# Integrate the interaction with a half space

• Nov 14th 2006, 08:49 AM
BearniX
Integrate the interaction with a half space
Hi, I'm having trouble with a problem for a molecule that is interacting with a planar half space.

Could someone help me walk through this integral?

$\phi_{12}(D) = -C_{12}\rho_2\int_0^{\infty }dx \int_0^{\infty}\frac{2\pi rdr}{((D+x)^2+r^2)^3}$
• Nov 14th 2006, 11:20 AM
topsquark
Quote:

Originally Posted by BearniX
Hi, I'm having trouble with a problem for a molecule that is interacting with a planar half space.

Could someone help me walk through this integral?

$\phi_{12}(D) = -C_{12}\rho_2\int_0^{\infty }dx \int_0^{\infty}\frac{2\pi rdr}{((D+x)^2+r^2)^3}$

As far as the r integration is concerned the x is a constant. So look at the integral:
$\int_0^{\infty} \frac{2r \, dr}{(A + r^2)^3}$

You can make the substitution $z = r^2$, $dz = 2rdr$

Thus:
$\int_0^{\infty} \frac{2r \, dr}{(A + r^2)^3} = \int_0^{\infty} \frac{dz}{(A + z)^3}$

Now let $\beta = z + A$, $d \beta = dz$:
$\int_0^{\infty} \frac{dz}{(A + z)^3} = \int_A^{\infty}\frac{d \beta}{\beta^3} = \frac{1}{2A^2}$

So, recalling that I called $A = (D + x)^2$
$\phi_{12}(D) = -C_{12}\rho_2\int_0^{\infty }dx \int_0^{\infty}\frac{2\pi rdr}{((D+x)^2+r^2)^3}$ $= -C_{12} \rho_2 \pi \int_0^{\infty}dx \int_0^{\infty} \frac{2r \, dr}{(A + r^2)^3}$

= $-C_{12} \rho_2 \pi \int_0^{\infty}dx \frac{1}{2A^2}$

= $-C_{12} \rho_2 \pi \int_0^{\infty}dx \frac{1}{2(D + x)^4}$

Again, let $\gamma = x + D$, $d \gamma = dx$. The integral becomes:
$-\frac{C_{12} \rho_2 \pi}{2} \int_D^{\infty}d \gamma \frac{1}{\gamma^4}$

= $-\frac{C_{12} \rho_2 \pi}{2} \frac{1}{3} \frac{1}{D^3}$

= $-\frac{C_{12} \rho_2 \pi}{6D^3}$

-Dan
• Nov 14th 2006, 12:39 PM
BearniX
Thanks topsquark.
Nice presentation of the substitutions.

Now I can sleap in peace...