Let $1 \le n_1 < n $ and $ n\ge 3$ be integers and let ${\bf \lambda}= \left( \lambda_j \right)_{j=1}^n \in {\mathbb R}$ such that $\lambda_j > 0 $ for $j=1,\cdots, n_1$ and $\lambda_j <0 $ for $j=n_1+1, \cdots, n$ and such that $1/{\bf \lambda}$ is ordered in descending order. In the following we will assume that $n_1=1$.
Now we define a function $ {\mathbb C} \ni u \rightarrow f(u) \in {\mathbb C}$:
\begin{equation} f(u):= \frac{1}{\sqrt{\prod\limits_{j=1}^n (1+ \imath u \lambda_j)}} \tag{1} \end{equation}
In addition we also make the following assumption below.
Assumption:
\begin{equation} max_{u \in {\mathbb R}} \left| \sum\limits_{j=1}^n Arg(1 + \imath u \lambda_j) \right| < \pi \tag{1a} \end{equation}
We have checked numerically that the assumption in question can be only satisfied when $n=3,4$ (not more than that). The assumption is a a necessary and sufficient condition for the function on the real axis to be continuous. In other words if the left hand side of $(1a)$ hits the value of $\pm \pi$ at some $u= u_0 \in {\mathbb R}$ then the imaginary part of our function has a jump ( a sign flip) and the real part has a cusp at this $u=u_0$ .
Now we consider the following improper integrals:
\begin{equation} \left( {\mathcal I}_-({\bf \lambda}), {\mathcal I}_+({\bf \lambda}) \right) := \left( \int\limits_{-\infty}^0 f(u) du, \int\limits_0^\infty f(u) du \right) \tag{2} \end{equation}
Note that the branching points of the integrand all lie on the imaginary axis.
Motivation:
It turns out that the integrals $(2)$ (or their generalizations to be exact) appear when computing the distribution function of the estimator of the mean reversion parameter in the Ornstein-Uhlenbeck model, see
Bao, Yong; Ullah, Aman; Wang, Yun, Distribution of the mean reversion estimator in the Ornstein-Uhlenbeck process, Econom. Rev. 36, No. 6-9, 1039-1056 (2017). ZBL1524.62406.
Results:
- By considering a contour in the upper half plane composed of, an interval along the real, negative half-axis then a rising vertical section parallel to the imaginary axis then a small half circle surrounding the branching point then a falling section parallel to the imaginary axis then an interval along the real, positive half axis and finally a big half circle in the upper half plane we have shown that the following identity holds true:
\begin{equation} {\mathcal I}_-({\bf \lambda}) - {\mathcal I}_+({\bf \lambda}) + 2 \int\limits_0^{1/\lambda_1} f(\imath \xi) \imath d\xi = 0 \tag{3a} \end{equation}
- Now, by considering a mirror reflection of the contour above with respect to the real axis modified accordingly to avoid the branching points we have shown that another identity holds true:
\begin{eqnarray} {\mathcal I}_-({\bf \lambda}) + {\mathcal I}_+({\bf \lambda}) + 2 \int\limits_{-1/\lambda_2}^{-1/\lambda_3} f(-\imath \xi) (-\imath) d\xi = 0 \quad & \quad \mbox{if $n=3$} \tag{4a} \\ {\mathcal I}_-({\bf \lambda}) - {\mathcal I}_+({\bf \lambda}) + 2 \int\limits_0^{-1/\lambda_2} f(-\imath \xi) (-\imath) d\xi + (\tau + \tau^3) \int\limits_{-1/\lambda_2}^{-1/\lambda_3} f(-\imath \xi) (-\imath) d\xi + (\tau^2 + \tau^2) \int\limits_{-1/\lambda_3}^{-1/\lambda_4} f(-\imath \xi) (-\imath) d\xi = 0 \quad & \quad \mbox{if $n=4$} \tag{4b} \\ \end{eqnarray}
with $\tau = -\imath$.
