray*} $

其中$\omega=h+H+\frac{1}{2}\int_{0}^{\pi}q(x){\rm d}x$.

本文研究Dirichlet-Robin边界条件和Dirichlet边界条件下积分-微分算子的迹公式.由于研究积分-微分算子迹公式的需要,在这篇文章中,我们还要假设$q\in C\left[0, \pi\right]$, $M\in W_2^{1}(D_0)$.

记$\lambda=\rho^{2}$, $\tau=\left|{\rm Im}\rho\right|.$根据文献[5],当$h=\infty$时,满足初始条件$\varphi(0, \lambda)=0$, $\varphi'(0, \lambda)=1$,方程(1.1)的解$\varphi(x, \lambda)$表示为

(2.1)$ \begin{equation} \varphi\left(x, \lambda\right)=\frac{1}{\rho}\sin (\rho x) +\int_{0}^{x}\varphi\left(t, \lambda\right)\left[\frac{\sin (\rho(x-t))}{\rho}q(t)+\int_{t}^{x}M(\xi, t)\frac{\sin (\rho(x-\xi))}{\rho}{\rm d}\xi\right]{\rm d}t. \end{equation} $

当$M\in W_{2}^{1}(D_0)$时,我们得到方程(1.1)的解$\varphi(x, \lambda)$的渐近式,这个引理在本文主要结果的证明过程中起着重要的作用.

引理2.1  当$M\in W_{2}^{1}(D_0)$时,则$\varphi(x, \lambda)$满足下面渐近式

(2.2)$\begin{eqnarray}\label{E2.2} \varphi(x, \lambda)&=&\frac{1}{\rho}\sin(\rho x)-\frac{\cos(\rho x)}{2\rho^2}\int_{0}^{x}q(t){\rm d}t+\frac{\cos(\rho x)}{2\rho^2}\int_{0}^{x}q(t)\cos(2\rho t){\rm d}t\nonumber\\ && +\frac{\sin(\rho x)}{2\rho^2}\int_{0}^{x}q(t)\sin(2\rho t){\rm d}t-\frac{\sin(\rho x)}{2\rho^3}\int_{0}^{x}M(t, t){\rm d}t \\&&-\frac{\sin(\rho x)}{8\rho^3}\left(\int_{0}^{x}q(t){\rm d}t\right)^2 +o\left(\frac{{\rm e}^{\tau x}}{\left|\rho\right|^{3}}\right). \end{eqnarray} $

设$\lambda_{n}(n=0, 1, 2, \cdots)$是Dirichlet-Robin边界条件下积分-微分算子的特征值,则对于充分大的$n$, $\lambda_{n}$满足下面的渐近式

(2.3)$ \begin{equation}\label{E2.3}\rho_{n}=\sqrt{\lambda_{n}}=n+\frac{1}{2}+\frac{\omega_1}{n\pi}+\frac{\kappa_{n, 1}}{n}, \quad H\in\mathbb{R} , \end{equation}$

其中$\omega_1=H+\frac{1}{2}\int_{0}^{\pi}q(x){\rm d}x$, $\{\kappa_{n, 1}\}\in l^2$.

设$\lambda_{n}(n=1, 2, \cdots)$是Dirichlet边界条件下积分-微分算子的特征值,则对于充分大的$n$, $\lambda_{n}$满足下面的渐近式

(2.4)$\begin{equation}\label{E2.4}\rho_{n}=\sqrt{\lambda_{n}}=n+\frac{\int_{0}^{\pi}q(t){\rm d}t}{n\pi}+\frac{\kappa_{n, 2}}{n}, \quad H=\infty, \end{equation}$

其中$\{\kappa_{n, 2}\}\in l^2$.关于(2.3)和(2.4)式的证明,读者可以参考文献[7].

