The first-passage-time moments for the Hougaard process and its Birnbaum–Saunders approximation

Abstract

Hougaard processes, which include gamma and inverse Gaussian processes as special cases, as well as the moments of the corresponding first-passage-time (FPT) distributions are commonly used in many applications. Because the density function of a Hougaard process involves an intractable infinite series, the Birnbaum–Saunders (BS) distribution is often used to approximate its FPT distribution. This article derives the finite moments of FPT distributions based on Hougaard processes and provides a theoretical justification for BS approximation in terms of convergence rates. Further, we show that the first moment of the FPT distribution for a Hougaard process approximated by the BS distribution is larger and provide a sharp upper bound for the difference using an exponential integral. The conditions for convergence coincidentally elucidate the classical convergence results of Hougaard distributions. Some numerical examples are proposed to support the validity and precision of the theoretical results.

Appendix

Lemma 5.1

For $c \in {\mathbb {R}}{\setminus } \{0\}$ and $0< \alpha < 1$, the real part of $(1+ c \imath )^\alpha -1$ is positive, i.e., ${{\textrm{Re}}}{(1+c \imath )^\alpha } > 1$.

Proof

By Euler formula and De Moivre formula, there is a unique principal argument $\vartheta = \arctan \left( c \right) \in (-\pi /2,\pi /2]$ such that $(1+ c \imath )^\alpha = (1+c^2)^\frac{\alpha }{2} \left( \cos (\alpha \vartheta ) + \imath \sin (\alpha \vartheta ) \right)$. Let $g( \alpha ) = \ln \left( {{\textrm{Re}}}(1+c\imath )^\alpha \right) = \frac{\alpha }{2} \ln (1+c^2) + \ln \left( \cos (\alpha \vartheta )\right)$ for $\alpha \in [0, 1]$. Hence, it is easy to check $g(0) = g(1) = 0$ and ${ \text{ d}^2 }g(\alpha )/{\text{ d } \alpha ^2} = - \vartheta ^2 \sec (\alpha \vartheta )^2 < 0$, indicating that $g(\alpha )$ is concave. Thus, we have $\ln \left( {{\textrm{Re}}}(1+c\imath )^\alpha \right) = g(\alpha ) > \min \{ g(0), g(1) \} = 0$ for $\alpha \in (0, 1)$, which is equivalent to ${{\textrm{Re}}}{(1+c \imath )^\alpha }>1$. $\square$

Lemma 5.2

• (i) For $n \in {\mathbb {N}}, s > 0$ and $0< \alpha < 1$,

  $$\begin{aligned}&\frac{ (1 -\exp (\alpha \pi \imath )s^\alpha )^{-n} - (1 -\exp (-\alpha \pi \imath ) s^\alpha )^{-n} }{2 \imath } \nonumber \\&\quad = \frac{\sin \left( n \arctan \left( \frac{\sin (\alpha \pi ) s^\alpha }{1- \cos (\alpha \pi ) s^\alpha } \right) \right) }{ \left( 1-2\cos (\alpha \pi )s^\alpha + s^{2\alpha }\right) ^\frac{n}{2} } {{\textrm{sgn}}} \left( \left( 1 - \cos (\alpha \pi ) s^\alpha \right) ^n \right) . \end{aligned}$$

  (13)

• (ii) When $\alpha = 0$, the above equation reduces to

  $$\begin{aligned}{} & {} \frac{ (-\pi \imath - \ln s)^{-n} - ( \pi \imath - \ln s )^{-n} }{2 \imath } \\{} & {} \quad = -\frac{\sin \left( n \arctan \left( \pi / \ln s \right) \right) }{ \left( \pi ^2 + (\ln s)^2 \right) ^\frac{n}{2} } {{\textrm{sgn}}} \left( \left( 1 - s \right) ^n \right) . \end{aligned}$$

