I am interested in computing the complex integral
$$
f (\omega) = \int_{-1}^{1} \!\! \mathrm{d} x \, \frac{\sqrt{1 – x^{2}}}{x – \omega} ,
$$
where ${ \omega \!=\! \omega_{0} + \mathrm{i} \eta }$ is a complex number.
This integral is very similar to integrals that appear in the plasma dispersion function,
up to two changes: (i) the range of integration is finite, ${ -1 \!\leq\! x \!\leq\! 1 }$; (ii) the integrand function ${ x \!\mapsto\! \sqrt{1 – x^{2}} }$ suffers from branch cuts when extended to the whole complex plane.
As is usually done in plasma physics, I want to compute this integral using Landau's prescription.
So, I first start with the case ${ \mathrm{Im} [\omega] > 0 }$. In that case, the integral can be done without any particular care. Using Mathematica, I got
$$
f (\omega) = -\pi (\omega_{0} + \mathrm{i} \eta) \bigg( 1 – \sqrt{1 + \frac{1}{(\eta – \mathrm{i} \omega_{0})^{2}}} \bigg) \quad \text{ for } \quad \mathrm{Im} [\omega] > 0
$$
Then, I compute the integral for ${ \mathrm{Im} [\omega] = 0 }$. In that case, I need to be careful with the pole that lies along the real line, noting that this pole only contributes if ${ |\omega_{0}| < 1 }$. And following Plemelj's formula, it only contributes ''half'' a residue, hence the factor $\mathrm{i} \pi$. Using once again Mathematica, I obtained
$$
f (\omega) =
\begin{cases}
\displaystyle – \pi \omega_{0} + \mathrm{Sign} [\omega_{0}] \, \pi \sqrt{\omega_{0}^{2} – 1} & \text{if} \quad |\omega_{0}| > 1 ,
\\
\displaystyle – \pi \omega_{0} + \mathrm{i} \pi \sqrt{1 – \omega_{0}^{2}} & \text{if} \quad |\omega_{0}| < 1 ,
\end{cases}
\quad \text{for} \quad \mathrm{Im}[\omega] = 0 .
$$
Finally, I tried to compute the same integral, this time for ${ \mathrm{Im} [\omega] < 0 }$. Naively, I thought that I could re-use the expression from the case ${ \mathrm{Im}[\omega] > 0 }$,
and simply add the contribution from ${ 2 \mathrm{i} \pi }$ times the residue of the pole in $\omega$. As such, I wrote
$$
f(\omega) =
\begin{cases}
\displaystyle -\pi (\omega_{0} + \mathrm{i} \eta) \bigg( 1 – \sqrt{1 + \frac{1}{(\eta – \mathrm{i} \omega_{0})^{2}}} \bigg) & \text{if} \quad |\omega_{0}| > 1
\\
\displaystyle -\pi (\omega_{0} + \mathrm{i} \eta) \bigg( 1 – \sqrt{1 + \frac{1}{(\eta – \mathrm{i} \omega_{0})^{2}}} \bigg) + 2 \mathrm{i} \pi \sqrt{1 – (\omega_{0} + \mathrm{i} \eta)^{2}} & \text{if} \quad |\omega_{0}| < 1
\end{cases}
\quad \text{for} \quad \mathrm{Im}[\omega] < 0 .
$$
Unfortunately, when I look at the behaviour of the function ${ (\omega_{0} , \eta) \!\mapsto\! f (\omega_{0} + \mathrm{i} \eta)}$, I notice that my function is not continuous. Indeed, it suffers from a discontinuity along the lines ${ \omega_{0} \!=\! \pm 1 }$ and ${ \eta < 0 }$. I believe that the issue comes from missing contributions associated the branch cuts of the function ${ (\omega_{0} , \eta) \!\mapsto\! \sqrt{1 – (\omega_{0} + \mathrm{i} \eta)^{2}} }$ that is discontinuous along the lines ${ |\omega_{0}| > 1 }$ and ${ \eta = 0 }$. But, it is unclear to me how those should be accounted for.