Numerical verification:
mprec = 15; prec = 10;
SetOptions[NIntegrate, WorkingPrecision -> mprec,
PrecisionGoal -> prec];
(*Note: We need n\[GreaterEqual]3 otherwise integral over big half \
circle diverges.*)
n = RandomInteger[{3, 4}];
lmb = RandomReal[{0, 10}, n, WorkingPrecision -> 50];
lmb = lmb Table[If[j == 1, 1, -1], {j, 1, n}];
(*Note: Here 1/lmb is sorted in a descending order.*)
lmb = Flatten[{lmb[[1]], Sort[Drop[lmb, 1], #1 < #2 &]}];
f[u_] := 1/Sqrt[Product[1 + I u lmb[[j]], {j, 1, Length[lmb]}]];
res = {NIntegrate[f[u], {u, -Infinity, 0}],
NIntegrate[f[u], {u, 0, Infinity}]};
(*The identities below holds true iff Abs[Sum[Arg[1+I u \
lmb[[j]]],{j,1,Length[lmb]}]]< Pi for all u \in (-Infinity,Infinity) *)
res[[1]] - res[[2]] + 2 NIntegrate[f[I xi] I, {xi, 0, 1/lmb[[1]]}]
ph1 = -I;
Which[
n == 3,
res[[1]] +
res[[2]] + (2) NIntegrate[
f[-I xi] (-I), {xi, -1/lmb[[2]], -1/lmb[[3]]}],
n == 4,
res[[1]] - res[[2]] +
2 NIntegrate[f[-I xi] (-I), {xi, 0, -1/lmb[[2]]}] +
(ph1 + ph1^3) NIntegrate[
f[-I xi] (-I), {xi, -1/lmb[[2]], -1/lmb[[3]]}] +
(ph1^2 + ph1^2) NIntegrate[
f[-I xi] (-I), {xi, -1/lmb[[3]], -1/lmb[[4]]}]
]
Question:
Now, having said all this my question is whether we can generalize the results $(4a),(4b)$ to the case when the assumption is not satisfied. In particular how would we proceed when $n > 4$?
First of all it is clear that the assumption can only be satisfied if the number of positive entries in ${\bf \lambda}$ is similar to that of negative entries. Indeed $\lim_{u \rightarrow \infty } | \sum\limits_{j=1}^n Arg(1 + \imath u \lambda_j) | = \pi/2 | \sum\limits_{j=1}^n (1_{\lambda_j >0} - 1_{\lambda_j < 0})| $ and therefore the numbers in question can only differ by at most two for the assumption to be satisfied. As such we analyzed the case where $n = 2 n_1 + \xi $ with $n_1 = 1,2,3,\cdots $ and $\xi=0,1,2$ (this is the case where there is $n_1$ and $n_1+\xi$ positive and negative entries respectively). Then by applying the Cauchy theorem to the contours described above and bearing in mind that each encircling of a branch point by an angle $\phi \in (0,2 \pi) $ induces a multiplicative phase factor $\exp(\imath \phi/2)$ we obtained the following identities:
\begin{eqnarray} &&(+\imath) \left({\mathcal I}_-({\bf \lambda}) + {\mathcal I}_+({\bf \lambda}) \right) 1_{n_1 \% 2=0} + \left({\mathcal I}_-({\bf \lambda}) - {\mathcal I}_+({\bf \lambda}) \right) 1_{n_1 \% 2=1} + \imath \sum\limits_{j=0}^{n_1-1} (\tau^j + \tau^{2 (n_1-1)-j}) \cdot \underline{\int\limits_{ \frac{1}{\lambda_{n_1-j+1}} }^{ \frac{1}{\lambda_{n_1-j}} } f(\imath \xi) d\xi} = 0 \\ % &&(-\imath) \left({\mathcal I}_-({\bf \lambda}) + {\mathcal I}_+({\bf \lambda}) \right) 1_{(n-n_1) \% 2=0} + \left({\mathcal I}_-({\bf \lambda}) - {\mathcal I}_+({\bf \lambda}) \right) 1_{(n-n_1) \% 2=1} + \imath \sum\limits_{j=0}^{n-n_1-1} (\tau^j + \tau^{2 (n-1-n_1)-j}) \cdot \underline{\int\limits_{ \frac{1}{\lambda_{n_1+j}} }^{ \frac{1}{\lambda_{n_1+j+1}} } f(\imath \xi) d\xi} = 0 \end{eqnarray}
Note 1:
The identities above hold true only if the assumption is satisfied.
Note 2:
The underlined quantities , i.e. the integrals over the finite intervals, are proportional to hypergeometric functions of a vector argument where the vector in question depends on ${\bf \lambda}$.
Note 3:
The Mathematica code snippet below verifies the identities in question numerically.