Proof

(i) By the identity $\exp ( \alpha \pi \imath ) = \cos (\alpha \pi ) + \sin (\alpha \pi ) \imath$ and binomial Theorem, we have

$$\begin{aligned}&\frac{ \left( 1- \exp (\alpha \pi \imath )s^\alpha \right) ^{-n} - \left( 1- \exp (-\alpha \pi \imath ) s^\alpha \right) ^{-n}}{2 \imath } \nonumber \\&\quad =\frac{ ( 1 - \cos (\alpha \pi ) s^\alpha + \sin (\alpha \pi ) \imath s^\alpha )^n -( 1 - \cos (\alpha \pi ) s^\alpha - \sin (\alpha \pi )\imath s^\alpha )^n }{2 \imath ( 1 - 2 \cos (\alpha \pi ) s^\alpha + s^{2\alpha } )^n } \nonumber \\&\quad =\frac{ \sum \nolimits _{j=0}^n {n \atopwithdelims ()j} ( 1 - \cos (\alpha \pi ) s^\alpha )^{n-j} (\sin (\alpha \pi ) s^\alpha )^{j} \left( \imath ^j - ( -\imath )^{j} \right) }{2 \imath ( 1 - 2 \cos (\alpha \pi ) s^\alpha + s^{2\alpha } )^n } \nonumber \\&\quad =\frac{ \sum \nolimits _{j=0}^{ \lfloor \frac{n-1}{2} \rfloor } {n \atopwithdelims ()2j+1} (-1)^j ( 1 - \cos (\alpha \pi ) s^\alpha )^{n-2j-1} (\sin (\alpha \pi ) s^\alpha )^{2j+1} }{ ( 1 - 2 \cos (\alpha \pi ) s^\alpha + s^{2\alpha } )^n }, \end{aligned}$$

(14)

where $\lfloor \cdot \rfloor$ denotes the floor function. Let $x = 1 - \cos (\alpha \pi ) s^\alpha$ and $y = \sin (\alpha \pi ) s^\alpha >0$. Use the multi-angle formula to get

$$\begin{aligned}&\sin ( n \arctan ( y / x ) ) \\&\quad = \sum \limits _{j=0}^{ \lfloor \frac{n-1}{2} \rfloor } {n \atopwithdelims ()2j+1} (-1)^j \left( \cos ( \arctan ( y / x ) ) \right) ^{n-2j-1} \\&\qquad \left( \sin (\arctan ( y / x ) ) \right) ^{2j+1} \\&\quad = \left( 1+ \frac{y^2}{x^2} \right) ^\frac{-n}{2} x^{-n} \sum \limits _{j=0}^{ \lfloor \frac{n-1}{2} \rfloor } {n \atopwithdelims ()2j+1} (-1)^j x^{n-2j-1} y^{2j+1}. \end{aligned}$$

Substituting the above identity to the numerator of (14) gives the desired result. (ii) For $\alpha = 0$, multiplying $\alpha ^n$ on both sides in (13) and applying L’Hôpital rule, the result follows. $\square$

Lemma 5.3

• (i) For $0<\alpha <1$,

  $$\begin{aligned}&\int _{0}^{\infty } \frac{ 1 }{ s + 1} \frac{ \sin \left( n \arctan \left( \frac{\sin (\alpha \pi ) s^\alpha }{1 - \cos (\alpha \pi ) s^\alpha } \right) \right) }{ ( 1 - 2 \cos (\alpha \pi ) s^\alpha + s^{2\alpha } )^\frac{n}{2} }\nonumber \\&\quad {{\textrm{sgn}}} \left( \left( 1 - \cos (\alpha \pi ) s^\alpha \right) ^n \right) {{\textrm{d}}} s \nonumber \\&\quad = \pi {{\textrm{Res}}}\left( \frac{(1-s^\alpha )^{-n}}{1-s},s=1 \right) . \end{aligned}$$

  (15)

