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Math Help - Integrate the interaction with a half space

  1. #1
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    Question 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}
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  2. #2
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    Quote Originally Posted by BearniX View Post
    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
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  3. #3
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    Thanks topsquark.
    Nice presentation of the substitutions.

    Now I can sleap in peace...
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