Appendix 1: Proof of Theorem 1
Using the explicit forms of \(A_{sd}\), \(B_{sdt}\), and D in (12), (13), and (28), respectively, one can rewrite (31) as
$$\begin{aligned} {{\mathcal{Q}}_1} = {\mathrm{{E}}_{{{\left| {{h_{sr}}} \right| }^2}}}\left\{ {\frac{{{{\bar{B}}_{sdt}}{P_s}{e^{ - \frac{{{2^{{C_0}}} - 1}}{{{\lambda _{sd}}{P_s}}}}}{2^{ - {C_0}}}}}{{{{\bar{B}}_{swt}}{P_s} - {2^{ - {C_0}}}\left( {{{\bar{B}}_{sdt}}{P_s} + {2^{{C_0}}} - 1} \right) }}} \right\} \end{aligned}$$
(66)
It is recalled that \({P_s} = \min \left( {\frac{{{I_m}}}{{{{\left| {{h_{sr}}} \right| }^2}}},{P_m}} \right) \) and hence, by the law of total probability, one can simplify (66) as
$$\begin{aligned} \begin{aligned} {{\mathcal{Q}}_1} &=\int \limits _0^\infty {\frac{{{{\bar{B}}_{sdt}}\min \left( {\frac{{{I_m}}}{x},{P_m}} \right) {e^{ - \frac{{{2^{{C_0}}} - 1}}{{{\lambda _{sd}}\min \left( {\frac{{{I_m}}}{x},{P_m}} \right) }}}}{2^{ - {C_0}}}}}{{{{\bar{B}}_{swt}}\min \left( {\frac{{{I_m}}}{x},{P_m}} \right) - {2^{ - {C_0}}}\left( {{{\bar{B}}_{sdt}}\min \left( {\frac{{{I_m}}}{x},{P_m}} \right) + {2^{{C_0}}} - 1} \right) }}{f_{{{\left| {{h_{sr}}} \right| }^2}}}\left( x \right) dx} \\ &=\int \limits _0^{\frac{{{I_m}}}{{{P_m}}}} {\frac{{{{\bar{B}}_{sdt}}{P_m}{e^{ - \frac{{{2^{{C_0}}} - 1}}{{{\lambda _{sd}}{P_m}}}}}{2^{ - {C_0}}}}}{{{{\bar{B}}_{swt}}{P_m} - {2^{ - {C_0}}}\left( {{{\bar{B}}_{sdt}}{P_m} + {2^{{C_0}}} - 1} \right) }}\frac{1}{{{\lambda _{sr}}}}{e^{ - \frac{x}{{{\lambda _{sr}}}}}}dx} \\&\quad + \int \limits _{\frac{{{I_m}}}{{{P_m}}}}^\infty {\frac{{{{\bar{B}}_{sdt}}\frac{{{I_m}}}{x}{e^{ - \frac{{{2^{{C_0}}} - 1}}{{{\lambda _{sd}}\frac{{{I_m}}}{x}}}}}{2^{ - {C_0}}}}}{{{{\bar{B}}_{swt}}\frac{{{I_m}}}{x} - {2^{ - {C_0}}}\left( {{{\bar{B}}_{sdt}}\frac{{{I_m}}}{x} + {2^{{C_0}}} - 1} \right) }}\frac{1}{{{\lambda _{sr}}}}{e^{ - \frac{x}{{{\lambda _{sr}}}}}}dx}\\ &=\frac{{{{\bar{B}}_{sdt}}{P_m}{e^{ - \frac{{{2^{{C_0}}} - 1}}{{{\lambda _{sd}}{P_m}}}}}{2^{ - {C_0}}}}}{{{{\bar{B}}_{swt}}{P_m} - {2^{ - {C_0}}}\left( {{{\bar{B}}_{sdt}}{P_m} + {2^{{C_0}}} - 1} \right) }}\left( {1 - {e^{ - \frac{{{I_m}}}{{{P_m}{\lambda _{sr}}}}}}} \right) \\&\quad + \frac{{{{\bar{B}}_{sdt}}{I_m}{2^{ - {C_0}}}}}{{{\lambda _{sr}}\left( {{2^{ - {C_0}}} - 1} \right) }}\int \limits _{\frac{{{I_m}}}{{{P_m}}}}^\infty {\frac{{{e^{ - \left( {\frac{{{2^{{C_0}}} - 1}}{{{\lambda _{sd}}{I_m}}} + \frac{1}{{{\lambda _{sr}}}}} \right) x}}}}{{x + \frac{{{{\bar{B}}_{swt}} - {2^{ - {C_0}}}{{\bar{B}}_{sdt}}}}{{{2^{ - {C_0}}} - 1}}{I_m}}}dx} \end{aligned} \end{aligned}$$
(67)
The last integral follows the general form of
$$\begin{aligned} \mathcal{L}(a,b,c)= \int \limits _a^\infty {\frac{{{e^{ - by}}}}{{y + c}}dy} \end{aligned}$$
(68)
and hence, can be solved with the aid of [32, Eq. (358.2)] as shown in (36), completing the proof.
