A_1 + A_2 +A_3.$$ By using Itô’s formula again we have $ {\mathrm{d}}(x_k-z)^{-1}=-(x_k-z)^{-2}{\mathrm{d}}x_k+\frac{1}{N}(x_k-z)^{-3}{\mathrm{d}}t $. Thus, we can now decompose the term $ A_3 $ as $ I_1+[I_2+I_3+I_4+I_5]{\mathrm{d}}t $, where $$\begin{aligned}
I_1 &=
-\dfrac{e^{-\frac{t}{2\xi}}}{\sqrt{N}}\sum_{-N {\leqslant}k {\leqslant}N}\dfrac{{\mathfrak{v}}_k}{(x_k-z)^2}{\mathrm{d}}B_k, \\
I_2 &=
\dfrac{e^{-\frac{t}{2\xi}}}{2\xi}\sum_{-N {\leqslant}k {\leqslant}N}\dfrac{{\mathfrak{v}}_k x_k}{(x_k-z)^2}, \\
I_3 &=
\dfrac{1}{2} \left( 1-\dfrac{1}{\xi} \right) e^{-\frac{t}{2\xi}} \sum_{-N {\leqslant}k {\leqslant}N} \dfrac{{\mathfrak{v}}_k}{x_k(x_k-z)^2}, \\
I_4 &= e^{-\frac{t}{2\xi}}\sum_{-N {\leqslant}k {\leqslant}N}{\mathfrak{v}}_k\left(-\dfrac{1}{(x_k-z)^2}\right)\dfrac{1}{2N}\sum_{l \neq \pm k}\dfrac{1}{x_k-x_l}=\dfrac{e^{-\frac{t}{2\xi}}}{2N}\sum_{l \neq \pm k}\dfrac{{\mathfrak{v}}_k}{(x_k-z)^2(x_l-x_k)}, \\
I_5 &= \dfrac{e^{-\frac{t}{2\xi}}}{N}\sum_{-N {\leqslant}k {\leqslant}N}\dfrac{{\mathfrak{v}}_k}{(x_k-z)^3}.\end{aligned}$$ Note that $$\partial_z f=e^{-\frac{t}{2\xi}}\sum_{-N {\leqslant}k {\leqslant}N}\dfrac{{\mathfrak{v}}_k}{(x_k-z)^2},\ \ \ \partial_{zz}f=2e^{-\frac{t}{2\xi}}\sum_{-N {\leqslant}k {\leqslant}N}\dfrac{{\mathfrak{v}}_k}{(x_k-z)^3}.$$ For the term $ A_2 $, by the equation , we have $$A_2 = e^{-\frac{t}{2\xi}}\sum_{-N {\leqslant}k {\leqslant}N}\dfrac{1}{x_k-z} \left[ \dfrac{1}{2} \left( 1-\dfrac{1}{\xi} \right) \dfrac{{\mathfrak{v}}_k}{x_k^2}  + \dfrac{1}{2N}\sum_{l \neq \pm k}\dfrac{{\mathfrak{v}}_l-{\mathfrak{v}}_k}{(x_k-x_l)^2} \right]=: B_1+B_2.$$ Note that $$\begin{gathered}
B_2 = \dfrac{e^{-\frac{t}{2\xi}}}{4N}\sum_{l \neq \pm k}\dfrac{{\mathfrak{v}}_l-{\mathfrak{v}}_k}{(x_l-x_k)^2}\left(\dfrac{1}{x_k-z}-\dfrac{1}{x_l-z}\right)=\dfrac{e^{-\frac{t}{2\xi}}}{4N}\sum_{l \neq \pm k}\dfrac{{\mathfrak{v}}_l-{\mathfrak{v}}_k}{x_l-x_k}\dfrac{1}{x_k-z}\dfrac{1}{x_l-z}\\
=-\dfrac{e^{-\frac{t}{2\xi}}}{2N}\sum_{l \neq \pm k}\dfrac{{\mathfrak{v}}_k}{x_l-x_k}\dfrac{1}{x_k-z}\dfrac{1}{x_l-z}.\end{gathered}$$ Combining with $ I_4 $, we obtain $$B_2 + I_4 =\dfrac{e^{-\frac{t}{2\xi}}}{2N}\sum_{l \neq \pm k}\dfrac{{\mathfrak{v}}_k}{x_l-x_k}\dfrac{1}{x_k-z}\left(\dfrac{1}{x_k-z}-\dfrac{1}{x_l-z}\right)=\dfrac{e^{-\frac{t}{2\xi}}}{2N}\sum_{l \neq \pm k}\dfrac{{\mathfrak{v}}_k}{(x_k-z)^2}\dfrac{1}{x_l-z}.$$ Moreover, we have that $$B_2+I_4+I_5 = s(z)\partial_z f+\dfrac{e^{-\frac{t}{2\xi}}}{2N}\sum_{-N {\leqslant}k {\leqslant}N}\dfrac{{\mathfrak{v}}_k}{(x_k-z)^3}+\dfrac{e^{-\frac{t}{2\xi}}}{2N}\sum_{-N {\leqslant}k {\leqslant}N}\dfrac{{\mathfrak{v}}_k}{(x_k-z)^2(x_k+z)}.$$ Then it suffices to calculate $ B_1 + I_3 $, and note that $$\begin{aligned}
B_1 + I_3 &= \dfrac{1}{2} \left( 1- \dfrac{1}{\xi} \right) e^{-\frac{t}{2\xi}} \sum_{k=1}^N \left( \dfrac{{\mathfrak{v}}_k}{x_k -z}\dfrac{1}{x_k^2} - \dfrac{{\mathfrak{v}}_k}{x_k +z}\dfrac{1}{x_k^2} + \dfrac{{\mathfrak{v}}_k}{(x_k - z)^2}\dfrac{1}{x_k} - \dfrac{{\mathfrak{v}}_k}{(x_k +z)^2}\dfrac{1}{x_k} \right)\\
&= \left( 1- 