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I've seen double integrals of the type

$$\int d^3r\,d^3r'\, f(\vec{r})g(\vec{r}-\vec{r}')$$

being solved by making the substitution $\vec{u}=\vec{r}-\vec{r}'$:

$$\int d^3r\,d^3u \,\, f(\vec{r})g(\vec{u})=-\left (\int d^3r \,\, f(\vec{r}) \,\right)\,\,\left( \int d^3u \,\,g(\vec{u})\,\right)$$

I don't understand how can one possibly regard $\vec{r}$ and $\vec{u}$ as independent variables (which is done in this last equal sign).

Note: I post this here because I've seen this done by physicists.

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    $\begingroup$ Would Mathematics be a better home for this question? $\endgroup$
    – Qmechanic ♦
    Commented Jan 9, 2017 at 12:39
  • $\begingroup$ Maybe! I've only seen this done by physicists, though: that's why I posted it here. $\endgroup$
    – soap
    Commented Jan 9, 2017 at 12:49
  • $\begingroup$ Have you tried calculating the value of the Jacobian for this coordinate change? $\endgroup$
    – Jahan Claes
    Commented Jan 9, 2017 at 19:04

2 Answers 2

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This is the integral of the convolution $f*g$ of functions $f$ and $g$. You have: $$ \begin{align} \int d^3r'\,d^3r\, f(\vec{r})g(\vec{r}-\vec{r}') &= \int d^3r'\,\left(\int d^3r\, f(\vec{r})g(\vec{r}-\vec{r}') \right)\\ &\overset{\text{Fubini}}= \int d^3r\,\left(\int d^3r'\, f(\vec{r})g(\vec{r}-\vec{r}') \right)\\ &\overset{f(r) \text{ does not depend on }r'}= \int d^3r\,f(\vec{r})\left(\int d^3r'\, g(\vec{r}-\vec{r}') \right)\\ &\overset{\vec{u}=\vec{r}-\vec{r}'}= \int d^3r\,f(\vec{r})\left(\int d^3u\, g(\vec{u}) \right)\\ &= \left(\int d^3r\,f(\vec{r}) \right) \left(\int d^3u\, g(\vec{u}) \right).\\ \end{align} $$ Notes:

  • the inner integral of lines 2, 3, 4 is computed for fixed $r$,

  • as mentioned by kryomaxim, the change of variable with fixed $r$ in the fourth line provides an inner integral independent on $r$ because the integration w.r.t. $r'$ (and $u$) is done on the whole space,

  • there is no minus sign in the change of variable $\vec{u}=\vec{r}-\vec{r}'$ because $d^3u = |\text{Jacobian}|d^3r'$ (for $x'=-x$, think also about $\int_{\mathbb{R}} dxf(x)=\int_{\mathbb{R}} dx'f(-x')$ which is a property of the Lebesgue integral, to compare to the Riemann integral where $\int_{-\infty}^{+\infty} dxf(x)=-\int_{+\infty}^{-\infty} dx'f(-x')=\int_{-\infty}^{+\infty} dx'f(-x')$).

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  • $\begingroup$ Perhaps I'm misinterpreting what $d^3\bar r$ means (or something else entirely), but there clearly is something wrong in your reasoning. Take the simple case $f(q) = g(q) = q$. Then, replacing $d^3 q$ with $dq$ without loss of generality (I hope), it's straightforward to work out that $$\int \int f(\tau)g(t - \tau) dt d\tau = C t + \frac{t^2\tau^2}4 - \frac{t\tau^3}3$$ while $$\left(\int f(\tau)d\tau\right)\left(\int g(t - \tau)dt\right) = \frac{t^3\tau}2 - \frac{t^2\tau^2}4 - \frac{C_1}2\tau^2 + C_1t\tau + \frac{C_2}2 t^2 + C_1C_2$$ So, the two are rather different. $\endgroup$
    – Rody Oldenhuis
    Commented Jan 10, 2017 at 23:07
  • $\begingroup$ @Rody Oldenhuis : your integrals do not exist, $q$ is not integrable on $\mathbb{R}$. Remind that the result holds for integrals on the whole space $\mathbb{R}^n$. Besides, your integrals, if they existed, could only be constants (except the last one), not depending on $t$ nor $\tau$, you are making a confusion between integral and primitive. BTW, $d^3r$ is the elementary volume $dx \, dy \, dz$ in $\mathbb{R}^3$. $\endgroup$
    – user130529
    Commented Jan 11, 2017 at 7:59
  • $\begingroup$ I see the problem now. I was doing indefinite integration while you are doing improper integration. The OP did not state that explicitly, but in the context of convolution that would be the sensible thing to do. And in that case, you are correct, because Fubini's theorem does not hold if either integral is non-finite. $\endgroup$
    – Rody Oldenhuis
    Commented Jan 11, 2017 at 13:08
  • $\begingroup$ @Rody Oldenhuis : agreed. The OP should have stated that explicitly, as it is it is confusing. $\endgroup$
    – user130529
    Commented Jan 11, 2017 at 14:49
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Initially, $r'$ and $r$ are Independent variables. After Substitution, $u$ is Independent on the other variable $r$, because $u$ is the variable $r'$, but shifted by a value $r$.

In General, this formula is true if the Integration bounds are set from $- \infty$ to $\infty$ in each direction. Integrals over an infinite Region are Translation invariant, therefore the shift by $r$ makes $u$ still an Independent variables.

If you want to know more of it mathematically you can look for "Theorem of Fubini".

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