原式为：

\iint_{x^2 + y^2<R^2} \frac{1}{a}\hat{a}dS\\

可以化简为：

\int_{0}^{R} \int_{0}^{2\pi} \frac{L - rcos\theta}{L^{2} + r^{2} - 2Lrcos\theta} r drd\theta\ \hat{x}\\

将式子分为两部分求解：

\frac{1}{\frac{L}{r} + \frac{r}{L} - 2cos\theta}\ \ \ \ and \ \ \ \frac{cos\theta}{(\frac{L}{r})^{2} + 1 - 2\frac{L}{r}cos\theta}\\

利用三角函数倍角公式有(令 t = tan\frac{\theta}{2} )：

\begin{align*} cos\theta & = \frac{cos^{2}\frac{\theta}{2} - sin^{2}\frac{\theta}{2}}{cos^{2}\frac{\theta}{2} + sin^{2}\frac{\theta}{2}} = \frac{1-tan^{2}\frac{\theta}{2}}{1+tan^{2}\frac{\theta}{2}}= \frac{1-t^{2}}{1 + t^2} \end{align*}\\

d\theta = 2d(arctan(t)) = \frac{2}{1+t^{2}}dt\\

先求解第一个部分，令 \alpha = \frac{L}{r} + \frac{r}{L} ：

\begin{align*} \int_{0}^{2\pi} \frac{1}{\alpha - 2cos\theta} d\theta & = \int_{-\infty}^{+\infty} \frac{2}{\alpha - 2(\frac{1-t^{2}}{1+t^{2}})} \frac{1}{1 + t^{2}}dt \\ &= \int_{-\infty}^{+\infty} \frac{2}{\alpha(1+t^{2}) - 2(1-t^{2})}dt\\ &= \int_{-\infty}^{+\infty} \frac{2}{\alpha-2 + (\alpha +2)t^{2}}dt\\ \\ & = \frac{2}{\alpha + 2}\int_{-\infty}^{+\infty}\frac{1}{\frac{\alpha - 2}{\alpha + 2} + t^{2}} dt\\ & = \frac{2\pi}{\sqrt{(\alpha + 2)(\alpha - 2)}}\\ & = \frac{2\pi Lr}{L^2 - r^2} \end{align*}\\

第二部分，令 \beta = (\frac{L}{r})^2 + 1\ ,\ \gamma = \frac{L}{r} ：

\begin{align*} \int_{0}^{2\pi} \frac{cos\theta}{(\frac{L}{r})^{2} + 1 - 2\frac{L}{r}cos\theta} d\theta & = \int_{0}^{2\pi} \frac{cos\theta}{\beta - 2\gamma cos\theta}d\theta\\ \end{align*}\\

用围道积分的方法来计算，设 z = e^{i\theta} ：

cos\theta = \frac{z^2 + 1}{2z}\ \ \ and\ \ \ d\theta = \frac{1}{ie^{i\theta}}dz = \frac{1}{iz}dz\\

\begin{align*} \int_{0}^{2\pi} \frac{cos\theta}{\beta - 2\gamma cos\theta}d\theta & =\frac{1}{i} \oint_{|z| < 1} \frac{z^2 + 1}{2z^2(\beta - 2\gamma (\frac{z^2 + 1}{2z}))} dz\\ & = \frac{1}{i} \oint_{|z| < 1} \frac{z^2 + 1}{2\beta z^2 - 2\gamma z(z^2 + 1)} dz\\ & = -\frac{1}{i} \oint_{|z| < 1} \frac{z^2 + 1}{2z(\gamma z^2 - \beta z+\gamma)}dz\\ & = -\frac{1}{i} \oint_{|z| < 1} \frac{z^2 + 1}{2z \gamma (z - \frac{\beta + \sqrt{\beta^2 - 4\gamma^2}}{2\gamma})(z - \frac{\beta - \sqrt{\beta^2 - 4\gamma^2}}{2\gamma})} dz \end{align*}\\

可以发现被积式有三个极点 z_{1} = 0\ ,\ z_{2}=\frac{\beta + \sqrt{\beta^2 - 4\gamma^2}}{2\gamma}\ \ \ and \ \ \ z_{3} = \frac{\beta - \sqrt{\beta^2 - 4\gamma^2}}{2\gamma} ，继续将 \beta,\ \gamma 的值带入后，我们可得：

z_2 = \frac{(\frac{L}{r})^2 + 1 + (\frac{L}{r})^2 - 1}{2(\frac{L}{r})} = \frac{L}{r}\\ z_3 = \frac{(\frac{L}{r})^2 + 1 - (\frac{L}{r})^2 + 1}{2(\frac{L}{r})} = \frac{r}{L}\\

依据留数定理，计算该积分值：

\begin{align*} \oint_{|z| < 1} \frac{z^2 + 1}{2z^2(\beta - 2\gamma (\frac{z^2 + 1}{2z}))} dz & = 2\pi i \sum Res[f(z_k), |z_k|<1]\\ & = 2\pi i\lim_{z \rightarrow 0} z\frac{z^2 + 1}{2z \gamma (z - \frac{\beta + \sqrt{\beta^2 - 4\gamma^2}}{2\gamma})(z - \frac{\beta - \sqrt{\beta^2 - 4\gamma^2}}{2\gamma})}\\ & \ \ \ \ + 2\pi i\lim_{z \rightarrow z_3}(z - \frac{r}{L}) \frac{z^2 + 1}{2z \frac{L}{r}(z-\frac{r}{L})(z-\frac{L}{r})}\\ & = 2\pi i[\frac{r}{2L} + \frac{(\frac{r}{L})^2 + 1}{2(\frac{r}{L} - \frac{L}{r})}]\\ & = \pi i\frac{r}{L}(1 - \frac{L^2 + r^2}{L^2 - r^2})\\ \end{align*}\\

将两部分的结果相结合：

\begin{align*} \int_{0}^{2\pi} \frac{L - rcos\theta}{L^{2} + r^{2} - 2Lrcos\theta} r d\theta & = \frac{2\pi Lr}{L^2 - r^2} + \pi \frac{r}{L}(1 - \frac{L^2 + r^2}{L^2 - r^2}) \\ &=\frac{2\pi Lr}{L^2 - r^2} - \frac{2\pi r^3}{L(L^2 - r^2)}\\ & = \frac{2\pi r}{L^2 - r^2}(L - \frac{r^2}{L})\\ & = \frac{2\pi r}{L} \end{align*}\\

最终结果为：

\int_{0}^{R} \int_{0}^{2\pi} \frac{L - rcos\theta}{L^{2} + r^{2} - 2Lrcos\theta} r drd\theta = \int_{0}^{R} \frac{2\pi r}{L} dr = \frac{\pi R^2}{L}\\