Here's something that has me a little perplexed. I'm missing some step in making a change of variables under a multiple integral, where the new variable depends on more than one of the original ones.

I can not think of a simplified example that demonstrates the problem, so I'll just list the actual integral I'm trying to do:

$$\int_0^1\mathrm{d}\xi'\frac{1 + \xi'^2}{(1 - \xi')_+}\Biggl(\iint_{\mathbb{R}^2}\frac{\mathrm{d}^2\vec{t}_\perp} {(2\pi)^2}\exp\biggl[-\frac{t_\perp^2Q_s^2}{4}\biggr]\frac{e^{i\xi'\vec{k}_\perp\cdot\vec{t}_\perp}}{t_\perp^2} - \frac{1}{(2\pi)^2}\iint_{\mathbb{R}^2}\mathrm{d}^2\vec{r}'_\perp\frac{e^{-i\vec{k}_\perp\cdot \vec{r}'_\perp}}{r'^2_\perp}\Biggr)$$

Each of the two integrals over $\mathbb{R}^2$ is individually divergent because of a singularity at 0. But when I take the difference, the divergences will cancel out (perhaps only in some limiting sense, but that's really what I'm after).

Accordingly, the way I tried to do this is to change variables in the second (inner) integral from $\vec{r}'_\perp$ to $-\xi'\vec{t}_\perp$ so that I could then subtract the integrands. I compute the Jacobian as

$$\mathrm{d}\xi'\mathrm{d}^2\vec{r}' = \begin{vmatrix}\frac{\partial\xi'}{\partial\xi'} & \frac{\partial\xi'}{\partial t_x} & \frac{\partial\xi'}{\partial t_y} \\ \frac{\partial r'_x}{\partial\xi'} & \frac{\partial r'_x}{\partial t_x} & \frac{\partial r'_x}{\partial t_y} \\ \frac{\partial r'_y}{\partial\xi'} & \frac{\partial r'_y}{\partial t_x} & \frac{\partial r'_y}{\partial t_y}\end{vmatrix}\mathrm{d}\xi'\mathrm{d}^2\vec{t} = \begin{vmatrix}1 & 0 & 0 \\ -t_x & -\xi' & 0 \\ -t_y & 0 & -\xi'\end{vmatrix}\mathrm{d}\xi'\mathrm{d}^2\vec{t} = \xi'^2\mathrm{d}\xi'\mathrm{d}^2\vec{t}$$

I notice that by making this change of variables, I change the second term from something independent of $\xi'$ to something dependent on $\xi'$. But leaving that possible issue aside for now... I get

$$\begin{gather}\int_0^1\mathrm{d}\xi'\frac{1 + \xi'^2}{(1 - \xi')_+}\Biggl(\iint_{\mathbb{R}^2}\frac{\mathrm{d}^2\vec{t}_\perp}{(2\pi)^2}\exp\biggl[-\frac{t_\perp^2Q_s^2}{4}\biggr]\frac{e^{i\xi'\vec{k}_\perp\cdot\vec{t}_\perp}}{t_\perp^2} - \frac{1}{(2\pi)^2}\iint_{\mathbb{R}^2}\mathrm{d}^2\vec{t}'_\perp\frac{e^{i\xi'\vec{k}_\perp\cdot \vec{t}'_\perp}}{t'^2_\perp}\Biggr)\\ =\int_0^1\mathrm{d}\xi'\frac{1 + \xi'^2}{(1 - \xi')_+}\iint_{\mathbb{R}^2}\frac{\mathrm{d}^2\vec{t}_\perp}{(2\pi)^2}\biggl(\exp\biggl[-\frac{t_\perp^2Q_s^2}{4}\biggr] - 1\biggr)\frac{e^{i\xi'\vec{k}_\perp\cdot\vec{t}_\perp}}{t_\perp^2}\end{gather}$$

I can express the inner integral in polar coordinates as

$$\begin{gather}&\int_0^1\mathrm{d}\xi'\frac{1 + \xi'^2}{(1 - \xi')_+}\int_0^{2\pi}\mathrm{d}\theta_t\int_0^\infty\frac{\mathrm{d}t_\perp\,t_\perp}{(2\pi)^2}\biggl(\exp\biggl[-\frac{t_\perp^2Q_s^2}{4}\biggr] - 1\biggr)\frac{e^{i\xi' k_\perp t_\perp\cos\theta_t}}{t_\perp^2}\\ &=\int_0^1\mathrm{d}\xi'\frac{1 + \xi'^2}{(1 - \xi')_+}\int_0^\infty\frac{\mathrm{d}t_\perp}{2\pi t_\perp}\biggl(\exp\biggl[-\frac{t_\perp^2Q_s^2}{4}\biggr] - 1\biggr)J_0(\xi' k_\perp t_\perp)\\ &=-\int_0^1\mathrm{d}\xi'\frac{1 + \xi'^2}{(1 - \xi')_+}\frac{1}{4\pi}\Gamma\biggl(0, \frac{k^2 \xi'^2}{Q_s^2}\biggr) \end{gather}$$