Appendix 2: Proof of Lemma 3
With appropriate variable changes in (34), one can rewrite (37) as
$$\begin{aligned} \begin{aligned} \Phi \left( {a,b,c,g,l,m} \right)&= - \int \limits _a^\infty {\frac{{{e^{ - bx}}}}{{x + c}}{e^{gx\left( {\frac{l}{x} + m} \right) }}Ei\left( { - gx\left[ {\frac{l}{x} + m} \right] } \right) dx}\\&= - {e^{gl}}\underbrace{\int \limits _a^\infty {\frac{{{e^{ - \left( {b - mg} \right) x}}}}{{x + c}}Ei\left( { - mgx - gl} \right) dx} }_{\Upsilon \left( {a,b - mg,c,gm,gl} \right) } \end{aligned} \end{aligned}$$
(69)
which coincides with (38), where
$$\begin{aligned} \Upsilon \left( {a,b,c,g,l} \right) = \int \limits _a^\infty {\frac{{{e^{ - bx}}}}{{x + c}}Ei\left( { - gx - l} \right) dx} \end{aligned}$$
(70)
Therefore, in order to complete the proof of Lemma 3, we must prove that (70) is represented in closed-form as (39). Towards this end, performing the variable change and then applying the series representation of Ei(x) in [32, Eq. (8.214.1)] results in
$$\begin{aligned} \begin{aligned} \Upsilon \left( {a,b,c,g,l} \right)&= - \frac{1}{g}\int \limits _{ - ag - l}^{ - \infty } {\frac{{{e^{b\frac{{y + l}}{g}}}}}{{ - \frac{{y + l}}{g} + c}}Ei\left( y \right) dy} \\&= - {e^{\frac{{bl}}{g}}}\int \limits _{ - \infty }^{ - ag - l} {\frac{{{e^{\frac{b}{g}y}}}}{{y + l - cg}}\left( {\texttt {C} + \ln \left( { - y} \right) + \sum \limits _{k = 1}^\infty {\frac{{{y^k}}}{{k \cdot k!}}} } \right) dy} \end{aligned} \end{aligned}$$
(71)
By denoting
$$\begin{aligned} {{\mathcal{K}}_1}&= \int \limits _{ - \infty }^{ - ag - l} {\frac{{{e^{\frac{b}{g}y}}}}{{y + l - cg}}dy} \end{aligned}$$
(72)
$$\begin{aligned} {{\mathcal{K}}_2}&= \int \limits _{ - \infty }^{ - ag - l} {\frac{{{e^{\frac{b}{g}y}}}}{{y + l - cg}}\ln \left( { - y} \right) dy} \end{aligned}$$
(73)
$$\begin{aligned} {{\mathcal{K}}_3}&= \int \limits _{ - \infty }^{ - ga - l} {\frac{{{y^k}}}{{y - cg + l}}{e^{\frac{b}{g}y}}dy}, \end{aligned}$$
(74)
it is apparent that (71) perfectly matches (39). As such, the proof is completed after solving integrands in (72), (73), and (74) as (40), (41), and (47), respectively.
