I am looking at this integral:

$$I(x_1^2, x_2^2) = \frac{1}{4}x_1^2 x_2^2 \int\limits_{-\infty}^{\infty} d\tau_3 \int\limits_{-\infty}^{\tau_3} d\tau_4 \int\limits_{-\infty}^{\tau_4} d\tau_5 \int\limits_{-\infty}^{\tau_5} d\tau_6\ (I_{13}I_{25} + I_{15}I_{23})(I_{14}I_{26} + I_{16}I_{24}), \tag{1}$$


$$I_{ij} := \frac{1}{(2\pi)^2} \frac{1}{x_i^2 + \tau_j^2}. \tag{2}$$

I would like to expand $I(1,x_2^2)$ as a power series centered at $x_2 \sim 0^+$, i.e.:

$$I(1,x_2^2) = c_0 + c_1 x_2 + c_2 x_2^2 + ... \tag{3}$$

with $x_2 > 0$. I know from numerical integration that:

$$I(1,0) = \frac{1}{2^{12} \pi^4} = c_0. \tag{4}$$

I would like to find as many coefficients as possible in eq. (3). My naive attempt was to power expand the integrand of eq. (1), which gives:

$$\begin{align} I(1,x_2^2) = \frac{x_2^2}{4(2\pi)^8} \sum_{k=0}^\infty \sum_{l=0}^\infty (-1)^{k+l} x_2^{2(k+l)} \int\limits_{-\infty}^{\infty} d\tau_3 \int\limits_{-\infty}^{\tau_3} d\tau_4 \int\limits_{-\infty}^{\tau_4} d\tau_5\\ \int\limits_{-\infty}^{\tau_5} d\tau_6\ \left( \frac{\tau_5^{-2(k+1)}}{1+\tau_3^2} + \frac{\tau_3^{-2(k+1)}}{1+\tau_5^2} \right) \left( \frac{\tau_6^{-2(l+1)}}{1+\tau_4^2} + \frac{\tau_4^{-2(l+1)}}{1+\tau_6^2} \right). \end{align} \tag{5}$$

The idea was to numerically integrate the integrals for given $k$ and $l$. Unfortunately this cannot be right, since there is no $c_0$ term, and the integrals seem to diverge for all $k,l$.

So can such an expansion as a power series be done, and if yes how?


So I have managed to reduce $(1)$ to a one-dimensional integral analytically (recall that I assume $x_1,x_2>0$):

$$\begin{align} I(x_1^2,x_2^2) = \frac{1}{256\pi^6} \int\limits_{-\infty}^\infty d\tau_3\ \Biggl\lbrace x_1 I_{13} \left(\tan^{-1} \frac{\tau_3}{x_2} \right)^2 \left( 2\tan^{-1} \frac{\tau_3}{x_1} + \pi \right)\\ + x_2 I_{23} \left(\tan^{-1} \frac{\tau_3}{x_1} \right)^2 \left( 2\tan^{-1} \frac{\tau_3}{x_2} + \pi \right) \Biggr\rbrace. \end{align}\tag{6}$$

From there I should be able to compute the coefficients. For example the coefficient $c_1$ should obey:

$$c_1 = \left. \frac{\partial}{\partial x_2} I(1,x_2^2) \right|_{x_2 = 0}. \tag{7}$$

However when I differentiate $(6)$ and then integrate for decreasing values of $x_2$, the result does not seem to converge. I am moderately confident that this is not a numerical artefact. I observe the same thing for the second derivative.

Any idea why this is happening, or what it could mean?


1 Answer 1


For $x_1, x_2 > 0$, we have

$$ I(x_1^2,x_2^2) = \frac{F(x_1/x_2) + F(x_2/x_1)}{2^{10}\pi^7}, $$


$$ F(x) := \int_{-\infty}^{\infty} \mathrm{d}t \, \frac{\arctan^2(xt)}{t^2+1}. $$

1. The behavior of $F(x)$ as $x \to 0^+$ is easier to study. Indeed, $F(0) = 0$ is obvious, and

$$ \frac{F(x)}{x} \stackrel{(u=xt)}= \int_{-\infty}^{\infty} \mathrm{d}u \, \frac{\arctan^2(u)}{u^2+x^2} \xrightarrow{x \to 0^+} \int_{-\infty}^{\infty} \mathrm{d}u \, \frac{\arctan^2(u)}{u^2} = 2\pi \log 2 $$

shows that

$$F(x) = (2\pi \log 2)x + o(x) \quad \text{as} \quad x \to 0^+.$$

In fact, it can be shown that $F(x)$ extends to an analytic function on all of $\mathbb{C}\setminus(-\infty, 1]$ that agrees with the integral definition along the positive line $(0, \infty)$.

2. The behavior of $F(1/x)$ as $x \to 0^+$ is much trickier. First, we have $F(+\infty) = \frac{\pi^3}{4}$. Then

\begin{align*} F(1/x) - F(+\infty) &= 2 \int_{0}^{\infty} \mathrm{d}t \, \frac{\arctan^2(t/x) - (\pi/2)^2}{t^2+1} \\ &= -4x \int_{0}^{\infty} \mathrm{d}t \, \frac{\arctan(t/x)\arctan(t)}{t^2+x^2} \tag{IbP} \\ &= -4x \int_{0}^{\infty} \mathrm{d}t \, \frac{\arctan(t/x)(\arctan(t) - t \mathbf{1}_{[0,1]}(t))}{t^2+x^2} \\ &\quad - 4x \int_{0}^{1} \mathrm{d}t \, \frac{\arctan(t/x)t}{t^2+x^2} . \end{align*}

In the last line, the first integral without the prefactor $4x$ converges as $x \to 0^+$ by the Dominated Convergence Theorem. Next,

$$ \int_{0}^{1} \mathrm{d}t \, \frac{\arctan(t/x)t}{t^2+x^2} = \int_{0}^{1/x} \mathrm{d}u \, \frac{\arctan(u)u}{u^2+1} \sim \frac{\pi}{2} \log(1/x) $$

as $x \to 0^+$, where the asymptotic equivalence in the last step follows from the L'Hopital's Rule. Combining altogether,

$$ F(1/x) = \frac{\pi^3}{4} + (2\pi + o(1)) x \log x \quad\text{as}\quad x \to 0^+. $$

Conclusion. Using the above estimates, we get

$$ I(1,x_2) = \frac{1}{2^{12}\pi^4} + \frac{1 + o(1)}{2^9\pi^6} x_2 \log x_2 \quad\text{as}\quad x_2 \to 0^+. $$

  • $\begingroup$ Wow, this is amazing! I tried to reproduce your conclusions numerically, and it is pretty close: computing $(I(1,x_2)-c_0)/x_2 \log x_2$ for $x_2 = 10^{-25}$, I obtain $2.01793 \cdot 10^{-6}$, while $1/2^9\pi^6$ gives $2.03157 \cdot 10^{-6}$. Quite convincing! Would you say that this means the integral cannot be expanded as a power series around $0$? $\endgroup$
    – Pxx
    Jul 9, 2020 at 9:00
  • 1
    $\begingroup$ @Jxx, Indeed, this shows that your function develops a logarithmic singularity around $x_2=0$, which prevents power series expansion from happening. $\endgroup$ Jul 9, 2020 at 9:06

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