Asymptotic behavior of an integral I am interested in the integral
\begin{align*}
\int_{\epsilon}^{\infty}dx_1\int_{\epsilon}^{\infty}dx_2\int_{\epsilon}^{\infty}dx_3\int_{\epsilon}^{\infty}dx_4\,\frac{1}{(x_1+x_3)(x_1+x_4)(x_2+x_3)(x_2+x_4)}e^{-x_1-x_2-x_3-x_4}.
\end{align*}
This diverges as $\epsilon\to 0$. I don't really care about the finite part of the integral, but I would like to know how it behaves as $\epsilon\to 0$. By dimensional analysis it diverges logarithmically, but there could be terms like $(\log\epsilon)^n$ also. Are there some standard tricks to deal with such questions?
 A: If you integrate by parts successively in the $x_1,x_2,x_3$, and $x_4$ variables, each time integrating the exponential factor and differentiating the denominator, then the resulting integral becomes absolutely integrable. You get 4 endpoint terms each of which is a similar integral, but in one fewer variable. You can then similarly do integrations by parts on each of the endpoint terms, and so on. It isn't pretty, but if you just want to get the order of magnitude as $\epsilon \rightarrow 0$ this should give it to you.  
A: I spent a few hours developing some tricks for this integral and similar ones, so I thought I would post them here. 
As an example consider a simpler integral,
\begin{align*}
f(\mu)=\int_{\epsilon}^{\infty}dx_1\int_{\epsilon}^{\infty}dx_2\int_{\epsilon}^{\infty}dx_3\, \frac{1}{(x_1+x_2)(x_1+x_3)(x_2+x_3)}e^{-\mu(x_1+x_2+x_3)}.
\end{align*}
The basic idea is to compute the derivatives of $f$ with respect to $\mu$ and then integrate. The first derivative of $f$ is 
\begin{align*}
\frac{\partial f}{\partial \mu}=-\frac{3}{2}\int_{\epsilon}^{\infty}dx_1\int_{\epsilon}^{\infty}dx_2\int_{\epsilon}^{\infty}dx_3\, \frac{1}{(x_1+x_3)(x_2+x_3)}e^{-\mu(x_1+x_2+x_3)}.
\end{align*}
This is a convergent integral, so we can set $\epsilon=0$ if we are only interested in the asymptotics. Making the change of variables $y_1=x_1/\ell$, $y_2=x_2/\ell$, $\ell=x_1+x_2+x_3$, we get
\begin{align*}
\frac{\partial f}{\partial \mu}=-\frac{3}{2}\int_{0}^{\infty} d\ell\, e^{-\mu \ell}\int_{0}^{1}dy_1\int_{0}^{1-y_1}dy_2\,\frac{1}{(1-y_1)(1-y_2)}=-\frac{\pi^2}{4\mu}.
\end{align*}
Integrating then gives $f=-\pi^2 \log (\mu)/4$, which by dimensional analysis yields $f\sim -\pi^2\log(\epsilon)/4$.
One can do a similar thing for the integral I originally posted, you get $-2\pi^2\log(\epsilon)/3$. It seems somewhat surprising from the form of the original integral that it's just a single log, but I checked this answer numerically and it agrees.
A: Make the substitution $y_1 = x_1 + x_3; y_2 = x_1 + x_4;  y_3 = x_2 + x_3; y_4 = x_1 + x_2 + x_3 + x_4$ Then, your integrand becomes 
$$\frac{e^{-{y_4}}}{y_1 y_2 y_3 (y_4 - y_2)}.$$
The first three $y$s integrate out (the limit does change from $\infty$ to linear functions of the other $y,$ but you just get a sum of logs, and you are left with a fairly disgusting one-dimensional integral.
EDIT Firstly, the "right substitution is:
$y_1 = x_1 + x_2+x_3 + x_4,  y_2 = x_1 + x_3;  y_3 = x_1 + x_4; y_4 = x_1 + x_2 - x_3 -x_4.$
Secondly, what you see, then, is that the integrand is of the form
$$\frac{e^{-{y_1}}}{y_2 (y_1- y_2 )y_3 (y_1-y_3)}.$$ Which is wonderful, but the fact that it does not depend on $y_4$ (whose range is from $-\infty$ to $\infty$) means that the integral actually diverges, no matter what $\epsilon$ is, so the question seems to be moot.