Starting with \({{\mathcal{K}}_1}\). Performing the variable change, one can rewrite \({{\mathcal{K}}_1}\) as
$$\begin{aligned} {{\mathcal{K}}_1} = - \int \limits _\infty ^{ag + l} {\frac{{{e^{ - \frac{b}{g}x}}}}{{ - x - cg + l}}dx} = - \int \limits _{ag + l}^\infty {\frac{{{e^{ - \frac{b}{g}x}}}}{{x + cg - l}}dx} \end{aligned}$$
(75)
Using (68), one can represent (75) as (40).
Performing the variable change and then applying the series representation of \({\ln \left( x \right) }\) in [32, Eq. (1.512.2)], one can simplify \({{\mathcal{K}}_2}\) as
$$\begin{aligned} \begin{aligned} {{\mathcal{K}}_2}&= - \int \limits _\infty ^{ag + l} {\frac{{{e^{ - \frac{b}{g}x}}}}{{ - x - cg + l}}\ln \left( x \right) dx} \\&= - \int \limits _{ag + l}^\infty {\frac{{{e^{ - \frac{b}{g}x}}}}{{x + cg - l}}\left[ {2\sum \limits _{k = 1}^\infty {\frac{1}{{2k - 1}}{{\left( {\frac{{x - 1}}{{x + 1}}} \right) }^{2k - 1}}} } \right] dx} \\&= - \sum \limits _{k = 1}^\infty {\frac{2}{{2k - 1}}} \int \limits _{ag + l}^\infty {\frac{{{e^{ - \frac{b}{g}x}}}}{{x + cg - l}}{{\left( {\frac{{x - 1}}{{x + 1}}} \right) }^{2k - 1}}dx} \end{aligned} \end{aligned}$$
(76)
Applying the binomial expansion in [32, Eq. (1.111)] to \((x-1)^{2k-1}\) of the above, one obtains
$$\begin{aligned} \begin{aligned} {{\mathcal{K}}_2}&= - \sum \nolimits _{k = 1}^\infty {\frac{2}{{2k - 1}}} \int \limits _{ag + l}^\infty {\frac{{{e^{ - \frac{b}{g}x}}}}{{x + cg - l}}\frac{{\sum \limits _{i = 0}^{2k - 1} {\left( {\begin{array}{c} {2k - 1} \\ i \\ \end{array}} \right) {x^i}{{\left( { - 1} \right) }^{2k - 1 - i}}} }}{{{{\left( {x + 1} \right) }^{2k - 1}}}}dx} \\&= - \sum \limits _{k = 1}^\infty {\frac{2}{{2k - 1}}\sum \limits _{i = 0}^{2k - 1} {\left( {\begin{array}{c} {2k - 1} \\ i \\ \end{array}} \right) {{\left( { - 1} \right) }^{2k - 1 - i}}} } \int \limits _{ag + l}^\infty {\frac{{{x^i}{e^{ - \frac{b}{g}x}}}}{{\left( {x + cg - l} \right) {{\left( {x + 1} \right) }^{2k - 1}}}}dx}, \end{aligned} \end{aligned}$$
(77)
where the notation \(\left( {\begin{array}{c} {2k - 1} \\ i \\ \end{array}}\right) = \frac{{\left( {2k - 1} \right) !}}{{i!\left( {2k - 1 - i} \right) !}}\) is the binomial coefficient.
