I have a complex function $\psi(\rho,z)$ in cylindrical coordinates, which has the conditions $\psi=0$ as $z\rightarrow\pm\infty$ and $\rho\rightarrow +\infty$. The quantity $E$ given by the double integral $$ E = 2\pi\int_{z=-\infty}^{z=+\infty}\int_{\rho=0}^{\rho=+\infty} \Bigg( \psi^*\frac{\partial^2\psi}{\partial\rho^2} + \psi^*\frac{1}{\rho}\frac{\partial\psi}{\partial\rho} + \psi^*\frac{\partial^2\psi}{\partial z^2} \Bigg) \rho \;\textrm{d}\rho\, \textrm{d}z \tag{1} $$ where $\psi^*$ denotes the complex conjugate of $\psi$, and the $2\pi$ appears because I have integrated out the angular coordinate, since $\psi$ is cylindrically-symmetric. (The quantity $E$ expresses the kinetic energy of a quantum mechanical wavefunction).
I would like to simplify this expression so that it contains first derivatives instead of second derivatives.
I know that in one-dimension in a variable $x$, I can use integration by parts to write the following $$ \int_{x=-\infty}^{x=+\infty} \psi^* \frac{\partial^2\psi}{\partial x^2} \,\textrm{d}x = \Bigg[ \psi^* \frac{\partial\psi}{\partial x} \Bigg]^{+\infty}_{-\infty} - \int_{x=-\infty}^{x=+\infty} \frac{\partial\psi^*}{\partial x} \frac{\partial\psi}{\partial x} \,\textrm{d}x $$ and if $\psi=0$ at $\pm\infty$ the first term vanishes to give $$ \begin{align} \int_{x=-\infty}^{x=+\infty} \psi^* \frac{\partial^2\psi}{\partial x^2} \,\textrm{d}x &= - \int_{x=-\infty}^{x=+\infty} \frac{\partial\psi^*}{\partial x} \frac{\partial\psi}{\partial x} \,\textrm{d}x\\ &= - \int_{x=-\infty}^{x=+\infty} \bigg(\frac{\partial\psi^*}{\partial x}\bigg)^* \bigg(\frac{\partial\psi}{\partial x}\bigg) \,\textrm{d}x \\ &= - \int_{x=-\infty}^{x=+\infty} \Bigg|\frac{\partial\psi}{\partial x}\Bigg|^2\,\textrm{d}x . \end{align} $$ Is it possible to use a similar trick to write Eq. (1) in terms of first derivatives? I am struggling with the order of integration and how to apply the integration by parts in 2D here, given that there is an extra factor of $\rho$ from the volume element in cylindrical coordinates.
If someone could take me through it that would be great, thank you!