Interesting identity for the value of an integral involving complex-valued square root.

Question

I am looking for a simple proof of the following identity which holds by numerical tests:

$$ \int_{-\infty}^\infty x\left(\sqrt{1+\frac{a}{1-x^2-2i\gamma x}}-\sqrt{1+\frac{a}{1-x^2+2i\gamma x} }\right)dx={i\pi a}\operatorname{sgn}\gamma, $$ where $a$ and $\gamma$ are real numbers. The result suggests using the residue theorem, but the residues at singularities of the integrand seem to be zero...

Any hint or suggestion are appreciated.

Answer 1

The result depends on the choice of branch cut. Indeed, with the principal square root, the result is false if $a$ is negative with large modulus:

In this answer, we will prove a corrected version of OP's identity. To this end, define

$$ f(z) := 1+\frac{a}{1-z^2-2i\gamma z} \tag{1}. $$

This functions has poles at $z = -i\gamma \pm \sqrt{1-\gamma^2}$ and zeros at $z = -i\gamma \pm \sqrt{1+a-\gamma^2}$. Now if $a > -1$ and $\gamma > 0$, then all of these points lie in the lower half plane, and so, there exists a unique analytic function $g$ on the closed upper half plane $\overline{\mathbb{H}} = \{z \in \mathbb{C} : \operatorname{Im}(z) \geq 0\}$ such that

$$ \lim_{\overline{\mathbb{H}} \ni z \to \infty} g(z) = 1 \qquad \text{and} \qquad g(z)^2 = f(z). \tag{2} $$

Moreover, we can check that $g(x) = \sqrt{f(x)}$ holds for $x \in \mathbb{R}$, where $\sqrt{\cdot}$ is the principal square root. Now we claim

Proposition. For each $a > -1$ and $\gamma > 0$, we have

$$ \int_{-\infty}^{\infty} x \left( \sqrt{f(x)} - \sqrt{f(-x)} \right) \, \mathrm{d}x = i \pi a. \tag{3} $$

Note that this claim is enough to establish OP's identity for $a > -1$.

Proof of Proposition. Write $I$ for the left-hand side of $\text{(3)}$. Then

\begin{align*} I &= \lim_{R\to\infty} \int_{-R}^{R} x \left( \sqrt{f(x)} - \sqrt{f(-x)} \right) \, \mathrm{d}x \\ &= 2 \lim_{R\to\infty} \int_{-R}^{R} x \left( \sqrt{f(x)} - 1 \right) \, \mathrm{d}x \\ &= 2 \lim_{R\to\infty} \int_{-\Gamma_{R}} z \left( g(z) - 1 \right) \, \mathrm{d}z, \end{align*}

where $-\Gamma_{R}$ is the clockwise-oriented upper semicircular contour from $-R$ to $R$. Parametrizing $-\Gamma_{R}$ by $z = Re^{i\theta}$ with $\theta$ from $\pi$ to $0$, we get

\begin{align*} I &= -2i \lim_{R\to\infty} \int_{0}^{\pi} R^2e^{2i\theta} \left( g(Re^{i\theta}) - 1 \right) \, \mathrm{d}\theta. \end{align*}

But in view of $\text{(2)}$, it follows that

$$ R^2e^{2i\theta} \left( g(Re^{i\theta}) - 1 \right) = \frac{R^2e^{2i\theta} \left( f(Re^{i\theta}) - 1 \right)}{g(Re^{i\theta}) + 1} \xrightarrow[R\to\infty]{} -\frac{a}{2} $$

holds uniformly in $\theta \in [0, \pi]$. Therefore we can interchange the order of limit and integration to obtain

\begin{align*} I &= -2i \int_{0}^{\pi} \lim_{R\to\infty} R^2e^{2i\theta} \left( g(Re^{i\theta}) - 1 \right) \, \mathrm{d}\theta \\ &= -2i \int_{0}^{\pi} \left( -\frac{a}{2} \right) \, \mathrm{d}\theta \\ &= a\pi i. \end{align*}

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Addendum. By the same technique, we have

Proposition. For each $a < -1$ and $\gamma > 0$, set $\beta = -\gamma+\sqrt{\gamma^2-a-1}$. Then we have

$$ \int_{-\infty}^{\infty} x \left( \sqrt{f(x)} - \sqrt{f(-x)} \right) \, \mathrm{d}x = i \pi a + 4i \int_{0}^{\beta} t \left( \frac{-a}{t^2+2\gamma t+1}-1 \right)^{1/2} \, \mathrm{d}t. $$

The extra integral on the right-hand side comes from part of the branch cut $[0,i\beta]$ lying inside the upper half plane. And Mathematica seems to suggest that this integral reduces to elliptic integral.