Again performing the variable change and then applying the binomial expansion to simplify the above as
$$\begin{aligned} \begin{aligned} {{\mathcal{K}}_2} &=- \sum \limits _{k = 1}^\infty {\frac{2}{{2k - 1}}\sum \limits _{i = 0}^{2k - 1} {\left( {\begin{array}{c} {2k - 1} \\ i \\ \end{array}} \right) {{\left( { - 1} \right) }^{2k - 1 - i}}} } \int \limits _{ag + l + 1}^\infty {\frac{{{{\left( {y - 1} \right) }^i}{e^{ - \frac{b}{g}\left( {y - 1} \right) }}}}{{\left( {y - 1 + cg - l} \right) {y^{2k - 1}}}}dy} \\ &=- \sum \limits _{k = 1}^\infty {\frac{{{e^{\frac{b}{g}}}2}}{{2k - 1}}\sum \limits _{i = 0}^{2k - 1} {\left( {\begin{array}{c} {2k - 1} \\ i \\ \end{array}} \right) {{\left( { - 1} \right) }^{2k - 1 - i}}} } \int \limits _{ag + l + 1}^\infty \frac{{{e^{ - \frac{b}{g}y}}}}{{\left( {y + cg - l - 1} \right) {y^{2k - 1}}}}\\&\quad \left[ {\sum \limits _{j = 0}^i {\left( {\begin{array}{c} i \\ j \\ \end{array}} \right) {y^j}{{\left( { - 1} \right) }^{i - j}}} } \right] dy\\ &=2{e^{\frac{b}{g}}}\sum \limits _{k = 1}^\infty {\sum \limits _{i = 0}^{2k - 1} {\sum \limits _{j = 0}^i {\left( {\begin{array}{c} {2k - 1} \\ i \\ \end{array}} \right) \left( {\begin{array}{c} i \\ j \\ \end{array}} \right) } \frac{{{{\left( { - 1} \right) }^j}}}{{2k - 1}}} } \int \limits _{ag + l + 1}^\infty {\frac{{{y^j}{e^{ - \frac{b}{g}y}}}}{{\left( {y + cg - l - 1} \right) {y^{2k - 1}}}}dy} \end{aligned} \end{aligned}$$
(78)
By denoting
$$\begin{aligned} {\mathcal{M}} = \int \limits _{ag + l + 1}^\infty {\frac{{{y^j}{e^{ - \frac{b}{g}y}}}}{{\left( {y + cg - l - 1} \right) {y^{2k - 1}}}}dy}, \end{aligned}$$
(79)
it is apparent that (78) coincides (41). Therefore, we must prove that the integrand in (79) can be represented as (42) to complete the proof of (41). Toward this end, we consider two cases: \({j = 2k - 1}\) (Case 1) and \({j \ne 2k - 1}\) (Case 2). For Case 1, \({\mathcal{M}}\) is rewritten as \({\int \nolimits _{ag + l + 1}^\infty {\frac{{{e^{ - \frac{b}{g}y}}}}{{y + cg - l - 1}}dy} }\), which becomes \({{\mathcal{L}}\left( {ag + l + 1,\frac{b}{g},cg - l - 1} \right) }\) as shown in (42). For Case 2, we denote \({\mathcal{M}} = {\int \nolimits _{ag + l + 1}^\infty {\frac{{{e^{ - \frac{b}{g}y}}}}{{\left( {y + cg - l - 1} \right) {y^{2k - j - 1}}}}dy} }\) as \({\mathcal{W}}\). Combining two cases shows that (79) coincides (42). Therefore, the remaining work is to prove that \({\mathcal{W}}\) is given by (43).
Applying the partial fraction decomposition, \({\mathcal{W}}\) can be simplified as
$$\begin{aligned} {\mathcal{W}} = G\int \limits _{ag + l + 1}^\infty {\frac{{{e^{ - \frac{b}{g}x}}}}{{y + cg - l - 1}}dy} + \sum \limits _{v = 1}^{2k - j - 1} {{K_{2k - j - 1 - v + 1}}} \int \limits _{ag + l + 1}^\infty {\frac{{{e^{ - \frac{b}{g}y}}}}{{{y^v}}}dy}, \end{aligned}$$
(80)
where G and \(K_m\) are defined in (44) and (45), respectively.
By denoting
$$\begin{aligned} {\mathcal{B}}\left( {a,b,v} \right) = \int \limits _a^\infty {\frac{{{e^{ - bx}}}}{{{x^v}}}dx}, \end{aligned}$$
(81)
it is obvious that (80) becomes (43). The function \({\mathcal{B}}\left( {a,b,v} \right) \) can be represented as (46) by firstly performing the variable change and then using [32, Eq. (3.381.6)].