• (ii) For $\alpha = 0$,

  $$\begin{aligned}{} & {} \int _{0}^{\infty } \frac{ -1 }{ s + 1} \frac{ \sin \left( n \arctan \left( \pi / \ln s \right) \right) }{ ( \pi ^2 + (\ln s)^2 )^\frac{n}{2} } {{\textrm{sgn}}} \left( \left( 1 - s \right) ^n \right) {{\textrm{d}}} s \\{} & {} \quad = \pi {{\textrm{Res}}}\left( \frac{(- \ln s)^{-n}}{1-s},s=1 \right) . \end{aligned}$$

Proof

(i) Define

$$\begin{aligned} q(s) = \frac{1}{2\imath } \frac{1}{1-s} \frac{1}{(1-s^\alpha )^n}, s \in {\mathbb {C}}. \end{aligned}$$

Then $s=1$ is a single pole of q(s) with order $n+1$. Consider the following closed keyhole contour IJKLI plotted in Fig. 5.

Fig. 5

Keyhole contour IJKLI and a pole at (1, 0)

We show that

$$\begin{aligned}&\left| \int _{\overset{\frown }{{\textrm{JK}}}} q(s) {{\textrm{d}}}s\right| \\&\quad = \frac{1}{2} \left| \int _{-\pi }^\pi \frac{r\exp ( \imath \vartheta )}{1- r\exp ( \imath \vartheta ) } \frac{1}{\left( 1- r^\alpha \exp ( \alpha \imath \vartheta )\right) ^n} {{\textrm{d}}} \vartheta \right| \\&\quad \le \frac{1}{2} \int _{-\pi }^\pi \frac{ \left| r\exp ( \imath \vartheta ) \right| }{\left| 1 - |r\exp ( \imath \vartheta )| \right| ^n }\frac{1}{\left| 1- |r^\alpha \exp ( \alpha \imath \vartheta )| \right| } {{\textrm{d}}} \vartheta \\&\quad = \frac{1}{2} \int _{-\pi }^\pi \frac{ r }{\left| 1 - r \right| } \frac{1}{\left| 1- r^\alpha \right| ^n} {{\textrm{d}}} \vartheta \rightarrow 0 \text{ as } r \rightarrow 0 \end{aligned}$$

and

$$\begin{aligned}&\left| \int _{\overset{\frown }{{\textrm{LI}}}} q(s) {{\textrm{d}}}s\right| \\&\quad =\frac{1}{2} \left| \int _{-\pi }^\pi \frac{R\exp ( \imath \vartheta )}{1- R\exp ( \imath \vartheta ) } \frac{1}{\left( 1- R^\alpha \exp ( \alpha \imath \vartheta )\right) ^n} {{\textrm{d}}} \vartheta \right| \\&\quad \le \frac{1}{2} \int _{-\pi }^\pi \frac{ R }{\left| R - 1 \right| } \frac{1}{\left| R^\alpha -1 \right| ^n} {{\textrm{d}}} \vartheta \rightarrow 0 \text{ as } R \rightarrow \infty . \end{aligned}$$

By the Euler formula, we have

$$\begin{aligned}&\left( \int _{\overline{{\textrm{IJ}}} } + \int _{\overline{{\textrm{KL}}} } \right) q(s) \text{ d } s \\&\quad = \exp ( \pi \imath ) \int _\infty ^0 q( r \exp ( \pi \imath ) ) \text{ d } r +\exp ( - \pi \imath ) \\&\qquad \int _0^\infty q( r \exp (-\pi \imath ) ) \text{ d } r \\&\quad = \int _0^ \infty q( r \exp ( \pi \imath ) ) \text{ d } r - \int _0^\infty q( r \exp (-\pi \imath ) ) \text{ d } r \\&\quad = \int _0^ \infty \frac{1}{2\imath } \frac{ 1 }{1-r \exp ( \pi \imath ) } \frac{1}{\left( 1-r^\alpha \exp (\alpha \pi \imath )\right) ^n} \text{ d } r \\&\qquad - \int _0^ \infty \frac{1}{2\imath } \frac{ 1 }{1-r \exp ( -\pi \imath ) } \frac{1}{\left( 1-r^\alpha \exp (-\alpha \pi \imath )\right) ^n} \text{ d } r \\&\quad = \int _0^ \infty \frac{\left( 1-r^\alpha \exp (\alpha \pi \imath )\right) ^{-n}-\left( 1-r^\alpha \exp (-\alpha \pi \imath )\right) ^{-n} }{2 (1 + r)\imath } \text{ d } r. \end{aligned}$$

