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I want to calculate the following integral $$ \int_0^{\infty}\int_0^{\infty}\frac{f(r_1,r_2)}{\lvert r_1-r_2\rvert}\; r_1^2 r_2^2\; dr_1 dr_2 \tag{1} $$ also I intend to use the following change variables $$ r=\lvert r_1-r_2\rvert \tag{2} $$ and $$ R=\frac{m_1r_1+m_2r_2}{m_1+m_2} \tag{3} $$

to transform $(1)$ using $(2)$ and $(3)$, assuming we know the form of $f(r,R)$ as $$ f(r,R)=\text{exp}(-ar-br^2) $$ where $a$ and $b$ are positive parameters. Well we know from $(2)$ and $(3)$ that $$ r_1=R-\frac{m_2 r}{(m_1+m_2)} \tag{4} $$ and $$ r_2=R-\frac{m_1 r}{(m_1+m_2)} \tag{5} $$ Should I simply replace $r_1^2$ and $r_2^2$ by $(4)$ and $(5)$? and what about differentials $dr_1$ and $dr_2$?

Wisdom
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  • At first glance, unless $f(r_1,r_2)=O(|r_1-r_2|^a)$ for some $a>0$ the integral is not well-defined (it diverges). As for changing variables, notice that the integral on the region $r_1>r_2$ equals that on the region $r_2>r_1$ by symmetry, so you can drop the absolute value. – Medo Aug 20 '21 at 13:04
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    Most importantly, how can we help you compute this without an explicit formula for $f$ ? – Medo Aug 20 '21 at 13:04
  • @Medo I added the form of $f(r,R)$ – Wisdom Aug 20 '21 at 13:14
  • @Wisdom Do you have that $b>0$ so that's a decaying exponential? – Alan Aug 20 '21 at 14:02
  • @Alan Yes, I added this to the question – Wisdom Aug 20 '21 at 14:07
  • @Wisdom. Do you need an exact computation, or an estimate in terms of $a,b$ ? – Medo Aug 20 '21 at 16:45
  • @Alan absolutely exact. However I didn't get your mean about "an estimate in terms of a,b" – Wisdom Aug 20 '21 at 17:25
  • That last question was from @Medo – Alan Aug 20 '21 at 17:58

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