Finally, we process \({{\mathcal{K}}_3}\). By performing the variable change and applying the binomial expansion, (74) is simplified as
$$\begin{aligned} \begin{aligned} {{\mathcal{K}}_3} &=- \int \limits _\infty ^{ag + l} {\frac{{{{\left( { - x} \right) }^k}}}{{ - x - cg + l}}{e^{ - \frac{b}{g}x}}dx} \\ &={\left( { - 1} \right) ^{k + 1}}\int \limits _{ag + l + cg - l}^\infty {\frac{{{{\left( {y + l - cg} \right) }^k}}}{y}{e^{ - \frac{b}{g}\left( {y + l - cg} \right) }}dy} \\ &={\left( { - 1} \right) ^{k + 1}}{e^{ - \frac{b}{g}\left( {l - cg} \right) }}\int \limits _{ag + cg}^\infty {\left[ {\sum \limits _{i = 0}^k {\left( {\begin{array}{c} k \\ i \\ \end{array}} \right) {y^i}{{\left( {l - cg} \right) }^{k - i}}} } \right] \frac{1}{y}{e^{ - \frac{b}{g}y}}dy} \\ &={\left( { - 1} \right) ^{k + 1}}{e^{ - \frac{b}{g}\left( {l - cg} \right) }}\sum \limits _{i = 0}^k {\left( {\begin{array}{c} k \\ i \\ \end{array}} \right) {{\left( {l - cg} \right) }^{k - i}}} \int \limits _{\left( {a + c} \right) g}^\infty {{y^{i - 1}}{e^{ - \frac{b}{g}y}}dy}\\ &={\left( { - 1} \right) ^{k + 1}}{e^{ - \frac{b}{g}\left( {l - cg} \right) }}\\& \quad \times \left[ {{{\left( {l - cg} \right) }^k}\int \limits _{\left( {a + c} \right) g}^\infty {\frac{{{e^{ - \frac{b}{g}y}}}}{y}dy} + \sum \limits _{i = 1}^k {\left( {\begin{array}{c} k \\ i \\ \end{array}} \right) {{\left( {l - cg} \right) }^{k - i}}} \int \limits _{\left( {a + c} \right) g}^\infty {{y^{i - 1}}{e^{ - \frac{b}{g}y}}dy} } \right] \end{aligned} \end{aligned}$$
(82)
By denoting
$$\begin{aligned} {\mathcal{H}}\left( {a,b,i} \right) = \int \limits _a^\infty {{y^i}{e^{ - by}}dy}, \end{aligned}$$
(83)
whose exact closed form is given by (48) with the aid of [32, Eq. (3.351.2)], it is obvious that (82) exactly matches (47), completing the proof.
Appendix 3: Proof of Lemma 4
With appropriate variable changes in (34), one can rewrite (49) as
$$\begin{aligned} \begin{aligned} \Omega \left( {a,b,c,g,l,m} \right)&= - \int \limits _a^\infty {\frac{{{e^{ - bx}}}}{{{{\left( {x + c} \right) }^2}}}{e^{gx\left( {\frac{l}{x} + m} \right) }}Ei\left( { - gx\left[ {\frac{l}{x} + m} \right] } \right) dx} \\&= - {e^{gl}}\underbrace{\int \limits _a^\infty {\frac{{{e^{ - \left( {b - mg} \right) x}}}}{{{{\left( {x + c} \right) }^2}}}Ei\left( { - mgx - gl} \right) dx} }_{\Lambda \left( {a,b - mg,c,gm,gl} \right) } \end{aligned} \end{aligned}$$
(84)
which coincides with (50), where
$$\begin{aligned} \Lambda \left( {a,b,c,g,l} \right) = \int \limits _a^\infty {\frac{{{e^{ - bx}}}}{{{{\left( {x + c} \right) }^2}}}Ei\left( { - gx - l} \right) dx} \end{aligned}$$
(85)
Next, we must prove that (85) matches with (51). Toward this end, we firstly perform the variable change and then apply the series representation of Ei(x) in [32, Eq. (8.214.1)] as
$$\begin{aligned} \begin{aligned} \Lambda \left( {a,b,c,g,l} \right)&= - \frac{1}{g}\int \limits _{ - ga - l}^{ - \infty } {\frac{{{e^{b\frac{{y + l}}{g}}}}}{{{{\left( { - \frac{{y + l}}{g} + c} \right) }^2}}}Ei\left( y \right) dy} \\&= - g\int \limits _{ - ga - l}^{ - \infty } {\frac{{{e^{b\frac{{y + l}}{g}}}}}{{{{\left( {y + l - cg} \right) }^2}}}\left( {\texttt {C} + \ln \left( { - y} \right) + \sum \limits _{k = 1}^\infty {\frac{{{y^k}}}{{k \cdot k!}}} } \right) dy} \end{aligned} \end{aligned}$$
(86)
By denoting
$$\begin{aligned} {{\mathcal{O}}_1}&= \int \limits _{ - ga - l}^{ - \infty } {\frac{{{e^{\frac{b}{g}y}}}}{{{{\left( {y + l - cg} \right) }^2}}}dy} \end{aligned}$$
(87)
$$\begin{aligned} {{\mathcal{O}}_2}&= \int \limits _{ - ga - l}^{ - \infty } {\frac{{{e^{\frac{b}{g}y}}}}{{{{\left( {y + l - cg} \right) }^2}}}\ln \left( { - y} \right) dy} \end{aligned}$$
(88)
$$\begin{aligned} {{\mathcal{O}}_3}&= {\int \limits _{ - ga - l}^{ - \infty } {\frac{{{y^k}{e^{\frac{b}{g}y}}}}{{{{\left( {y + l - cg} \right) }^2}}}dy} }, \end{aligned}$$
(89)
it is apparent that (86) matches (51). Therefore, in order to complete the proof, we must represent (87), (88), and (89) as (52), (53), and (59), correspondingly.
