How prove this hypersingular operator $(T\psi)((z(t))$

Let $\Omega\subset R^2$ be a simply connected bounded domain with inﬁnitely diﬀerentiable boundary $\partial\Omega$and unit normal vector $v$ directed into the exterior of $\Omega$ $$\Phi{(x,y)}=\dfrac{i}{4}H^{(1)}_{0}(k|x-y|),x\neq y$$ we denote the fundamental solution to the two-dimensional Helmholtz equation in terms of the ﬁrst kind Hankel function of order zero

where the Helmholtz equation $$\Delta u+k^2u=0, \mbox{in} R^2\overline{\Omega}$$ and $$(T\psi)(x):=\dfrac{\partial}{\partial v(x)}\int_{\partial\Omega}\dfrac{\partial\Phi{(x,y)}}{\partial v(y)}\psi{(y)}ds(y),x\in\partial\Omega.$$

let: $$x=z(t)=(z_{1}(t),z_{2}(t)),y=z(\tau)=(z_{1}(\tau),z_{2}(\tau)),-1\le t,\tau\le 1$$ show that: $$(T\psi)((z(t))=\dfrac{1}{|z'(t)|}\int_{-1}^{1}\left(\dfrac{1}{2\pi}\cdot\dfrac{1}{\tau-t}\dfrac{d}{d\tau}\psi(z(\tau))+L(t,\tau)\psi(z(\tau))\right)d\tau$$

where $$L(t,\tau)=-\dfrac{i}{2}\dfrac{z'(t)\{z(t)-z(\tau)\}z'(\tau)\{z(t)-z(\tau)\}}{|z(t)-z(\tau)|^2}\left(k^2H^{(1)}_{0}(k|z(t)-z(\tau)|)-\dfrac{2kH^{(1)}_{1}(k|z(t)-z(\tau)|)}{ |z(t)-z(\tau)|}\right)-\dfrac{ik}{2}\dfrac{z'(t)z'(\tau)}{|z(t)-z(\tau)|}H^{(1)}_{1}(k|z(t)-z(\tau)|)-\dfrac{1}{\pi}\dfrac{1}{(\tau-t)^2}+\dfrac{ik^2}{2}H^{(1)}_{0}(k|z(t)-z(\tau)|)z'(t)z'(\tau)$$ This results is from this paper:http://www.sciencedirect.com/science/article/pii/S0377042703005867

and I have Calculate for a few days,at last I failure

since I have post this question:The Helmholtz equation: How prove this $T\psi{(x)}\in\Omega$.

I want use this results $$(T\psi)(x)=\dfrac{\partial}{\partial s(x)}\int_{\partial\Omega}\Phi{(x,y)}\dfrac{\partial \psi}{\partial s}(y)ds(y)+k^2v(x)\cdot\int_{\partial \Omega}\Phi{(x,y)}v(y)\psi{(y)}ds(y),x\in\partial\Omega$$ I know this problem is limit problem,But I feel the pressure to prove this