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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?

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  • $\begingroup$ Please clarify the bounds and order of integration. Is it $$\int_\epsilon^\infty \int_\epsilon^\infty \int_\epsilon^\infty \int_\epsilon^\infty \ldots dx_1 dx_2 dx_3 dx_4$$ or what do you want to express? $\endgroup$ – AlexR Jan 12 '14 at 18:59
  • $\begingroup$ Thanks for the tip, I simplified this. $\endgroup$ – Matthew Dodelson Jan 12 '14 at 19:01
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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.

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  • $\begingroup$ Actually, the integral can't be evaluated in closed form. $\endgroup$ – Igor Rivin Jan 12 '14 at 19:48
  • $\begingroup$ I didn't say it could.. but you can integrate by parts in a given variable with the others fixed, as I indicated. $\endgroup$ – Zarrax Jan 12 '14 at 19:55
  • $\begingroup$ This method looks like it will work. I'll give it a shot and accept this answer if it does. $\endgroup$ – Matthew Dodelson Jan 12 '14 at 21:20
  • $\begingroup$ It might also be easier to change variables first to $y_1 = {x_1\over \epsilon}$, $y_2 = {x_2 \over \epsilon}$ etc since then the integrals are over a fixed set and the epsilon only appears in the exponential. $\endgroup$ – Zarrax Jan 12 '14 at 22:01
  • $\begingroup$ @Zarrax, I tried this, but doesn't differentiating the denominator actually make the divergence worse instead of better? Am I missing something? $\endgroup$ – Matthew Dodelson Jan 13 '14 at 2:50
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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.

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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.

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  • $\begingroup$ The bounds of the integral, @IgorRivin, the bounds of the integral... $\endgroup$ – Did Jan 12 '14 at 19:01
  • $\begingroup$ I tried that, but unfortunately the Jacobian determinant is zero for that transformation. $\endgroup$ – Matthew Dodelson Jan 12 '14 at 19:02
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    $\begingroup$ @Did yes, what is your problem, exactly? $\endgroup$ – Igor Rivin Jan 12 '14 at 19:02
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    $\begingroup$ That should reduce your dimension a lot. $\endgroup$ – Igor Rivin Jan 12 '14 at 19:07
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    $\begingroup$ @Did you started with the tone, so heal thyself first. $\endgroup$ – Igor Rivin Jan 12 '14 at 20:56

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