Proof that $G=e^{-zs}-k^2z\int_z^\infty \frac{{J}_1\left( k\sqrt{t^2-z^2} \right)}{ k\sqrt{t^2-z^2}}e^{-st}dt$ verifies $ \partial_{zz}G-s^2G- k^2G=0$

Question

I want to verify that $$G(s,z) = e^{-zs} - k^2 z \int_z^\infty \dfrac{{J}_1\left( k\sqrt{t^2 - z^2} \right)}{ k\sqrt{t^2 - z^2}} e^{-st}\, dt$$ verifies $$ \partial_{zz} G - s^2 G - k^2 G = 0. \label{1}\tag{1} $$ This must be true because (see below) \begin{align} G(s,z) &= e^{-zs} - k^2 z \int_z^\infty \dfrac{{J}_1\left( k\sqrt{t^2 - z^2} \right)}{ k\sqrt{t^2 - z^2}} e^{-st}\, dt,\\ &= \int_0^\infty \left[ - kz \dfrac{{J}_1\left( k\sqrt{t^2 - z^2} \right)}{\sqrt{t^2 - z^2}}\, \theta (t-z) + \delta(t-z) \right] e^{-st}\, dt = e^{-z \sqrt{s^2 + k^2}}, \end{align} and clearly $G(s,z) = e^{-z \sqrt{s^2 + k^2}}$ verifies \eqref{1}.

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My attempt:

Calling $$ I := \int_z^\infty \dfrac{{J}_1\left( k\sqrt{t^2 - z^2} \right)}{ k\sqrt{t^2 - z^2}} e^{-st}\, dt $$ we can compute that ($\frac{d}{dz} \equiv '$) $$ G'(s,z) = - s e^{-zs} - k^2 I + \dfrac{k^2 z}{2} e^{-zs} - k^2 z \int_z^\infty \left[\dfrac{{J}_1\left( k\sqrt{t^2 - z^2} \right)}{ k\sqrt{t^2 - z^2}} \right]' e^{-st}\, dt, $$ and \begin{align} G''(s,z) &= s^2 e^{-zs} - k^2 \int_z^\infty \left[\dfrac{{J}_1\left( k\sqrt{t^2 - z^2} \right)}{ k\sqrt{t^2 - z^2}} \right]' e^{-st}\, dt + \dfrac{k^2}{2} e^{-zs} + \dfrac{k^2}{2} e^{-zs} - s \dfrac{k^2 z}{2} e^{-zs} + \\ & - k^2 \int_z^\infty \left[\dfrac{{J}_1\left( k\sqrt{t^2 - z^2} \right)}{ k\sqrt{t^2 - z^2}} \right]' e^{-st}\, dt - k^2 \dfrac{k^2 z}{8} e^{-zs} - k^2 z \int_z^\infty \left[\dfrac{{J}_1\left( k\sqrt{t^2 - z^2} \right)}{ k\sqrt{t^2 - z^2}} \right]'' e^{-st}\, dt, \end{align} which is simplified to \begin{equation} G''(s,z) = \left[ s^2 + k^2 - s \dfrac{k^2 z}{2} - k^2 \dfrac{k^2 z}{8} \right] e^{-zs} - k^2 \int_z^\infty (2\partial_{z} + z \partial_{zz}) \left[\dfrac{{J}_1\left( k\sqrt{t^2 - z^2} \right)}{ k\sqrt{t^2 - z^2}} \right] e^{-st}\, dt. \end{equation}

