# Integrate the interaction with a half space

#### 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?

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

#### topsquark

Forum Staff
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

#### BearniX

Thanks topsquark.
Nice presentation of the substitutions.

Now I can sleap in peace...