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.