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Hello! Imagine I have a function

$f(r_1, r_2) = \int_{-\infty}^\infty g(|r_1 - r_3|) g(|r_1 - r_4|) g(|r_2 - r_3|) g(|r_2 - r_4|) g(|r_3 - r_4|) \; d r_3 d r_4$

Is there a way I could get rid of the integration via some (integral) transform?

so transformed $f$, that is $\hat f (k_1, k_2)$, was some algebraic combination of $\hat g (k_1, k_2)$?

Actually I need a bit more complicated case of $r$ being 3D vectrors, but I think the solution for a scalar version should be just the same

$f(\boldsymbol r_1, \boldsymbol r_2) = \int g(|\boldsymbol r_1 - \boldsymbol r_3|) g(|\boldsymbol r_1 - \boldsymbol r_4|) g(|\boldsymbol r_2 - \boldsymbol r_3|) g(|\boldsymbol r_2 - \boldsymbol r_4|) g(|\boldsymbol r_3 - \boldsymbol r_4|) \; d \boldsymbol r_3 d \boldsymbol r_4$

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The integral $f(\mathbf{r}_1,\mathbf{r}_2)$ depends on $|\mathbf{r}_1-\mathbf{r}_2|$ only. To go further, it might help to specify the function $g$ somewhat. –  Did Apr 24 '11 at 9:08
    
>To go further, it might help to specify the function g somewhat. Unfortunately this integral is the part of an equation for g. Indeed there is a lot of integrals of the same structure in that equation. g itself is a radial distribution function encountered in statistical mechanics, but that fact is really not very much to give. –  Yrogirg Apr 24 '11 at 17:09
    
$f$ is not a true function of two variables, and neither is $g$. The change of dummy variables $r_3 \to r_3 + r_1$ and $r_4 \to r_4+r_1$ shows that $f$ depends on $|r_2-r_1|$ only. So you should write it in the form $$ f(x)=\int_{R^2} g(z)g(y)g(y-z)g(x-y)g(x-z) dy dz $$ with $x>0$ and $g$ even. –  Jung Wen Chen Jan 31 at 14:36

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