(the last equality comes from Mathematica, I haven't gotten to show it myself). But the true result is supposed to be

$$-\int_0^1\mathrm{d}\xi'\frac{1 + \xi'^2}{(1 - \xi')_+}\frac{1}{4\pi}\biggl[\Gamma\biggl(0, \frac{k^2 \xi'^2}{Q_s^2}\biggr) {\color{red}{+ \ln\xi'^2}}\biggr]$$

I can't figure out where in this procedure that extra logarithm (in red) is supposed to come in. I guess it's supposed to be something in the change of variables that does it, but given that I've computed the Jacobian determinant and also the new region of integration (still $\mathbb{R}^2$), I don't know what else there would be. And I would very much like to know how I can avoid getting caught by this issue in the future.


1 Answer 1


Subtracting one divergent integral from another is not, without additional information, going to yield a meaningful result (even if the singularities are in some sense "the same"). To take a silly example, suppose you're asked to calculate $$ \int_{0}^{1}\frac{dx}{x}-\int_{0}^{1}\frac{dt}{t}. $$ Obviously this should be zero. Right? But by making the change of variable $t=ax$ in the second integral, it is seen to also equal $$ \int_{0}^{1}\frac{dx}{x}-\int_{0}^{1/a}\frac{dx}{x}=\int_{1/a}^{1}\frac{dx}{x}=\ln x\Big\vert_{1/a}^{1}=\ln a, $$ where $a$ is completely arbitrary. Hopefully the relevance to your problem is clear: changes of variables can change the overall result if the individual integrals are improper. In order to nail down a particular result, you need to specify the regularization procedure. Here, you might make the individual integrals meaningful by calculating $$ \int_{\varepsilon}^{1}\frac{dx}{x}-\int_{\varepsilon}^{1}\frac{dt}{t} $$ and then letting $\varepsilon\rightarrow 0$. (The result would then be zero no matter what changes of variable were carried out midstream.) If you adopt the same procedure in your problem (i.e., restrict the first integral to $|t_\perp|>\varepsilon$ and the second to $|r'_{\perp}|>\varepsilon$, then make your change of variable and subtract the results, and finally let $\varepsilon$ go to $0$), you will get an additional contribution from the annular region between radii $\varepsilon$ and $\xi' \varepsilon$. Specifically, the new term is $$ \int_{\varepsilon}^{\xi'\varepsilon}\frac{dt_\perp}{2\pi t_\perp} J_0\left(\xi'k_\perp t_\perp\right). $$ In the limit as $\varepsilon\rightarrow 0$, the Bessel function is irrelevant (it just goes to $1$), and the rest integrates to $$ \frac{1}{2\pi}\ln \xi' = \frac{1}{4\pi}\ln\xi'^2, $$ which is the term you were missing.

  • 1
    $\begingroup$ Great, that helps! I had noticed that regularization procedure gives the right answer, but I wasn't sure whether it was a valid way to proceed. Although... it seems like this transfers the arbitrariness to the regularization procedure, i.e. why would it be invalid to cut off one integral at $\epsilon$ and the other at $2\epsilon$, or $\xi'\epsilon$, etc.? (Maybe that should be a whole separate question?) $\endgroup$
    – David Z
    Sep 7, 2012 at 17:38
  • 2
    $\begingroup$ Yes, the fact that the regularization procedure affects the answer shows that the initial expression isn't well-defined as it stands. But there can be physical reasons (assuming the motivating problem is physics-based?) to choose one regularization procedure over another. Often it's possible to embed the regularization into the original problem somehow... e.g., through an unknown but large energy scale... so that it will carry through to all of the divergent integrals that appear in the middle of the calculation. And if you're lucky (as here), you can still let it go to $\infty$ in the end. $\endgroup$
    – mjqxxxx
    Sep 7, 2012 at 20:25
  • $\begingroup$ Yep, this is a physics-based problem. Unfortunately I don't think there is a physically relevant scale $\varepsilon$ to use for regularizing these integrals, but I will have to look through the paper I'm getting this from in some more detail. Thanks! $\endgroup$
    – David Z
    Sep 10, 2012 at 20:00
  • $\begingroup$ @DavidZ: Could you please provide the paper this integral comes from or explain your investigation why this particular regularization procedure is adopted? Thanks. $\endgroup$
    – Hans
    Nov 10, 2015 at 20:44
  • $\begingroup$ @Hans the source is arxiv:1203.6139, equation (45). $\endgroup$
    – David Z
    Nov 11, 2015 at 4:12

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