Mathematica returns incorrect evaluation of integral

Question

Consider the integral $$ \begin{equation} u(x):= \lim_{\eta \to 0^+}\int_{-\infty}^\infty \frac{d s}{s^2 + 1} \left( \frac{1}{x - i \eta - \sqrt{s^2 + 1}} - \frac{1}{x - i \eta + \sqrt{s^2 + 1}} \right) . \end{equation} $$ for $x \in \mathbb{R}$.

Mathematica will evaluate this integral without complaint

In[] : = Assuming[x \[Element] Reals && \[Eta] > 0, Limit[Integrate[1/(s^2 + 1) (1/(x - I \[Eta] - Sqrt[s^2 + 1]) - 1/(x - I \[Eta] + Sqrt[s^2 + 1])), {s, -\[Infinity], \[Infinity]}], \[Eta] -> 0, Direction -> "FromAbove"]]

and returns $$ u_\mathrm{MMA}(x) = -\frac{4 \arccos\left(\sqrt{1-x^2}\right)}{\left| x\right| \sqrt{1-x^2} } $$ However this answer cannot be correct, it is an even function ($u(-x) = u(x)$), whereas the definition manifestly has the propery $u(-x) = u(x)^*$ (this can be confirmed using NIntegrate for small finite values of $\eta$). I.e. both the real and imaginary parts of $u_\mathrm{MMA}(x)$ are even, when the imaginary part should be odd.

Is there a way of getting Mathematica to return the correct answer for these types of integral?

Answer 1

The issue appears to be that Mathematica is handling the assumptions concerning x incorrectly. If you explicitly tell it that $x < -1$, $|x| \leq 1$, or $x > 1$, it produces three slightly different functional forms:

ulow[x_] = Assuming[x < -1 && \[Eta] > 0, ... ]
umid[x_] = Assuming[1 > x > -1 && \[Eta] > 0, ... ]
uhigh[x_] = Assuming[1 < x && \[Eta] > 0, ... ]

(* (4 I (\[Pi] + ArcSin[x]))/(x Sqrt[-1 + x^2]) *)
(* -((4 ArcCos[Sqrt[1 - x^2]])/Sqrt[x^2 - x^4]) *)
(* (4 I ArcSin[x])/(x Sqrt[-1 + x^2]) *)

So if desired, we can assemble these into a Piecewise function that works for all x:

u[x_] := Piecewise[{{ulow[x], x< -1}, {umid[x], -1 <= x <= 1}, {uhigh[x], x > 1}}] 
Plot[Evaluate[ReIm[u[x]]], {x, -Pi, Pi}]

This does appear to have the property that $\Re(u(x))$ is even in $x$ and $\Im(u(x))$ is odd in $x$.