# Can this integral be expanded as a power series?

I am looking at this integral:

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

with:

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

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?

EDIT:

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?

For $$x_1, x_2 > 0$$, we have

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

where

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

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

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

• 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 $0$?
– Pxx
Jul 9, 2020 at 9:00
• @Jxx, Indeed, this shows that your function develops a logarithmic singularity around $x_2=0$, which prevents power series expansion from happening. Jul 9, 2020 at 9:06