本文的主要结果是两类边界条件下积分-微分算子$L$的迹公式,即

定理2.1  Dirichlet-Robin边界条件下积分-微分算子$L$的迹公式为

(2.5)$\begin{equation}\label{E2.5}\sum\limits_{n=0}^{\infty}\left[\lambda_{n}-\left(n+\frac{1}{2}\right)^{2}-\frac{2\omega_1}{\pi}\right]\\=-\frac{1}{4}\left(q(0)-q(\pi)\right)-\frac{\omega_1}{\pi}+\frac{1}{2}\int_{0}^{\pi}M(t, t){\rm d}t-\frac{H^{2}}{2}.\end{equation}$

定理2.2  Dirichlet边界条件下积分-微分算子$L$的迹公式为

(2.6)$\begin{equation}\label{E2.6}\sum\limits_{n=1}^{\infty}\left(\lambda_{n}-n^2-\frac{\int_{0}^{\pi}q(t){\rm d}t}{\pi}\right)=-\frac{1}{4}\left(q(0)+q(\pi)\right)+\frac{1}{\pi}\int_{0}^{\pi}q(t){\rm d}t+\frac{1}{2}\int_{0}^{\pi}M(t, t){\rm d}t.\end{equation}$

显然, Dirichlet边界条件下积分-微分算子$L$的迹公式(2.6)是Dirichlet边界条件下Sturm-Liouville算子的迹公式(1.4)的本质推广.

本文主要结论的证明如下.

引理2.1的证明  由(2.1)式,得

(2.7)$\begin{equation}\label{E2.7}\varphi(x, \lambda)=\frac{1}{\rho}\sin\left(\rho x\right)+O\left(\frac{{\rm e}^{\tau x}}{\rho^2}\right).\end{equation}$

由(2.7)式,可得

(2.8)$\begin{eqnarray}\label{E2.8}&&\int_{0}^{x}\varphi(t, \lambda)\frac{\sin(\rho(x-t))}{\rho}q(t){\rm d}t \\&=&\int_{0}^{x}\left(\frac{1}{\rho}\sin\left(\rho t\right)+O\left(\frac{{\rm e}^{\tau t}}{\rho^2}\right)\right)\frac{\sin(\rho(x-t))}{\rho}q(t){\rm d}t \\&=&\frac{1}{\rho^2}\int_{0}^{x}q(t)\cdot\frac{1}{2}\left[\cos\left(2\rho t-\rho x\right)-\cos(\rho x)\right]{\rm d}t+O\left(\frac{{\rm e}^{\tau x}}{\rho^3}\right) \\&=&\frac{1}{2\rho^2}\int_{0}^{x}q(t)\cos\left(2\rho t-\rho x\right){\rm d}t-\frac{1}{2\rho^2}\int_{0}^{x}q(t)\cos(\rho x){\rm d}t+O\left(\frac{{\rm e}^{\tau t}}{\rho^2}\right)+O\left(\frac{{\rm e}^{\tau x}}{\rho^3}\right) \\&=&\frac{1}{2\rho^2}\int_{0}^{x}q(t)\left(\cos(2\rho t)\cos(\rho x)+\sin(2\rho t)\sin(\rho x)\right){\rm d}t-\frac{\cos(\rho x)}{2\rho^2}\int_{0}^{x}q(t){\rm d}t+O\left(\frac{{\rm e}^{\tau x}}{\rho^3}\right) \\&=&\frac{\cos(\rho x)}{2\rho^2}\int_{0}^{x}q(t)\cos(2\rho t){\rm d}t+\frac{\sin(\rho x)}{2\rho^2}\int_{0}^{x}q(t)\sin(2\rho t){\rm d}t \\&&-\frac{\cos(\rho x)}{2\rho^2}\int_{0}^{x}q(t){\rm d}t+O\left(\frac{{\rm e}^{\tau x}}{\rho^3}\right).\end{eqnarray}$

因为$M\in W_2^{1}(D_0)$,所以通过分部积分,得

(2.9)$\begin{eqnarray}\label{E2.9}&&\int_{t}^{x}M(\xi, t)\frac{\sin(\rho(x-\xi))}{\rho}{\rm d}\xi \\&=&\frac{1}{\rho^{2}}M(\xi, t)\cos(\rho(x-\xi))\biggr|_{t}^{x}-\frac{1}{\rho^{2}}\int_{t}^{x}M_{\xi}'(\xi, t)\cos(\rho(x-\xi)){\rm d}\xi \\&=&\frac{1}{\rho^{2}}M(x, t)-\frac{1}{\rho^{2}}M(t, t)\cos(\rho(x-t))-\frac{1}{\rho^{2}}\int_{t}^{x}M_{\xi}'(\xi, t)\cos(\rho(x-\xi)){\rm d}\xi.\end{eqnarray}$