Starting with \(\mathcal{O}_1\). By performing the variable change, one obtains
$$\begin{aligned} \begin{aligned} {{\mathcal{O}}_1}&= - \int \limits _{ga + l}^\infty {\frac{{{e^{ - \frac{b}{g}x}}}}{{{{\left( { - x + l - cg} \right) }^2}}}dx} \\&= - {e^{ - \frac{{b\left( {l - cg} \right) }}{g}}}\int \limits _{ga + cg}^\infty {\frac{{{e^{ - \frac{b}{g}y}}}}{{{y^2}}}dy}, \end{aligned} \end{aligned}$$
(90)
which coincides with (52) after using (81).
We simplify \({{\mathcal{O}}_2}\) through the variable changes, the series representation of \({\ln \left( x \right) }\), and the binomial expansion as
$$\begin{aligned} \begin{aligned} {{\mathcal{O}}_2} &=- \int \limits _{ga + l}^\infty {\frac{{{e^{ - \frac{b}{g}x}}}}{{{{\left( { - x + l - cg} \right) }^2}}}\ln \left( x \right) dx} \\ &=- \int \limits _{ga + l}^\infty {\frac{{{e^{ - \frac{b}{g}x}}}}{{{{\left( {x + cg - l} \right) }^2}}}\left[ {2\sum \limits _{k = 1}^\infty {\frac{1}{{2k - 1}}{{\left( {\frac{{x - 1}}{{x + 1}}} \right) }^{2k - 1}}} } \right] dx} \\ &=- \sum \limits _{k = 1}^\infty {\frac{2}{{2k - 1}}} \int \limits _{ga + l}^\infty \frac{{{e^{ - \frac{b}{g}x}}}}{{{{\left( {x + cg - l} \right) }^2}{{\left( {x + 1} \right) }^{2k - 1}}}}\\& \quad \times \left[ {\sum \limits _{i = 0}^{2k - 1} {\left( {\begin{array}{c} {2k - 1} \\ i \\ \end{array}} \right) {x^i}{{\left( { - 1} \right) }^{2k - 1 - i}}} } \right] dx \\ &=- \sum \limits _{k = 1}^\infty {\frac{2}{{2k - 1}}} \sum \limits _{i = 0}^{2k - 1} {\left( {\begin{array}{c} {2k - 1} \\ i \\ \end{array}} \right) } {\left( { - 1} \right) ^{2k - 1 - i}}\int \limits _{ga + l + 1}^\infty {\frac{{{{\left( {y - 1} \right) }^i}{e^{ - \frac{b}{g}\left( {y - 1} \right) }}}}{{{{\left( {y - 1 + cg - l} \right) }^2}{y^{2k - 1}}}}dy} \\ &=- \sum \limits _{k = 1}^\infty {\frac{2}{{2k - 1}}} \sum \limits _{i = 0}^{2k - 1} {\left( {\begin{array}{c} {2k - 1} \\ i \\ \end{array}} \right) } {\left( { - 1} \right) ^{2k - 1 - i}}\int \limits _{ga + l + 1}^\infty \frac{{{e^{ - \frac{b}{g}\left( {y - 1} \right) }}}}{{{{\left( {y - 1 + cg - l} \right) }^2}{y^{2k - 1}}}}\\&\quad \times\left[ {\sum \limits _{j = 0}^i {\left( {\begin{array}{c} i \\ j \\ \end{array}} \right) {y^j}{{\left( { - 1} \right) }^{i - j}}} } \right] dy. \end{aligned} \end{aligned}$$
(91)
By denoting
$$\begin{aligned} {\mathcal{U}}={\int \limits _{ga + l + 1}^\infty {\frac{{{y^j}{e^{ - \frac{b}{g}y}}}}{{{{\left( {y - 1 + cg - l} \right) }^2}{y^{2k - 1}}}}dy} }, \end{aligned}$$
(92)
and after simplification, (91) matches (53).