We can see that $$ \partial_z \left[\dfrac{{J}_1\left( k\sqrt{t^2 - z^2} \right)}{ k\sqrt{t^2 - z^2}} \right] = - \dfrac{z}{t} \partial_t \left[\dfrac{{J}_1\left( k\sqrt{t^2 - z^2} \right)}{ k\sqrt{t^2 - z^2}} \right] $$ and $$ \partial_{zz} \left[\dfrac{{J}_1\left( k\sqrt{t^2 - z^2} \right)}{ k\sqrt{t^2 - z^2}} \right] = \left( \dfrac{z^2}{t^2} \partial_{tt} + \left( \dfrac{z^2}{t^2} - 1 \right) \dfrac{1}{t} \partial_{t} \right) \left[\dfrac{{J}_1\left( k\sqrt{t^2 - z^2} \right)}{ k\sqrt{t^2 - z^2}} \right] $$ so

\begin{equation} G''(s,z) = \left[ s^2 + k^2 - s \dfrac{k^2 z}{2} - k^2 \dfrac{k^2 z}{8} \right] e^{-zs} - k^2 z \int_z^\infty \left[ \dfrac{z^2}{t^2} \partial_{tt} + \left( \dfrac{z^2}{t^2} - 3 \right) \dfrac{1}{t} \partial_{t} \right] \left[\dfrac{{J}_1\left( k\sqrt{t^2 - z^2} \right)}{ k\sqrt{t^2 - z^2}} \right] e^{-st}\, dt. \end{equation}

Putting this into (\ref{1}) we find that

\begin{equation} \left[ - s \dfrac{k^2 z}{2} - k^2 \dfrac{k^2 z}{8} \right] e^{-zs} - k^2 z \int_z^\infty \left[ \dfrac{z^2}{t^2} \partial_{tt} + \left( \dfrac{z^2}{t^2} - 3 \right) \dfrac{1}{t} \partial_{t} - s^2 - k^2 \right] \left[\dfrac{{J}_1\left( k\sqrt{t^2 - z^2} \right)}{ k\sqrt{t^2 - z^2}} \right] e^{-st}\, dt \stackrel{?}{=} 0,\label{eq:objectivo}\tag{2} \end{equation} which isn't a trivial equation by any means., and I am unsure about how to proceed from here.

Can someone give me a hint about how to continue with the proof? I have thought about using integration by parts and the Laplace transform properties of the derivative and of $1/t f(t)$, but without much success...

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Let the Laplace transform (see here) \begin{equation} \mathcal{L} \left\{{J}_0 \left( k \sqrt{t^2 - z^2} \right)\, \theta (t-z) \right\}(s) = \dfrac{e^{-z \sqrt{s^2 + k^2}}}{\sqrt{s^2 + k^2}}, \end{equation} which is a well known result. Then, as \begin{equation} \dfrac{{J}_1\left( k\sqrt{t^2 - z^2} \right)}{\sqrt{t^2 - z^2}}\, \theta (t-z) - \dfrac{\delta(t-z)}{kz} = -\dfrac{1}{kt} \left[{J}_0\left( k \sqrt{t^2 - z^2} \right)\, \theta (t-z)\right]', \end{equation} and using the fact that \begin{equation} \mathcal{L} \left\{ \dfrac{1}{t} f(t)\right\}(s) = \int_s^\infty F(\sigma) \,d\sigma, \end{equation} and \begin{equation} \mathcal{L} \left\{ f'(t)\right\}(s) = s F(s) - f(0^-), \end{equation} then \begin{equation} \mathcal{L} \left\{ \dfrac{1}{t} \left[{J}_0\left( k \sqrt{t^2 - z^2} \right)\, \theta (t-z)\right]'\right\}(s) = \int_s^\infty \sigma \dfrac{e^{-z \sqrt{\sigma^2 + k^2}}}{\sqrt{\sigma^2 + k^2}} \,d\sigma = \dfrac{1}{z} e^{-z \sqrt{s^2 + k^2}}. \end{equation} And putting everything together we find that

\begin{equation} \mathcal{L} \left\{ - kz \dfrac{{J}_1\left( k\sqrt{t^2 - z^2} \right)}{\sqrt{t^2 - z^2}}\, \theta (t-z) + \delta(t-z) \right\}(s) = e^{-z \sqrt{s^2 + k^2}}. \end{equation}