# Singularity structure of integrals of rational functions

Suppose I have a convergent integral of the form $\int_0^1dx_1\dots\int_0^1 dx_n \frac{P(x_i)}{Q(x_i)}$, where $P$ and $Q$ are polynomial functions of $n$ nonnegative real variables $x_i$. Let the coefficients of the various monomials in $Q$ be $a_k$. Depending on the structure of $Q$, the integral could develop singularities in the limit that some of the coefficients $a_k$ vanish. For example, the integral $\int_0^1 dx\frac 1 {a+x}$ goes like $-\log a$ as $a\to 0$. Is there a systematic way to extract the structure of these singularities for general multidimensional integrals?

The particular example I'm interested in is the following: I'd like to determine the singularities of

$\begin{eqnarray} \int_0^1 d\lambda_1\int_0^1 d\lambda_2 \int_0^1d\alpha_1 \int_0^1d\alpha_2 \frac{\alpha_1\alpha_2}{(\lambda_2\alpha_1\alpha_2+\lambda_1\alpha_1\alpha_2+a\alpha_2\lambda_1^2+b\alpha_1\lambda_2^2+c\alpha_1\alpha_2)^2}, \end{eqnarray}$

as $a,b,$ and $c$ tend to zero. If necessary, it's alright to assume they all tend to zero at the same rate. I naively expect a leading singularity that looks like $\log a \log b \log c$, followed by sub-leading singularities that look like $\log a \log b$ times some convergent integral, $\log b \log c$ times some other convergent integral, etc.. For my purposes, these $\log$-squared terms suffice, but I'd be interested to know how to systematically go further.

Edit: The above integral is an example of a particular Feynman integral, written in terms of Feynman parameters.

• The Feynman integral is a hypothetical integral that should allow one to integrate over spaces of curves, very similar to the Wiener measure - the formal difference being that the "time parameter" is multiplied by a $\sqrt {-1}$. How is your integral an instance of a Feynman integral? – Alex M. May 20 '18 at 15:17
• @AlexM. The "Feynman integral" that you refer to is the Feynman path integral. By "Feynman integral" in his edit, David S-D means an integral arising from the evaluation of a Feynman diagram (which are related, as they appear in perturbative calculations of path integrals). See en.wikipedia.org/wiki/Feynman_diagram and en.wikipedia.org/wiki/Feynman_parametrization – j.c. May 20 '18 at 15:45
• The back of envelope computation (no rigor!) suggests that even the leading singularity is some constant times $\log a\log b\log c+\frac 12(\log a+\log b)\log^2 c+\frac 13\log^3 c$. To extract the lower order terms may be a real headache (alas, I don't know any systematic way of doing that). What is the minimal information you can get away with? – fedja May 20 '18 at 16:32
• What you ask for can be done for this particular integral (no systematic approach though, just some combination of trickery and luck) except $\log^2 c$ appears in the quadratic part as well. Let me know if you are still interested after 7 years (if you aren't, I'd rather spare a couple of hours of typing). – fedja May 20 '18 at 18:47
• @j.c. OK, I'll post in small morsels as usual, when I have some time for typing :-) – fedja May 22 '18 at 23:34