由(2.9)式,得

(2.10)$\begin{eqnarray}\label{E2.10}&&\int_{0}^{x}\varphi(t, \lambda)\left(\int_{t}^{x}M(\xi, t)\frac{\sin(\rho(x-\xi))}{\rho}{\rm d}\xi\right) {\rm d}t\nonumber\\&=&\int_{0}^{x}\frac{1}{\rho}\sin(\rho t)\left[\frac{1}{\rho^{2}}M(x, t)-\frac{1}{\rho^{2}}M(t, t)\cos(\rho(x-t))\right]{\rm d}t+o\left(\frac{{\rm e}^{\tau x}}{\rho^3}\right)\nonumber\\&=&\frac{1}{\rho^3}\int_{0}^{x}M(x, t)\sin(\rho t){\rm d}t-\frac{1}{\rho^3}\int_{0}^{x}M(t, t)\cdot\frac{1}{2}\left[\sin(\rho x)+\sin(2\rho t-\rho x)\right]{\rm d}t+o\left(\frac{{\rm e}^{\tau x}}{\rho^3}\right)\nonumber\\&=&\frac{1}{\rho^3}\int_{0}^{x}M(x, t)\sin(\rho t){\rm d}t-\frac{1}{2\rho^{3}}\int_{0}^{x}M(t, t)\sin(\rho x){\rm d}t\nonumber\\&&-\frac{1}{2\rho^{3}}\int_{0}^{x}M(t, t)\sin(\rho(2t-x)){\rm d}t+o\left(\frac{{\rm e}^{\tau x}}{\rho^3}\right).\end{eqnarray}$

因此(2.8)和(2.10)式证明下式成立

(2.11)$\begin{eqnarray}\label{E2.11}\varphi(x, \lambda)&=&\frac{1}{\rho}\sin(\rho x)-\frac{\cos(\rho x)}{2\rho^2}\int_{0}^{x}q(t){\rm d}t+\frac{\cos(\rho x)}{2\rho^2}\int_{0}^{x}q(t)\cos(2\rho t){\rm d}t\nonumber\\&&+\frac{\sin(\rho x)}{2\rho^2}\int_{0}^{x}q(t)\sin(2\rho t){\rm d}t+O\left(\frac{{\rm e}^{\tau x}}{\rho^3}\right).\end{eqnarray}$

再次将(2.11)式代入(2.1)式,通过计算,得$\varphi(x, \lambda)$的表达式为

(2.12)$\begin{eqnarray}\label{E2.12}\varphi(x, \lambda)&=&\frac{1}{\rho}\sin(\rho x)-\frac{\cos(\rho x)}{2\rho^2}\int_{0}^{x}q(t){\rm d}t+\frac{\cos(\rho x)}{2\rho^2}\int_{0}^{x}q(t)\cos(2\rho t){\rm d}t\nonumber\\&&+\frac{\sin(\rho x)}{2\rho^2}\int_{0}^{x}q(t)\sin(2\rho t){\rm d}t-\frac{\sin(\rho x)}{2\rho^3}\int_{0}^{x}M(t, t){\rm d}t\nonumber\\&&-\frac{\sin(\rho x)}{8\rho^3}\left(\int_{0}^{x}q(t){\rm d}t\right)^2+o\left(\frac{{\rm e}^{\tau x}}{\left|\rho\right|^{3}}\right).\end{eqnarray}$

证毕.

