1
$\begingroup$

Let $f$ be a function defined on $\mathbb{C}^2$ given by $$ f(s,t)=\int\limits_{-\infty}^{\infty}dk_1 \int\limits_{-\infty}^{\infty}dk_2 \int\limits_{-\infty}^{\infty}dk_3 \frac{1}{\left(\sqrt{s}-k_1\right)^2+k_2^2+k_3^2} \frac{1}{\left(\sqrt{t}-k_1\right)^2+k_2^2+k_3^2}\cdot \frac{1}{k_1^2+k_2^2+k_3^2+s}. $$ I would like to show that $f(s,t)$ is analytic for all $s,t$ satisfying $\text{Im}(s) > 0$ and $\text{Im}(t) > 0$ (the square root is defined with a branch cut on the positive real axis) or at least find the domain of analyticity if the this statement is not true.

My attempt:For any given $s_0,t_0$ we want to use Osgood's Lemma to prove analyticity in the two variables independently in an open poly disc around $s_0,t_0$ of radius $\epsilon_1,\epsilon_2$. We notice that poles of the integrand are at $k_1=\text{Re}(s), k_2^2+k_3^2=(\text{Im}(\sqrt{s}))^2$ and $k_1=\text{Re}(t), k_2^2+k_3^2=(\text{Im}(\sqrt{t}))^2$ for the first and second denominators respectively. For $\text{Im}(s) > 0$ and $\text{Im}(t) > 0$ the third denominator has no poles. We can therefore split the integrand into two pieces: $$ f(s,t)=\left(\int\limits_{-\infty}^{\text{Re}(\sqrt{t})}+ \int\limits_{\text{Re}(\sqrt{t})}^{\text{Re}(\sqrt{s})}+\int\limits_{\text{Re}(\sqrt{s})}^{\infty}\right)dk_1 \int\limits_{-\infty}^{\infty}dk_2 \int\limits_{-\infty}^{\infty}dk_3 \frac{1}{\left(\sqrt{s}-k_1\right)^2+k_2^2+k_3^2} \frac{1}{\left(\sqrt{t}-k_1\right)^2+k_2^2+k_3^2}\cdot \frac{1}{k_1^2+k_2^2+k_3^2+s}. $$

In this form, we only have a divergence at the end points of the three domain and the function is clearly integrable in each domain (since the divergence at the end point is power counting finite). We would now like to evaluate anti holomorphic derivative and since in each domain, the modulus of the function is also integrable, we use Fubini to switch the derivative and the integration. The holomorphicity of the integrand implies the holomorphicity the $f(s,t)$. Is this line of reasoning correct? If not, what would complete such a proof?

Context: An (infinite class) of such integrals appear in the evaluation of massless Feynman amplitudes. A natural question one asks when studying quantum field theory is wether such an amplitude admits an analytic continuation.

Edit: Any pointers to known results that might help prove analyticity of a complex function of two variables, defined under the integral sign, with singular and integrable subsurfaces will be very helpful.

$\endgroup$
4
  • $\begingroup$ What does Im(s,t) mean? $\endgroup$ Commented Aug 17, 2023 at 17:54
  • $\begingroup$ @DanielAsimov Thanks for your comment. I meant that as shorthand for the two inequalities Im(s)>0 and Im(t)>0, where Im(z)=Imaginary part of z. $\endgroup$ Commented Aug 17, 2023 at 18:12
  • $\begingroup$ I still don't know what Im(s,t) means. $\endgroup$ Commented Aug 17, 2023 at 18:22
  • $\begingroup$ @DanielAsimov Thanks again. I have edited the question to be clearer. $\endgroup$ Commented Aug 17, 2023 at 18:24

0

You must log in to answer this question.