Therefore, the remaining step for proving \({{\mathcal{O}}_2}\) is to show that (92) is (54). To compute \({\mathcal{U}}\), we consider two cases: \({j = 2k - 1}\) (Case 1) and \({j \ne 2k - 1}\) (Case 2). For Case 1, \({\mathcal{U}}\) becomes \({\int \nolimits _{ga + l + 1}^\infty {\frac{{{e^{ - \frac{b}{g}y}}}}{{{{\left( {y - 1 + cg - l} \right) }^2}}}dy} }\) and after performing the variable change, it reduces to \({\mathcal{U}} = {e^{\frac{b}{g}\left( {cg - l - 1} \right) }}\int \nolimits _{\left( {a + c} \right) g}^\infty {\frac{{{e^{ - \frac{b}{g}x}}}}{{{x^2}}}dx}\), which immediately follows (54) after using (81). For Case 2, we denote
$$\begin{aligned} {\mathcal{U}}={\int \limits _{ga + l + 1}^\infty {\frac{{{e^{ - \frac{b}{g}y}}}}{{{{\left( {y - 1 + cg - l} \right) }^2}{y^{2k - j - 1}}}}dy} }={\mathcal{G}} \end{aligned}$$
(93)
Combining results in two cases shows that (92) becomes (54). Therefore, the next step is to show that \({\mathcal{G}}\) can be represented as (55). To do that, we firstly apply the partial fraction decomposition as
$$\begin{aligned} {\mathcal{G}}&= {} {J_1}\int \limits _{ga + l + 1}^\infty {\frac{{{e^{ - \frac{b}{g}y}}}}{{y + cg - l - 1}}dy} + {J_2}\int \limits _{ga + l + 1}^\infty {\frac{{{e^{ - \frac{b}{g}y}}}}{{{{\left( {y + cg - l - 1} \right) }^2}}}dy} \nonumber \\&\quad + \sum \limits _{v = 1}^{2k - j - 1} {{L_{2k - j - 1 - v + 1}}\int \limits _{ag + l + 1}^\infty {\frac{{{e^{ - \frac{b}{g}y}}}}{{{y^v}}}dy} }, \end{aligned}$$
(94)
where \(J_1\), \(J_2\), and \(L_m\) are defined in (56), (57), and (58), correspondingly.
Then performing the variable changes for the second term in (94) and using (68) and (81), it is straightforward to show that (94) coincides (55).