定理2.1的证明  由引理2.1中的(2.2)式,得

(2.13)$\begin{eqnarray}\label{E2.13}\varphi'(x, \lambda)&=&\cos(\rho x)-\frac{\sin(\rho x)}{2\rho}\int_{0}^{x}q(t)\cos(2\rho t){\rm d}t+\frac{\cos(\rho x)}{2\rho}\int_{0}^{x}q(t)\sin(2\rho t){\rm d}t\nonumber\\&&+\frac{\sin(\rho x)}{2\rho}\int_{0}^{x}q(t){\rm d}t-\frac{\cos(\rho x)}{2\rho^2}\int_{0}^{x}M(t, t){\rm d}t \\&&-\frac{\cos(\rho x)}{8\rho^2}\left(\int_{0}^{x}q(t){\rm d}t\right)^2+o\left(\frac{{\rm e}^{\tau x}}{|\rho|^{2}}\right).\end{eqnarray}$

定义Dirichlet-Robin边界条件下积分-微分算子$L$特征值的示性函数为

$w(\lambda)=\varphi'(\pi, \lambda)+H\varphi(\pi, \lambda).$

由(2.2)和(2.13)式,得

(2.14)$\begin{eqnarray}\label{E2.14}w(\lambda)&=&\varphi'(\pi, \lambda)+H\varphi(\pi, \lambda)\nonumber\\&=&\cos(\rho\pi)-\frac{\sin(\rho\pi)}{2\rho}\int_{0}^{\pi}q(t)\cos(2\rho t){\rm d}t+\frac{\cos(\rho \pi)}{2\rho}\int_{0}^{\pi}q(t)\sin(2\rho t){\rm d}t\nonumber\\&&+\frac{\sin(\rho\pi)}{2\rho}\int_{0}^{\pi}q(t){\rm d}t-\frac{\cos(\rho\pi)}{2\rho^2}\int_{0}^{\pi}M(t, t){\rm d}t-\frac{\cos(\rho \pi)}{8\rho^2}\left(\int_{0}^{\pi}q(t){\rm d}t\right)^2\nonumber\\&&+\frac{H}{\rho}\sin(\rho\pi)-\frac{H\cos(\rho\pi)}{2\rho^2}\int_{0}^{\pi}q(t){\rm d}t+o\left(\frac{{\rm e}^{\tau\pi}}{\left|\rho\right|^{2}}\right).\end{eqnarray}$

则$\lambda_{n}\left(n=0, 1, 2, \cdots\right)$为$w(\lambda)$的零点.令$w_{0}(\lambda)=\cos\rho\pi$,则

(2.15)$\begin{eqnarray}\label{E2.15}\frac{w(\lambda)}{w_{0}(\lambda)}&=&1-\frac{\tan(\rho\pi)}{2\rho}\int_{0}^{\pi}q(t)\cos(2\rho t){\rm d}t+\frac{1}{2\rho}\int_{0}^{\pi}q(t)\sin(2\rho t){\rm d}t\nonumber\\&&+\frac{\tan(\rho\pi)}{2\rho}\int_{0}^{\pi}q(t){\rm d}t-\frac{1}{2\rho^2}\int_{0}^{\pi}M(t, t){\rm d}t-\frac{1}{8\rho^2}\left(\int_{0}^{\pi}q(t){\rm d}t\right)^2\nonumber\\&&+\frac{H}{\rho}\tan(\rho\pi)-\frac{H}{2\rho^2}\int_{0}^{\pi}q(t){\rm d}t+o\left(\frac{{\rm e}^{\tau\pi}}{\left|\rho\right|^{2}\cos(\rho\pi)}\right).\end{eqnarray}$

记$\Gamma_{N_{0}}:=\{\rho:\left|\rho\right|=N_{0}+\frac{3}{4}\}$, $G_{\delta}=\{\rho :\left|\rho-\left(k+\frac{1}{2}\right)\right|\geq\delta, k=0, \pm1, \pm2, \cdots\}$,其中$N_{0}$充分大, $\delta>0$充分小.则

$\begin{eqnarray*}\left|\cos\left(\rho\pi\right)\right|\geq C_{\delta}{\rm e}^{\tau\pi}, \quad \lambda\in C_{\delta}, \quad \left|\rho\right|\geq\rho^{*}, \end{eqnarray*}$

