Integrate the interaction with a half space

Nov 2006
3
0
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?


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

topsquark

Forum Staff
Jan 2006
11,575
3,454
Wellsville, NY
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?


\(\displaystyle \phi_{12}(D) = -C_{12}\rho_2\int_0^{\infty }dx \int_0^{\infty}\frac{2\pi rdr}{((D+x)^2+r^2)^3} \)
As far as the r integration is concerned the x is a constant. So look at the integral:
\(\displaystyle \int_0^{\infty} \frac{2r \, dr}{(A + r^2)^3}\)

You can make the substitution \(\displaystyle z = r^2\), \(\displaystyle dz = 2rdr\)

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

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

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

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

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

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

= \(\displaystyle -\frac{C_{12} \rho_2 \pi}{2} \frac{1}{3} \frac{1}{D^3}\)

= \(\displaystyle -\frac{C_{12} \rho_2 \pi}{6D^3}\)

-Dan
 
Nov 2006
3
0
Thanks topsquark.
Nice presentation of the substitutions.

Now I can sleap in peace...