How should one proceed to compute explicitly (and correctly!) the present integral?
PS: In practice, I must also compute a second integral, where one makes the replacement ${ \sqrt{1 – x^{2}} \!\to\! x \sqrt{1 – x^{2}}}$ in the integrand. It is likely that computing this second integral requires solving the exact same difficulties.
Best Answer
1. Focusing only on the integral
$$ f(\omega) = \int_{-1}^{1} \frac{\sqrt{1-x^2}}{x-\omega} \, \mathrm{d}x \tag{*} $$
mathematically, OP's first formula works for all $\omega \in \mathbb{C}\setminus[-1,1]$, i.e.,
$$ f(\omega) = \pi \omega \left( \sqrt{1 - \frac{1}{\omega^2}} - 1 \right), \tag{1} $$
where $\sqrt{\,\cdot\,}$ is the principal square root. Indeed, it is not hard to check that $\text{(1)}$ holds for $\omega \in (1, \infty)$. Since both sides of $\text{(1)}$ are analytic on the open connected set $\mathbb{C}\setminus[-1,1]$, the identity extends to all of $\mathbb{C}\setminus[-1,1]$ by the principle of analytic continuation. (Note that the branch cut of $\sqrt{1-\omega^{-2}}$ is precisely $[-1, 1]$.) Also by the continuity, $\text{(1)}$ continues to hold at $\omega = \pm 1$.
Figure 1. The following is a domain coloring of the function $f(\omega)$ in $\text{(1)}$. (Here, hue represents the argument of $f(z)$ and brightness represents the magnitude of $f(z)$. The discontinuities in brightness is introduced at every level of the form $2^n$ to better visualize the growth. The white line represents the branch cut of $f$.)
2. However, thinking that this question concerns the Landau prescription, I guess OP is actually interested in the integral
$$ f(\omega) = \int_{\mathcal{C}} \frac{\sqrt{1-\xi^2}}{\xi - \omega} \, \mathrm{d}\xi $$
where $\mathcal{C}$ is a Landau contour from $-1$ to $1$. Saying differently, $f(\omega)$ is the analytic continuation of the function $\text{(*)}$, initially defined on $\operatorname{Im}(\omega) > 0 $ and then extended across and beyond the slit $[-1, 1]$. The resulting function is
$$ f(\omega) = \begin{cases} - \pi \omega + \pi \omega \sqrt{1 - \omega^{-2}}, & \operatorname{Im}(\omega) > 0, \\ - \pi \omega + \pi i \sqrt{1 - \omega^2}, & \omega \in [-1, 1], \\ - \pi \omega + \pi \omega \sqrt{1 - \omega^{-2}} + 2\pi i \sqrt{1 - \omega^2}, & \operatorname{Im}(\omega) < 0. \end{cases} \tag{2} $$
This function is analytic outside $(-\infty,-1]\cup[1,\infty)$.
We also note that it is impossible to find an analytic continuation of $f$ on all of $\mathbb{C}$, since $\pm 1$ are branch points of $f$.
Figure 2. The following is a domain coloring of the function $f(\omega)$ in $\text{(2)}$. Again, the white line represents the branch cut of $f$.
3. Finally, we have
\begin{align*} \int_{-1}^{1} \frac{x\sqrt{1-x^2}}{x-\omega} \, \mathrm{d}x &= \int_{-1}^{1} \frac{(x - \omega + \omega)\sqrt{1-x^2}}{x-\omega} \, \mathrm{d}x \\ &= \int_{-1}^{1} \sqrt{1-x^2} \, \mathrm{d}x + \omega \int_{-1}^{1} \frac{\sqrt{1-x^2}}{x-\omega} \, \mathrm{d}x \\ &= \frac{\pi}{2} + \omega f(\omega). \end{align*}
This relation can be used to find the expression for $g$.