其中$\rho^{*}=\rho^{*}\left(\delta\right)$充分大, $C_{\delta}$连续地依赖于$\delta$.因此

(2.16)$\begin{equation}\label{E2.16}\left|o\left(\frac{{\rm e}^{\tau\pi}}{\left|\rho\right|^{2}\cos(\rho\pi)}\right)\right|=o\left(\frac{1}{\left|\rho\right|^2}\right), \quad \left|\rho\right|\geq\rho^{*}.\end{equation}$

令$\mu_{j}=j+\frac{1}{2}\left(j=0, \pm1, \pm2, \cdots\right)$是$w_0(\lambda)$的零点.则由留数定理,得

(2.17)$\begin{equation}\label{E2.17}\frac{1}{2\pi{\rm i}}\oint_{\Gamma_{N_{0}}}\lambda\left[\frac{w'(\lambda)}{w(\lambda)}-\frac{w_{0}'(\lambda)}{w_{0}(\lambda)}\right]{\rm d}\lambda=\sum\limits_{n=0}^{N_{0}}\left[2\lambda_{n}-2\left(n+\frac{1}{2}\right)^2\right].\end{equation}$

另一方面,由分部积分,得

(2.18)$\begin{eqnarray}\label{E2.18}\frac{1}{2\pi{\rm i}}\oint_{\Gamma_{N_{0}}}\lambda\left[\frac{w'(\lambda)}{w(\lambda)}-\frac{w_{0}'(\lambda)}{w_{0}(\lambda)}\right]{\rm d}\lambda&=&\frac{1}{2\pi{\rm i}}\oint_{\Gamma_{N_{0}}}\lambda {\rm d}\ln\left(\frac{w(\lambda)}{w_{0}(\lambda)}\right) \\&=&-\frac{1}{2\pi{\rm i}}\oint_{\Gamma_{N_{0}}}2\rho \ln\left(\frac{w(\lambda)}{w_{0}(\lambda)}\right){\rm d}\rho.\end{eqnarray}$

由(2.17)和(2.18)式,得

(2.19)$\begin{equation}\label{E2.19}\sum\limits_{n=0}^{N_{0}}\left[2\lambda_{n}-2\left(n+\frac{1}{2}\right)^2\right]=-\frac{1}{2\pi{\rm i}}\oint_{\Gamma_{N_{0}}}2\rho \ln\left(\frac{w(\lambda)}{w_{0}(\lambda)}\right){\rm d}\rho.\end{equation}$

根据泰勒展式

$\ln(1+x)=x-\frac{1}{2}x^2+o(x^2), $

则由(2.15)式,得

(2.20)$\begin{eqnarray}\label{E2.20}\ln\left(\frac{w(\lambda)}{w_0(\lambda)}\right)&=&-\frac{\tan(\rho\pi)}{2\rho}\int_{0}^{\pi}q(t)\cos(2\rho t){\rm d}t+\frac{1}{2\rho}\int_{0}^{\pi}q(t)\sin(2\rho t){\rm d}t\nonumber\\&&+\frac{\tan(\rho\pi)}{2\rho}\int_{0}^{\pi}q(t){\rm d}t-\frac{1}{2\rho^2}\int_{0}^{\pi}M(t, t){\rm d}t-\frac{1}{8\rho^2}\left(\int_{0}^{\pi}q(t){\rm d}t\right)^2\nonumber\\&&+\frac{H}{\rho}\tan(\rho\pi)-\frac{H}{2\rho^2}\int_{0}^{\pi}q(t){\rm d}t-\frac{1}{2}\cdot\frac{\tan^{2}\rho\pi}{4\rho^2}\left(\int_{0}^{\pi}q(t){\rm d}t\right)^2\nonumber\\&&-\frac{1}{2}\cdot\frac{H^2}{\rho^2}\cdot\tan^{2}(\rho\pi)-\frac{1}{2}\cdot\frac{H\tan^{2}(\rho\pi)}{\rho^2}\int_{0}^{\pi}q(t){\rm d}t+o\left(\frac{1}{\left|\rho\right|^2}\right).\end{eqnarray}$

