问题

做积分 \begin{align*} A =& \int_{-1}^1 \mathrm{d}x \int_0^1 \mathrm{d}p \cdot \delta (\omega-\delta+\frac{1}{2}p^2+\frac{1}{2}q^{2}-pqx)p^2 \\ =& \int_{-1}^1 \mathrm{d}x \int_0^1 \mathrm{d}p \cdot \delta (\frac{1}{2}p^2-pqx+g)p^2 \end{align*} \begin{align*} B =& \int_{-1}^1 \mathrm{d}x \int_0^1 \mathrm{d}p \cdot \mathcal{P} \frac{p^2}{\omega-\delta+\frac{1}{2}p^2+\frac{1}{2}q^{2}-pqx}\\ =& \int_{-1}^1 \mathrm{d}x \int_0^1 \mathrm{d}p \cdot \mathcal{P} \frac{p^2}{\frac{1}{2}p^2-pqx+g}\\ \end{align*} 其中 $g= \omega-\delta+\frac{1}{2}q^2$

结果

Dirac Delta 部分

\begin{align*} A = \int_{-1}^1\mathrm{d}x \cdot \frac{1}{|r_1-r_2|}\left\{ \theta(-g)\theta(\frac{1+2g}{2q}-x)r_1^2 +\theta(g)\theta(x-\frac{\sqrt{2g}}{q})\theta(x)\left[ \theta( \frac{1+2g}{2q}-x) r_1^2 +r_2^2\right] \right\} \end{align*} 其中 \begin{align*} r_1 =& qx + \sqrt{(qx)^2-2g}\\ r_2 =& qx - \sqrt{(qx)^2-2g} \end{align*}

Cauchy Principal Value 部分

化简主值部分 \begin{align*} \frac{p^2}{\frac{1}{2}p^2-pqx+g} =2 + \frac{-2g}{\frac{1}{2}p^2-pqx+g} + 2qx \frac{p}{\frac{1}{2}p^2-pqx+g} \end{align*} 则 \begin{align*} B =&\int_{-1}^1 \mathrm{d}x \cdot \left\{ 2 + \left[ \theta(g)\theta(\frac{\sqrt{2g}}{q}-x)\left( \frac{-4g}{2g-(qx)^2} \arctan t|_{\frac{-qx}{\sqrt{2g-(qx)^2}}} ^{\frac{1-qx}{\sqrt{2g-(qx)^2}}} + 2qx\ln\left[ \frac{1}{2}p^2-qxp + g \right]_{p0}^{p1} \right)\right]\right. \\ &\left. + \left[ \theta(-g) +\theta(g)\theta(x-\frac{\sqrt{2g}}{q}) \right]\cdot \left\{ \frac{-2g}{r_1-r_2}\ln \left|\frac{r_2(1-r_1)}{r_1(1-r_2)}\right| +\frac{2qx}{r_1-r_2}\left( r_1\ln\left|\frac{1-r_1}{r_1}\right|-r_2\ln\left|\frac{1-r_2}{r_2} \right| \right)\right\} \right\} \end{align*}

特例

\begin{align*} \int_{-1}^1 \mathrm{d}x \int_0^1\mathrm{d}p \cdot \mathcal{P} \frac{p}{p^2-x^2} =\ln 4 \end{align*}