Then the result follows by Lemma 5.2(i) and the Cauchy residue Theorem. (ii) For $\alpha = 0$, multiplying $\alpha ^n$ on both sides in (15) and applying L’Hôpital rule, the result follows. $\square$

Proof of Theorem 2.2

The following auxiliary Lemmas 5.4–5.5 are needed to prove Theorem 2.2.

Lemma 5.4

Let $n, \ell , m_\ell \in {\mathbb {N}}$, $h(x) = (1 - b \imath x)^\alpha$ for $0< \alpha < 1$, $b > 0$ and $x \in {\mathbb {C}}$. Then

$$\begin{aligned}&\lim \limits _{x \rightarrow 0} \frac{{{\textrm{d}}}^{ m_\ell } \left( 1 - h\left( \ell x \right) \right) ^n}{{{\textrm{d}}} x^{ m_\ell } } \\&\quad =1\{m_\ell \ge n \} (-1)^n \left( - b \ell \imath \right) ^{m_\ell } \Gamma (m_\ell +1) \\&\qquad \sum \limits _{ i=0 }^{ n-1 } { i-n-1 \atopwithdelims ()i} { (n-i) \alpha \atopwithdelims ()m_\ell }. \end{aligned}$$

Proof

Define

$$\begin{aligned} {\mathscr {D}}^n_w&= \left\{ (m_1^*,\ldots , m_n^*) \in \{{\mathbb {N}}\cup \{0\} \}^n \bigg | \sum \limits _{i=1}^{n} m_i^* = m_\ell , m_w^* = 0 \right\} , \\&\quad w = 1, \ldots , n,\\ {\mathscr {E}}_0^n&= \left\{ (m_1^*,\ldots , m_n^*) \in \{{\mathbb {N}}\cup \{0\} \}^n \bigg | \sum \limits _{i=1}^{n} m_i^* = m_\ell \right\} \end{aligned}$$

and ${\mathscr {E}}_i^n = \bigcup _{1 \le w_1< w_2< \cdots < w_i \le n} \bigcap _{j = 1}^i {\mathscr {D}}^n_{w_j}$ for $i = 1,2, \ldots , n-1$.

By the general Leibniz rule, we have

$$\begin{aligned}&\lim \limits _{x \rightarrow 0} \frac{{{\textrm{d}}}^{ m_\ell } \left( 1 - h\left( \ell x\right) \right) ^n}{{{\textrm{d}}} x^{ m_\ell } } \\&\quad = \lim \limits _{x \rightarrow 0} \Gamma (m_\ell +1)\sum \limits _{ {\mathscr {E}}_0^n } \prod \limits _{j=1}^n \frac{{{{\textrm{d}}}^{ m^*_{j} } \left( 1 - h\left( \ell x\right) \right) }/{{{\textrm{d}}} x^{ m^*_{j} } } }{ \Gamma (m_{j}^*+1) }. \end{aligned}$$