所以(2.20)式意味着下式成立

(2.21)$\begin{eqnarray}\label{E2.21}2\rho \ln\left(\frac{w(\lambda)}{w_{0}(\lambda)}\right)&=&-\tan(\rho\pi)\int_{0}^{\pi}q(t)\cos(2\rho t){\rm d}t+\int_{0}^{\pi}q(t)\sin(2\rho t){\rm d}t\nonumber\\&&+\tan(\rho\pi)\int_{0}^{\pi}q(t){\rm d}t-\frac{1}{\rho}\int_{0}^{\pi}M(t, t){\rm d}t-\frac{1}{4\rho}\left(\int_{0}^{\pi}q(t){\rm d}t\right)^2\nonumber\\&&+2H\tan(\rho\pi)-\frac{H}{\rho}\int_{0}^{\pi}q(t){\rm d}t-\frac{\tan^{2}(\rho\pi)}{4\rho}\left(\int_{0}^{\pi}q(t){\rm d}t\right)^2\nonumber\\&&-\frac{H^2}{\rho}\cdot\tan^{2}(\rho\pi)-\frac{H\tan^{2}(\rho\pi)}{\rho}\int_{0}^{\pi}q(t){\rm d}t+o\left(\frac{1}{\left|\rho\right|}\right).\end{eqnarray}$

将(2.21)式代入(2.19)式,得

(2.22)$\begin{eqnarray}\label{E2.22}&&-\frac{1}{2\pi{\rm i}}\oint_{\Gamma_{N_{0}}}2\rho \ln\left(\frac{w(\lambda)}{w_{0}(\lambda)}\right){\rm d}\rho \\&=&\frac{1}{2\pi{\rm i}}\oint_{\Gamma_{N_{0}}}\Biggr\{\tan(\rho\pi)\int_{0}^{\pi}q(t)\cos(2\rho t){\rm d}t-\int_{0}^{\pi}q(t)\sin(2\rho t){\rm d}t\nonumber\\&&-\tan(\rho\pi)\int_{0}^{\pi}q(t){\rm d}t+\frac{1}{\rho}\int_{0}^{\pi}M(t, t){\rm d}t+\frac{1}{4\rho}\left(\int_{0}^{\pi}q(t){\rm d}t\right)^2\nonumber\\&&-2H\tan(\rho\pi)+\frac{H}{\rho}\int_{0}^{\pi}q(t){\rm d}t+\frac{\tan^{2}(\rho\pi)}{4\rho}\left(\int_{0}^{\pi}q(t){\rm d}t\right)^2\nonumber\\&&+\frac{H^2}{\rho}\cdot\tan^{2}(\rho\pi)+\frac{H\tan^{2}(\rho\pi)}{\rho}\int_{0}^{\pi}q(t){\rm d}t+o\left(\frac{1}{|\rho|}\right)\Biggr\}{\rm d}\rho.\end{eqnarray}$

经过计算得

(2.23)$\begin{eqnarray}\label{E2.23}&&{\rm Res}\left[\tan(\rho\pi)\int_{0}^{\pi}q(t)\cos(2\rho t){\rm d}t\right]=\sum\limits_{n=-N_{0}}^{N_{0}}-\frac{1}{\pi}\int_{0}^{\pi}q(t)\cos((2n+1)t){\rm d}t, \nonumber\\&&{\rm Res}\left[\tan(\rho\pi)\right]=\sum\limits_{n=-N_{0}}^{N_{0}}-\frac{1}{\pi}, \quad{\rm Res}\left[\int_{0}^{\pi}q(t)\sin(2\rho t){\rm d}t\right]=0, \quad {\rm Res}\left[\frac{1}{\rho}\right]=1, \nonumber\\&&{\rm Res}\left[\frac{\tan^{2}(\rho\pi)}{\rho}\right]=-\sum\limits_{n=-N_{0}}^{N_{0}}\frac{4}{(2n+1)^{2}\pi^2}.\end{eqnarray}$