OK. It is going to be long and extremely boring, but, as promised, here goes. We'll get the precision $O(mL^2)$ where $m=\max(a,b,c), L=\max(\log\frac 1a,\log\frac 1b,\log\frac 1c)$, which should be enough for any reasonable interpretation of "$a,b,c$ going to $0$ at about the same rate".
We start with the approximation of an auxiliary integral: $$I(a,b)=\iint_{[a,+\infty)\times[b,+\infty)}\frac{1}{(1+x+y)^2}\frac{dx}{x}\frac{dy}{y}$$ as $a,b\to 0+$.
First, we have $$\frac{1}{(1+x+y)^2}-\frac 1{(1+x)^2}\frac 1{(1+y)^2}=\frac{xy(2+2x+2y+xy)}{(1+x+y)^2(1+x)^2(1+y)^2} \\ \le \frac{2xy}{(1+x)^{2}(1+y)^{2}}\,,$$ which means that $I(a,b)=J(a)J(b)+D_1+O(a+b)$ where $$J(a)=\int_{a}^\infty\frac{1}{(1+x)^2}\frac{dx}x\,,$$ and $$D_1=\iint_{(0,\infty)^2}\frac{(2+2x+2y+xy)dx\,dy}{(1+x+y)^2(1+x)^2(1+y)^2}$$ (below we will always denote by $D_k$ various numerical constants whose values are expressed by some convergent integrals. Some of them are easy to evaluate and some aren't, but that distinction won't bother us here).
Now $$J(a)=\int_a^\infty\left[\frac 1{(1+x)^2}-\chi_{[0,1]}(x)\right]\frac{dx}{x}+\log\frac 1a\,.$$ Since for $x<1$, we have $\frac 1{1+x}-1=-\frac x{1+x}$, we conclude that $$J(a)=\log\frac 1a+D_2+O(a)$$ with $$D_2=\int_0^\infty \left[\frac 1{(1+x)^2}-\chi_{[0,1]}(x)\right]\frac{dx}{x}$$ Multiplying out, we get $$I(a,b)=\log\frac 1a\log\frac 1b+D_2(\log\frac 1a+\log\frac 1b)+D_3 \\ +O\left((a+b)(\log\frac 1b+\log\frac 1a)\right)\,.$$ Now we can return to our quadruple integral. Let us rewrite the integrand in the form $$\frac{1}{(\lambda_1+\lambda_2+c)^2}\frac{1}{\left(1+\frac{a\lambda_1^2}{\lambda_1+\lambda_2+c}\frac 1{\alpha_1}+ \frac{b\lambda_2^2}{\lambda_1+\lambda_2+c}\frac 1{\alpha_2}\right)^2} \,d\lambda_1\,d\lambda_2\,\frac{d\alpha_1}{\alpha_1}\,\frac{d\alpha_2}{\alpha_2}$$ For fixed $\lambda_1,\lambda_2$, we can now happily integrate $\alpha_1,\alpha_2$ out using the computation above to reduce our integral to $$\iint_{[0,1]^2} \frac 1{(\lambda_1+\lambda_2+c)^2}\left[\log\frac{\lambda_1+\lambda_2+c}{a\lambda_1^2}\log\frac{\lambda_1+\lambda_2+c}{b\lambda_2^2} \\+ D_2\left(\log\frac{\lambda_1+\lambda_2+c}{a\lambda_1^2}+\log\frac{\lambda_1+\lambda_2+c}{b\lambda_2^2}\right)+D_3\right]\,d\lambda_1\,d\lambda_2+E_1$$ where the error term $E_1$ is controlled by $$\iint_{[0,1]^2}\frac{1}{(\lambda_1+\lambda_2+c)^2}\left[ \left(\frac{a\lambda_1^2}{\lambda_1+\lambda_2+c}+ \frac{b\lambda_2^2}{\lambda_1+\lambda_2+c}\right) \\ \left(\log\frac{\lambda_1+\lambda_2+c}{a\lambda_1^2}+\log\frac{\lambda_1+\lambda_2+c}{b\lambda_2^2}\right) \right]\,d\lambda_1\,d\lambda_2$$
Now we do not need a special investigation of $E_1$ because the first factor in the brackets is bounded by $a+b$ and if we look at the rest, it is just one of the terms in the main asymptotic integral. Next, once we open the parentheses and expand logarithms of products, we see that we just need to approximate three double integrals: $$I_2=\iint_{[0,1]^2} \frac {d\lambda_1\,d\lambda_2}{(\lambda_1+\lambda_2+c)^2}\log\frac{\lambda_1+\lambda_2+c}{\lambda_1^2}\log\frac{\lambda_1+\lambda_2+c}{\lambda_2^2}\,,$$ $$I_1=\iint_{[0,1]^2} \frac {d\lambda_1\,d\lambda_2}{(\lambda_1+\lambda_2+c)^2}\log\frac{\lambda_1+\lambda_2+c}{\lambda_1^2}\,,$$ and $$I_0=\iint_{[0,1]^2} \frac {d\lambda_1\,d\lambda_2}{(\lambda_1+\lambda_2+c)^2}$$ IT is not hard to convince yourself that in all $3$ cases the integral over the domain $\lambda_1+\lambda_2>1$ tends to its value at $c=0$ and the deviation from that value is $O(c)$, so we can replace those parts by appropriate constants. In the other part it is convenient to make the change of variable $\lambda_1=v\lambda,\lambda_2=(1-v)\lambda$ ($0<v,\lambda<1$) and to integrate $v$ out (again splitting logarithms of products into sums). We see that we need to approximate $$I_2'=\int_0^1\frac{\lambda\,d\lambda}{(\lambda+c)^2}\log^2\frac{\lambda+c}{\lambda^2}\,,$$ $$I_1'=\int_0^1\frac{\lambda\,d\lambda}{(\lambda+c)^2}\log\frac{\lambda+c}{\lambda^2}$$ $$I_0'=\int_0^1\frac{\lambda\,d\lambda}{(\lambda+c)^2}$$