Since

$$\begin{aligned} \frac{{{\textrm{d}}}^{ m^*_{j} } \left( 1 - h\left( \ell x\right) \right) }{{{\textrm{d}}} x^{ m^*_{j} } }= & {} 1\{ m^*_{j} \ge 1 \} (-1) \left( - b \ell \imath \right) ^{m^*_j}\\{} & {} \frac{\Gamma (\alpha +1)}{\Gamma (\alpha - m^*_{j} + 1 )} \end{aligned}$$

for each $j=1,2,\ldots , n$,

$$\begin{aligned}&\lim \limits _{x \rightarrow 0} \frac{{{\textrm{d}}}^{ m_\ell } \left( 1 - h\left( \ell x\right) \right) ^n}{{{\textrm{d}}} x^{ m_\ell } } \nonumber \\&\quad = 1\{ m_\ell \ge n \} \Gamma (m_\ell +1) \sum \limits _{ {\mathscr {C}}^n_{ m_\ell } }\prod \limits _{j=1}^n \frac{(-1) \left( - b \ell \imath \right) ^{m^*_j} \Gamma (\alpha +1)}{\Gamma (m_{j}^*+1) \Gamma (\alpha - m^*_{j} + 1 )} \nonumber \\&\quad = 1\{ m_\ell \ge n \} (-1)^n \left( - b \ell \imath \right) ^{m_\ell } \Gamma (m_\ell +1) \sum \limits _{ {\mathscr {C}}^n_{ m_\ell } } \prod \limits _{j=1}^n { \alpha \atopwithdelims ()m_j^* }. \end{aligned}$$

(16)

Then applying the exclusion-inclusion principle, we obtain

$$\begin{aligned} \sum \limits _{ {\mathscr {C}}^n_{ m_\ell } } \prod \limits _{j=1}^n { \alpha \atopwithdelims ()m_j^* }&= \sum \limits _{ i=0 }^{ n-1 } (-1)^i \sum \limits _{ {\mathscr {E}}^n_i } \prod \limits _{j=1}^n { \alpha \atopwithdelims ()m_j^* } \\&= \sum \limits _{ i=0 }^{ n-1 } (-1)^i {n \atopwithdelims ()i} \sum \limits _{ {\mathscr {E}}^{n-i}_0 } \prod \limits _{j=1}^{n-i} { \alpha \atopwithdelims ()m_j^* } \\&= \sum \limits _{ i=0 }^{ n-1 } {i-n-1 \atopwithdelims ()i} { (n-i) \alpha \atopwithdelims ()m_\ell }, \end{aligned}$$

where the last equality is due to the generalized Chu-Vandermonde identity (see Graham et al. 1994, page 248). By substituting the above identity into (16), we get the desired result. $\square$

Lemma 5.5

(Gould 1978)

• (i) If $f: {\mathbb {R}} \rightarrow {\mathbb {R}}$ is a function which nth derivative exists, then we get

  $$\begin{aligned} f^{(n)}(x) = \lim \limits _{h \rightarrow 0} \frac{1}{h^n} \sum \limits _{k=0}^n (-1)^{n-k} {n \atopwithdelims ()k} f\left( x + k h \right) . \end{aligned}$$

• (ii) For $n \in {\mathbb {N}}$, we have

  $$\begin{aligned} \sum \limits _{k=0}^n (-1)^k {n \atopwithdelims ()k} k^n = \Gamma (n+1). \end{aligned}$$

Now, we return to the proof of Theorem 2.2.

Proof

(i)-(a) Since the integrand, $\frac{1-\exp \left( -\omega \imath s\right) }{s(1-h(s))^n}$, has a pole of order n at $s=0$, we have

$$\begin{aligned}&\text{ Res }\left( \frac{1-\exp \left( -\omega \imath s\right) }{s(1-h(s))^n},s=0 \right) \nonumber \\&\quad = \frac{1}{\Gamma (n)} \lim \limits _{s \rightarrow 0}\left( \frac{{{\textrm{d}}}^{n-1}}{{{\textrm{d}}} s^{n-1} } \frac{ s^{n-1}\left( 1-\exp \left( -\omega \imath s\right) \right) }{(1-h(s))^n} \right) , \end{aligned}$$

