I have an integral like Eq. E.21 :
$I=\int d^2R \int d^2r (r\cdot\nabla_R \phi(R))^2 e^{-f(r)}$
where, $r=r_1-r_2, R=\dfrac{r_1+r_2}{2}.$
How do I expand this dot product of $r$ with gradient $\nabla_R \phi(R)$? Author(in attached link) has written it as:
$I=\dfrac{1}{2}\int d^2R (\partial_X\phi)^2+(\partial_Y\phi)^2) (\int d^2r r^2e^{-f(r)})$
I was thinking of it like $(r\cdot\nabla_R \phi(R))^2=|r|^2 |\nabla_R\phi|^2\cos^2(\theta)$ but how do I determine $\theta$. Looking at the result that author has got it should be $\theta=45^o$! How to find that?
I have figured it out, actually its quite easy:
$\int d^2R\int d^2r(r\cdot\nabla_R\phi(R))^2=\int d^2R\int d^2r (r_i r_j \nabla_{R_i}\phi \nabla_{R_j}\phi),$
Now, if $i\neq j$ then integral over $d^2r$ will be zero because integral will become odd (Note; $f(r)$ in question is even). So, when $i=j$:
$\int d^2R\int d^2r (r_i^2 (\nabla_{R_i}\phi)^2 )=\int d^2R\int d^2r (x^2 (\nabla_{X}\phi)^2+y^2 (\nabla_{Y}\phi)^2 ),$
But, we know $\int d^2r x^2=\int d^2r y^2=\int d^2r \dfrac{r^2}{2}$ so:
$I=\int d^2R ((\nabla_{X}\phi)^2+(\nabla_{Y}\phi)^2 )\int d^2r\dfrac{1}{2}r^2 e^{-f(r)}.$