# Constructing molecular integrals with derivative on the r_inv operator by partial integration

In quantum chemstry integrals of the following form are sometimes needed:

$F_x = \left<\phi_i\left|\nabla^x r^{-1}\right|\phi_j\right>$

But these integrals are often not directly available.

What instead is available is integrals of the form:

$f_x = \left<\nabla^x\phi_i\left|r^{-1}\right|\phi_j\right>$ $f_{x,y} = \left<\nabla^x\phi_i\left|r^{-1}\right|\nabla^y\phi_j\right>$

Let us therefore try to formulate $$F_x$$ in terms of $$f_x$$ and $$f_{x,y}$$. Now switching the integral notation to the more “regular” notation:

$F_x = \int_\Omega \phi_i \phi_j \nabla^x r^{-1} \mathrm{d}r$

Doing integration by parts:

$F_x = \left[ \phi_i \phi_j \nabla^{x-1} r^{-1} \right]_{\partial\Omega} - \int_\Omega \nabla \left(\phi_i \phi_j \right) \nabla^{x-1} r^{-1} \mathrm{d}r$

The wave function goes to zero at infinity, therefore the first term is zero:

$F_x = - \int_\Omega \nabla \phi_i \phi_j \nabla^{x-1} r^{-1} \mathrm{d}r - \int_\Omega \phi_i \nabla \phi_j \nabla^{x-1} r^{-1} \mathrm{d}r$

Doing integration by parts again:

$F_x = -\left(\left[ \nabla \phi_i \phi_j\nabla^{x-2} r^{-1} \right]_{\partial\Omega} - \int_\Omega \nabla \left(\nabla \phi_i \phi_j\right) \nabla^{x-2} r^{-1} \mathrm{d}r \right) -\left(\left[ \phi_i \nabla \phi_j\nabla^{x-2} r^{-1} \right]_{\partial\Omega} - \int_\Omega \nabla \left( \phi_i \nabla \phi_j \right) \nabla^{x-2} r^{-1} \mathrm{d}r \right)$

The wave function goes to zero at infinity, therefore the first term is zero:

$F_x = -\left(-\int_\Omega \nabla^2 \phi_i \phi_j \nabla^{x-2} r^{-1} \mathrm{d}r - \int_\Omega \nabla \phi_i \nabla \phi_j \nabla^{x-2} r^{-1} \mathrm{d}r\right) -\left(-\int_\Omega \nabla \phi_i \nabla \phi_j \nabla^{x-2} r^{-1} \mathrm{d}r - \int_\Omega \phi_i \nabla^2 \phi_j \nabla^{x-2} r^{-1} \mathrm{d}r \right)$

Simplifying:

$F_x = \int_\Omega \nabla^2 \phi_i \phi_j \nabla^{x-2} r^{-1} \mathrm{d}r + \int_\Omega \phi_i \nabla^2 \phi_j \nabla^{x-2} r^{-1} \mathrm{d}r + 2\int_\Omega \nabla \phi_i \nabla \phi_j \nabla^{x-2} r^{-1} \mathrm{d}r$

We therefore now have:

$F_1 = \left<\phi_i\left|\nabla r^{-1}\right|\phi_j\right> = - \int_\Omega \nabla \phi_i \phi_j r^{-1} \mathrm{d}r - \int_\Omega \phi_i \nabla \phi_j r^{-1} \mathrm{d}r$

and,

$F_2 = \left<\phi_i\left|\nabla^2 r^{-1}\right|\phi_j\right> = \int_\Omega \nabla^2 \phi_i \phi_j r^{-1} \mathrm{d}r + \int_\Omega \phi_i \nabla^2 \phi_j r^{-1} \mathrm{d}r + 2\int_\Omega \nabla \phi_i \nabla \phi_j r^{-1} \mathrm{d}r$

I.e. the integrals have been constructed without taking the derivative of $$r^{-1}$$. If hihger derivatives are needed, simply just do partial integration again.

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