(17)

where $h(s) = (1 - b \imath s)^\alpha$ is defined in Lemma 5.4. By Lemma 5.5 (i), the Eq. (17) can be expressed as

$$\begin{aligned}&\text{ Res }\left( \frac{1-\exp \left( -\omega \imath s\right) }{s(1-h(s))^n},s=0 \right) \nonumber \\&\quad = \frac{1}{\Gamma (n) } \lim _{s \rightarrow 0, x \rightarrow 0} \frac{1}{x^{n-1}} \sum _{k=0}^{n-1} (-1)^{n-1-k} {n-1\atopwithdelims ()k} \nonumber \\&\qquad \frac{ (s + kx)^{n-1} \left( 1 - \exp \left( - \omega \imath \left( s + kx\right) \right) \right) }{ \left( 1 - h\left( s + kx\right) \right) ^{n} } \nonumber \\&\quad = \frac{1}{\Gamma (n) } \lim \limits _{x \rightarrow 0} \bigg \{ \frac{ (-1)^{n-1} \omega \imath }{( \alpha b \imath x)^n}\nonumber \\&\qquad + \sum \limits _{k=1}^{n-1} (-1)^{n-1-k} {n-1\atopwithdelims ()k} \left( kx\right) ^{n-1} \frac{ 1 - \exp \left( - k \omega \imath x \right) }{ \left( 1 - h\left( k x\right) \right) ^{n} } \bigg \} \nonumber \\&\quad = \lim \limits _{x \rightarrow 0} \frac{1}{\Gamma (n) ( \alpha b \imath )^{ n } x^{n-1} \prod \limits _{k=1}^{n-1} \left( 1 - h\left( k x\right) \right) ^n } \nonumber \\&\qquad \bigg \{ (-1)^{n-1} \omega \imath \prod \limits _{\ell =1 }^{n-1} \left( 1 - h\left( \ell x\right) \right) ^n \nonumber \\&\qquad + \sum _{k=1 }^{n-1} (-1)^{n-1-k} {n-1\atopwithdelims ()k} \left( \alpha b \imath \right) ^{n} (k x)^{n-1} \nonumber \\&\qquad \left( 1 - \exp \left( - k \omega \imath x \right) \right) \prod \limits _{\ell =1, \ell \ne k }^{n-1} \left( 1 - h\left( \ell x\right) \right) ^n \bigg \}. \end{aligned}$$

(18)

To calculate the limit in (18) by L’Hôpital rule, taking the derivative $n^2-1$ times of the denominator of (18) and using Lemma 5.5(i), we get

$$\begin{aligned}&\lim \limits _{x \rightarrow 0} \frac{{{\textrm{d}}}^{ n^2-1 } }{{{\textrm{d}}} x^{ n^2-1 } } \bigg \{ \Gamma (n) ( \alpha b \imath )^{ n } x^{n-1} \prod \limits _{k=1}^{n-1} \left( 1 - h\left( k x\right) \right) ^n \bigg \} \nonumber \\&\quad = \Gamma (n) ( \alpha b \imath )^{ n } \lim \limits _{x \rightarrow 0, y \rightarrow 0} \frac{1}{y^{n^2-1}} \sum _{k_1=0}^{n^2-1} (-1)^{n^2-1-k_1} {n^2-1 \atopwithdelims ()k_1} \nonumber \\&\qquad \bigg \{ \left( x+ k_1y \right) ^{ n-1 } \prod \limits _{k=1}^{n-1} \left( 1 - h\left( k \left( x+ k_1y\right) \right) \right) ^n \bigg \} \nonumber \\&\quad = \Gamma (n) ( \alpha b \imath )^{ n } \lim \limits _{y \rightarrow 0} \sum _{k_1=0}^{n^2-1} (-1)^{n^2-1-k_1} {n^2-1 \atopwithdelims ()k_1} k_1^{n-1}\nonumber \\&\qquad \prod \limits _{k=1}^{n-1} \frac{ \left( 1 - h\left( k k_1y\right) \right) ^n}{y^n} \nonumber \\&\quad = \Gamma (n) ( \alpha b \imath )^{ n } \sum _{k_1=0}^{n^2-1} (-1)^{n^2-1-k_1} {n^2-1 \atopwithdelims ()k_1} k_1^{n-1}\nonumber \\&\qquad \prod \limits _{k=1}^{n-1} \left( \alpha b k k_1\imath \right) ^n \nonumber \\&\quad = \Gamma (n)^{n+1} \left( \alpha b \imath \right) ^{ n^2 } \Gamma (n^2), \end{aligned}$$