由(2.19), (2.22)和(2.23)式,得

$\begin{eqnarray*}&&\sum\limits_{n=0}^{N_{0}}\left[2\lambda_{n}-2\left(n+\frac{1}{2}\right)^2\right]=-\frac{1}{2\pi{\rm i}}\oint_{\Gamma_{N_{0}}}2\rho \ln\left(\frac{w(\lambda)}{w_{0}(\lambda)}\right){\rm d}\rho\\&=&\sum\limits_{n=-N_{0}}^{N_{0}}-\frac{1}{\pi}\int_{0}^{\pi}q(t)\cos((2n+1)t){\rm d}t+\sum\limits_{n=-N_{0}}^{N_{0}}\frac{1}{\pi}\cdot\int_{0}^{\pi}q(t){\rm d}t+\int_{0}^{\pi}M(t, t){\rm d}t\\&&+\frac{1}{4}\left(\int_{0}^{\pi}q(t){\rm d}t\right)^2+2H\sum\limits_{n=-N_{0}}^{N_{0}}\frac{1}{\pi}+H\int_{0}^{\pi}q(t){\rm d}t\\&&-\left[\frac{1}{4}\left(\int_{0}^{\pi}q(t){\rm d}t\right)^2+H^2+H\int_{0}^{\pi}q(t){\rm d}t\right]\sum\limits_{n=-N_{0}}^{N_{0}}\frac{4}{(2n+1)^{2}\pi^2}\\&=&-\frac{1}{\pi}\int_{0}^{\pi}q(t)\cos t{\rm d}t-\frac{2}{\pi}\sum\limits_{n=0}^{N_{0}}\int_{0}^{\pi}q(t)\cos (2nt)\cos t{\rm d}t\\&&+\frac{2N_{0}+1}{\pi}\int_{0}^{\pi}q(t){\rm d}t+\int_{0}^{\pi}M(t, t){\rm d}t+\frac{1}{4}\left(\int_{0}^{\pi}q(t){\rm d}t\right)^2+\frac{2H(2N_{0}+1)}{\pi}\\&&+H\int_{0}^{\pi}q(t){\rm d}t-\left[\frac{1}{4}\left(\int_{0}^{\pi}q(t){\rm d}t\right)^2+H^2+H\int_{0}^{\pi}q(t){\rm d}t\right]\sum\limits_{n=0}^{N_{0}}\frac{8}{(2n+1)^{2}\pi^2}.\end{eqnarray*}$

从而得到

(2.24)$\begin{eqnarray}\label{E2.24}&&\sum\limits_{n=0}^{N_{0}}\left[\lambda_{n}-\left(n+\frac{1}{2}\right)^2-\frac{2H+\int_{0}^{\pi}q(t){\rm d}t}{\pi}\right]\nonumber\\&=&-\frac{1}{2\pi}\int_{0}^{\pi}q(t)\cos t{\rm d}t-\frac{1}{\pi}\sum\limits_{n=0}^{N_{0}}\int_{0}^{\pi}q(t)\cos (2nt)\cos t{\rm d}t-\frac{H+\frac{1}{2}\int_{0}^{\pi}q(t){\rm d}t}{\pi}\nonumber\\&&+\frac{1}{2}\int_{0}^{\pi}M(t, t){\rm d}t+\frac{1}{8}\left(\int_{0}^{\pi}q(t){\rm d}t\right)^2+\frac{H}{2}\int_{0}^{\pi}q(t){\rm d}t\nonumber\\&&-\left[\frac{1}{4}\left(\int_{0}^{\pi}q(t){\rm d}t\right)^2+H^2+H\int_{0}^{\pi}q(t){\rm d}t\right]\sum\limits_{n=0}^{N_{0}}\frac{4}{(2n+1)^{2}\pi^2}.\end{eqnarray}$

由(2.3)式知,级数

$\sum\limits_{n=0}^{\infty}\left[\lambda_{n}-\left(n+\frac{1}{2}\right)^2-\frac{2H+\int_{0}^{\pi}q(t){\rm d}t}{\pi}\right]$

