0
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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 $f$.)

Graph of f, version 1

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 $f$.

Graph of f, version 2

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 $g$.

$\endgroup$
7
  • $\begingroup$ I am surprised that your solution does not involve the usual Landau's trick ? $$\int_{-\infty}^{+\infty} \!\!\!\! \mathrm{d} x \frac{F(x)}{x - \omega} = \begin{cases} \displaystyle \int_{-\infty}^{+\infty} \!\! \mathrm{d} x \frac{F(x)}{x - \omega} & \mathrm{Im}[\omega] > 0 \\ \displaystyle \mathcal{P} \int_{-\infty}^{+\infty} \!\!\!\! \mathrm{d} x \frac{F(x)}{x - \omega} + \mathrm{i} \pi F(\omega) & \mathrm{Im} [\omega] = 0 \\ \displaystyle \int_{- \infty}^{+ \infty} \!\!\!\! \mathrm{d} x \frac{F (x)}{x - \omega} + 2 \mathrm{i} \pi F (\omega) & \mathrm{Im} [\omega] < 0 \end{cases}$$ $\endgroup$ – jibe Nov 27 '20 at 22:35
  • $\begingroup$ @jibe The good thing here is that the "density" function $F(x)=\sqrt{1-x^2}\,\mathbf{1}_{[-1,1]}(x)$ is compactly supported, hence allowing the integral to "cross" the real line analytically by bypassing the slit $[-1,1]$. If the density function $F$ is analytic on an open neighborhood of $\mathbb{R}$ (hence, in particular, its restriction on $\mathbb{R}$ is supported on all of $\mathbb{R}$), such as in the case of $F(x)=e^{-x^2}$, then I agree that the integral will necessarily involve that trick. $\endgroup$ – Sangchul Lee Nov 27 '20 at 22:40
  • $\begingroup$ Ok, so now I'm confused... From the physics point of view, the identity ${ \overline{f (\omega)} = f(\overline{\omega}) }$ should not be satisfied. [In practice, ${ \mathrm{Im} [\omega] > 0 }$ corresponds to unstable modes, while ${ \mathrm{Im} [\omega] < 0 }$ corresponds to damped modes, which are very different]. In the case, ${ F (x) = e^{-x^2} }$, the conjugation identity is not satisfied, and this agrees with the physical intuition. Yet, I am surprised that in the case ${ F (x) = \sqrt{1 - x^{2}} \mathbb{1}_{[-1,1]} (x) }$, we find that this identify is satisfied. What am I missing here? $\endgroup$ – jibe Nov 27 '20 at 22:55
  • 1
    $\begingroup$ @jibe, I have to take back my last claim "... I agree that the integral will necessarily involve that trick." in the first comment. Now I see that the formalism that you mentioned in the first comment is not compatible with what we would obtain if we literally have computed the integral. Instead, what you mentioned in your first comment (esp. that piecewise expression in the right-hand side) is the recipe for computing the analytic continuation of $$\int_{-\infty}^{\infty}\frac{F(x)}{x-\omega}\,\mathrm{d}x\qquad\text{on }\operatorname{Im}(\omega)>0$$to all of $\mathbb{C}$. $\endgroup$ – Sangchul Lee Nov 27 '20 at 23:13
  • 1
    $\begingroup$ Yes, I think you got it spot on; I was indeed interested in the function obtained from the analytic continuation of the function computed in the upper half of the complex plane. I also see the mistake in my initial approach: for $\mathrm{Im}[\omega] < 0$, I should always have added the residue ${2 \mathrm{i} \pi F (\omega)}$, rather than adding it only for ${ |\mathrm{Re}[\omega]|<1 }$. On the physics side, the branch cuts along the lines ${ \mathrm{Re}[\omega] = 0 }$ are ok, they highlight, indeed, a drastic change between unstable vs. damped modes. Thanks for your kind help! $\endgroup$ – jibe Nov 28 '20 at 12:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.