(19)

where the last identity holds according to Lemma 5.5(ii). Taking the derivative $n^2-1$ times of the numerator of (18) and using the general Leibniz rule, we have

$$\begin{aligned}&\lim \limits _{x \rightarrow 0} \frac{{{\textrm{d}}}^{ n^2 -1} }{{{\textrm{d}}} x^{ n^2-1 } } \bigg \{ (-1)^{n-1} \omega \imath \prod \limits _{\ell =1 }^{n-1} \left( 1 - h\left( \ell x\right) \right) ^n\nonumber \\&\quad + \sum _{k=1 }^{n-1} (-1)^{n-1-k} {n-1\atopwithdelims ()k} \left( \alpha b \imath \right) ^{n} (k x)^{n-1} \nonumber \\&\quad \times \left( 1 - \exp \left( - k \omega \imath x \right) \right) \prod \limits _{\ell =1, \ell \ne k }^{n-1} \left( 1 - h\left( \ell x\right) \right) ^n \bigg \} \nonumber \\&=(-1)^{n-1} \omega \imath \sum _{{\mathscr {F}}^n_{0}} \frac{\Gamma (n^2)}{\prod \limits _{j=0}^{n-1} \Gamma (m_{j}+1) } \nonumber \\&\qquad \lim \limits _{x \rightarrow 0} \prod \limits _{\ell =1}^{n-1} \frac{{{\textrm{d}}}^{ m_\ell }}{{{\textrm{d}}} x^{ m_\ell } } \left( 1 - h\left( \ell x\right) \right) ^n\nonumber \\&\qquad +\sum _{k=1}^{n-1} (-1)^{n-1-k} {n-1\atopwithdelims ()k} (\alpha b \imath )^n k^{n-1} \nonumber \\&\qquad \times \sum \limits _{ {\mathscr {F}}^n } \frac{\Gamma (n^2+1)}{\prod \limits _{j=0}^{n-1} \Gamma (m_{j}+1) } \lim \limits _{x \rightarrow 0} \bigg \{ \frac{{{\textrm{d}}}^{ m_0 } }{{{\textrm{d}}} x^{ m_0 } } x^{ n - 1 } \frac{{{\textrm{d}}}^{m_k} }{{{\textrm{d}}} x^{m_k}} \nonumber \\&\qquad \left( 1 - \exp \left( - k \omega \imath x \right) \right) \prod \limits _{\ell =1 }^{n-1} \frac{{{\textrm{d}}}^{m_\ell } }{{{\textrm{d}}} x^{m_\ell }} \left( 1 - h\left( \ell x\right) \right) ^n \bigg \}, \end{aligned}$$

where

$$\begin{aligned} {\mathscr {F}}_0^n&= \left\{ (m_0, \ldots , m_{n-1}) \in \{{\mathbb {N}}\cup \{0\} \}^{n} \bigg | \sum \limits _{i=0}^{n-1}m_i = n^2, m_0 = 1 \right\} ,\\ {\mathscr {F}}^n&= \left\{ (m_0,\ldots , m_{n-1}) \in \{{\mathbb {N}}\cup \{0\} \}^n \bigg | \sum \limits _{i=0}^{n-1}m_i = n^2 - 1 \right\} . \end{aligned}$$