绝对收敛,所以在(2.24)式中,令$N_{0}\rightarrow\infty$,得

(2.25)$\begin{eqnarray}\label{E2.25}&&\sum\limits_{n=0}^{\infty}\left[\lambda_{n}-\left(n+\frac{1}{2}\right)^2-\frac{2H+\int_{0}^{\pi}q(t){\rm d}t}{\pi}\right]\nonumber\\&=&-\frac{1}{2\pi}\int_{0}^{\pi}q(t)\cos t{\rm d}t-\frac{1}{\pi}\sum\limits_{n=0}^{\infty}\int_{0}^{\pi}q(t)\cos (2nt)\cos t{\rm d}t-\frac{H+\frac{1}{2}\int_{0}^{\pi}q(t){\rm d}t}{\pi}\nonumber\\&&+\frac{1}{2}\int_{0}^{\pi}M(t, t){\rm d}t+\frac{1}{8}\left(\int_{0}^{\pi}q(t){\rm d}t\right)^2+\frac{H}{2}\int_{0}^{\pi}q(t){\rm d}t\nonumber\\&&-\left[\frac{1}{4}\left(\int_{0}^{\pi}q(t){\rm d}t\right)^2+H^2+H\int_{0}^{\pi}q(t){\rm d}t\right]\sum\limits_{n=0}^{\infty}\frac{4}{(2n+1)^{2}\pi^2}\nonumber\\&=&-\frac{1}{2\pi}\int_{0}^{\pi}q(t)\cos t{\rm d}t-\frac{1}{\pi}\sum\limits_{n=0}^{\infty}\int_{0}^{\pi}q(t)\cos (2nt)\cos t{\rm d}t-\frac{H+\frac{1}{2}\int_{0}^{\pi}q(t){\rm d}t}{\pi}\nonumber\\&&+\frac{1}{2}\int_{0}^{\pi}M(t, t){\rm d}t+\frac{1}{8}\left(\int_{0}^{\pi}q(t){\rm d}t\right)^2+\frac{H}{2}\int_{0}^{\pi}q(t){\rm d}t\nonumber\\&&-\frac{1}{2}\left[\frac{1}{4}\left(\int_{0}^{\pi}q(t){\rm d}t\right)^2+H^2+H\int_{0}^{\pi}q(t){\rm d}t\right]\nonumber\\&=&-\frac{1}{2\pi}\int_{0}^{\pi}q(t)\cos t{\rm d}t-\frac{1}{\pi}\sum\limits_{n=0}^{\infty}\int_{0}^{\pi}q(t)\cos (2nt)\cos t{\rm d}t\\&&-\frac{H+\frac{1}{2}\int_{0}^{\pi}q(t){\rm d}t}{\pi}+\frac{1}{2}\int_{0}^{\pi}M(t, t){\rm d}t-\frac{H^2}{2}.\end{eqnarray}$

记$b_{2n}=\frac{1}{\pi}\int_{0}^{\pi}q(t)\cos t\cos(2nt) {\rm d}t$,则$b_{0}=\frac{1}{\pi}\int_{0}^{\pi}q(t)\cos t {\rm d}t$,并且

(2.26)$\begin{equation}\label{E2.26}b_{2n}=\frac{1}{2}\cdot\frac{1}{\pi}\int_{0}^{2\pi}q\left(\frac{u}{2}\right)\cos\left(\frac{u}{2}\right)\cos (nu) {\rm d}u.\end{equation}$

由(2.26)式知, $2b_{2n}$就是$q\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)$的Fourier展开的系数.根据Fourier级数收敛定理,得到

(2.27)$\begin{equation}\label{E2.27}b_{0}+\sum\limits_{n=0}^{\infty}2b_{2n}=\frac{q(0)\cos0+q(\pi)\cos\pi}{2}=\frac{1}{2}\left(q(0)-q(\pi)\right).\end{equation}$

由(2.25)-(2.27)式,得

$\begin{eqnarray}&&\sum\limits_{n=0}^{\infty}\left[\lambda_{n}-\left(n+\frac{1}{2}\right)^{2}-\frac{2H+\int_{0}^{\pi}q(t){\rm d}t}{\pi}\right]\nonumber\\&=&-\frac{1}{4}\left(q (0)-q(\pi)\right)-\frac{H+\frac{1}{2}\int_{0}^{\pi}q(t){\rm d}t}{\pi}+\frac{1}{2}\int_{0}^{\pi}M(t, t){\rm d}t-\frac{H^{2}}{2}.\nonumber\end{eqnarray}$

证毕.

类似可证定理2.2.