Show that the steady-state temperature distribution problem $$\frac{\partial^2u}{\partial r^2}+\frac{1}{r}\frac{\partial u}{\partial r}+\frac{\partial^2u}{\partial z^2}=0$$ subject to $\displaystyle-\kappa \frac{\partial u}{\partial z}\bigg|_{(r,0)}=\frac{Q}{a^2}h(a-r)$ where $a>0$ and $h$ is Heaviside unit step function and $u(r,z)\to 0$ as $\sqrt{r^2+z^2}\to \infty$ where $z>0$ has the formal solution $$u(r,z)=\frac{Q}{\kappa a}\int_0^a e^{-z\rho}J_1(a\rho)J_1(r\rho)d\rho$$
I approached the problem by a Hankel transform of order $0$. Taking the transform on both sides w.r.t. $r$ of the equation gives $$\mathcal{H}_0\bigg(\frac{\partial^2u}{\partial r^2}+\frac{1}{r}\frac{\partial u}{\partial r};\rho\bigg)+\mathcal{H}_0\bigg(\frac{\partial^2u}{\partial z^2};\rho\bigg)=0 \\ \implies \rho\mathcal{H}_1\bigg(\frac{\partial u}{\partial r};\rho\bigg)+\mathcal{H}_0\bigg(\frac{\partial^2u}{\partial z^2};\rho\bigg)=0 \\ \implies -\rho^2\mathcal{H}_0(u(r,z);\rho)+\frac{d^2}{dz^2}\mathcal{H}_0(u(r,z);\rho)=0 \\ \implies -\rho^2U(\rho,z)+\frac{d^2}{dz^2}U(\rho,z)=0 \\ \implies U(\rho,z)=Ae^{\rho z}+Be^{-\rho z}$$ Where $U(\rho,z)=\mathcal{H}_0(u(r,z);\rho)$. Now using the $u(r,z)\to 0$ as $\sqrt{r^2+z^2}\to \infty$ gives $A=0$. So the partial solution becomes $$U(\rho,z)=Be^{-\rho z}$$ Now the first condition involving the Heaviside step function will give us $B$ after which we can inverse Hankel to get the complete solution. But since I don't have a formula of Hankel transform of unit step function, I am unable to proceed further. Can someone help me here to complete the solution in this regard? Thanks in advance.