Using Lemma 5.4 with

$$\begin{aligned}{} & {} \lim \limits _{x \rightarrow 0} \frac{{{\textrm{d}}}^{ m_0 } }{{{\textrm{d}}} x^{ m_0 }} x^{n} = 1 \{m_{ 0 } =n \} \Gamma (n+1)\\{} & {} \quad \text{ and }\quad \lim \limits _{x \rightarrow 0} \frac{{{\textrm{d}}}^{ m_k } }{{{\textrm{d}}} x^{ m_k } } \left( 1-\exp \left( - k \omega \imath x\right) \right) \\{} & {} \qquad = - 1 \{m_k \ge 1 \} \left( - k \omega \imath \right) ^{m_k}, \end{aligned}$$

we get

$$\begin{aligned}&(-1)^{n-1} \omega \imath \sum \limits _{ {\mathscr {A}}^n_0 } \frac{\Gamma (n^2)}{\prod \nolimits _{j=1}^{n-1} \Gamma (m_{j}+1) }\nonumber \\&\quad \prod \limits _{\ell =1 }^{n-1} \bigg \{ (- 1)^n \Gamma (m_\ell + 1) \left( - b \ell \imath \right) ^{m_\ell } \nonumber \\&\quad \sum \limits _{ i=0 }^{ n-1 } { i-n-1 \atopwithdelims ()i} { (n-i) \alpha \atopwithdelims ()m_\ell } \bigg \} \nonumber \\&\quad + \sum _{k=1}^{n-1} (-1)^{n-1-k} {n-1\atopwithdelims ()k} \left( \alpha b \imath \right) ^{n} k^{n-1}\nonumber \\&\quad \sum \limits _{ {\mathscr {A}}^n_{k} } \frac{\Gamma (n^2)}{\prod \nolimits _{j=1}^{n-1} \Gamma (m_{j}+1) } \Gamma (n) (-1) \left( - k \omega \imath \right) ^{m_k} \nonumber \\&\quad \times \prod \limits _{\ell =1, \ell \ne k }^{n-1} \bigg \{(-1)^n \left( - b \ell \imath \right) ^{m_\ell } \Gamma (m_\ell +1)\nonumber \\&\quad \sum \limits _{ i=0 }^{ n-1 } { i-n-1 \atopwithdelims ()i} { (n-i) \alpha \atopwithdelims ()m_\ell } \bigg \} \nonumber \\&\quad = \Gamma (n^2) b^{n^2} \imath ^{n^2} \Bigg \{ \sum \limits _{ {\mathscr {A}}^n_0 } (\omega /b) \prod \limits _{\ell =1 }^{n-1} \bigg \{ \ell ^{m_\ell }\sum \limits _{ i=0 }^{ n-1 } { i-n-1 \atopwithdelims ()i}\nonumber \\&\qquad { (n-i) \alpha \atopwithdelims ()m_\ell } \bigg \} \nonumber \\&\qquad + \sum _{k=1}^{n-1} (-1)^{k} {n-1\atopwithdelims ()k} \alpha ^n \sum \limits _{ {\mathscr {A}}^n_{k} } \frac{(\omega /b)^{m_k}}{\Gamma (m_{k}+1) } k^{n-1+m_k}\nonumber \\&\qquad \prod \limits _{\ell =1}^{n-1} \bigg \{ \ell ^{m_\ell } \sum \limits _{ i=0 }^{ n-1 } { i-n-1 \atopwithdelims ()i} { (n-i) \alpha \atopwithdelims ()m_\ell } \bigg \} \Bigg \}. \end{aligned}$$

(20)

By combining (19) and (20), we obtain the desired result of the residue.

(i)-(b) The derivation of the residue can be followed by similar procedure. (ii) The results can be established by L’Hôpital rule directly. This completes the proof. $\square$