We simplify \({{\mathcal{O}}_3}\) through the variable changes and the binomial expansion as
$$\begin{aligned} \begin{aligned} {{\mathcal{O}}_3} &= - {\left( { - 1} \right) ^k}\int \limits _{ag + l}^\infty {\frac{{{x^k}{e^{ - \frac{b}{g}x}}}}{{{{\left( {x + cg - l} \right) }^2}}}dx} \\ &={\left( { - 1} \right) ^{k + 1}}\int \limits _{ag + l + cg - l}^\infty {\frac{{{{\left( {y + l - cg} \right) }^k}{e^{ - \frac{b}{g}\left( {y + l - cg} \right) }}}}{{{y^2}}}dy} \\ &={e^{ - \frac{b}{g}\left( {l - cg} \right) }}{\left( { - 1} \right) ^{k + 1}}\int \limits _{\left( {a + c} \right) g}^\infty {\left[ {\sum \limits _{i = 0}^k {\left( {\begin{array}{c} k \\ i \\ \end{array}} \right) } {y^i}{{\left( {l - cg} \right) }^{k - i}}} \right] \frac{{{e^{ - \frac{b}{g}y}}}}{{{y^2}}}dy} \\ &={e^{ - \frac{b}{g}\left( {l - cg} \right) }}{\left( { - 1} \right) ^{k + 1}}\left[ {{{\left( {l - cg} \right) }^k}\int \limits _{\left( {a + c} \right) g}^\infty {\frac{{{e^{ - \frac{b}{g}y}}}}{{{y^2}}}dy} + {{\left( {l - cg} \right) }^{k - 1}}k\int \limits _{\left( {a + c} \right) g}^\infty {\frac{{{e^{ - \frac{b}{g}y}}}}{y}dy} + } \right. \\&\quad + \left. {\sum \limits _{i = 2}^k {\left( {\begin{array}{c} k \\ i \\ \end{array}} \right) } {{\left( {l - cg} \right) }^{k - i}}\int \limits _{\left( {a + c} \right) g}^\infty {{y^{i - 2}}{e^{ - \frac{b}{g}y}}dy} } \right] \end{aligned} \end{aligned}$$
(95)
Using (81) and (83) reduces (95) to (59), completing the proof.
Appendix 4: Proof of Theorem 2
Inserting \(B_{sdt}\) in (13), \(A_{sd}\) in (12), \(B_{swt}\) in (20), D in (28) and \(A_{sw}\) in (19) into (32), one can simplify (32) as
$$\begin{aligned} \begin{aligned} {{\mathcal{Q}}_2} &={\mathrm{{E}}_{{{\left| {{h_{sr}}} \right| }^2}}}\left\{ {{{\bar{B}}_{sdt}}{P_s}{e^{ - \frac{{{2^{{C_0}}} - 1}}{{{\lambda _{sd}}{P_s}}}}}{2^{ - {C_0}}}\left( {{A_{tw}} + \frac{{{{\bar{B}}_{swt}}{P_s}}}{{{{\bar{B}}_{swt}}{P_s} - {2^{ - {C_0}}}\left( {{{\bar{B}}_{sdt}}{P_s} + {2^{{C_0}}} - 1} \right) }}} \right) } \right. \\&\quad \times \left. {\frac{{{\mathcal{V}}\left( {\frac{{{2^{{C_0}}}}}{{{\lambda _{sd}}{P_s}}} + \frac{1}{{{\lambda _{sw}}{P_s}}},{2^{ - {C_0}}}\left( {{{\bar{B}}_{sdt}}{P_s} + {2^{{C_0}}} - 1} \right) } \right) }}{{{{\bar{B}}_{swt}}{P_s} - {2^{ - {C_0}}}\left( {{{\bar{B}}_{sdt}}{P_s} + {2^{{C_0}}} - 1} \right) }}} \right\} \end{aligned} \end{aligned}$$
(96)
Recalling \({P_s} = \min \left( {\frac{{{I_m}}}{{{{\left| {{h_{sr}}} \right| }^2}}},{P_m}} \right) \) to further simplify (96) as
$$\begin{aligned} \begin{aligned} {{\mathcal{Q}}_2} &=\int \limits _0^{\frac{{{I_m}}}{{{P_m}}}} {{{\bar{B}}_{sdt}}{P_m}{e^{ - \frac{{{2^{{C_0}}} - 1}}{{{\lambda _{sd}}{P_m}}}}}{2^{ - {C_0}}}\left( {{A_{tw}} + \frac{{{{\bar{B}}_{swt}}{P_m}}}{{{{\bar{B}}_{swt}}{P_m} - {2^{ - {C_0}}}\left( {{{\bar{B}}_{sdt}}{P_m} + {2^{{C_0}}} - 1} \right) }}} \right) } \\&\quad \times \frac{{{\mathcal{V}}\left( {\frac{{{2^{{C_0}}}}}{{{\lambda _{sd}}{P_m}}} + \frac{1}{{{\lambda _{sw}}{P_m}}},{2^{ - {C_0}}}\left( {{{\bar{B}}_{sdt}}{P_m} + {2^{{C_0}}} - 1} \right) } \right) }}{{{{\bar{B}}_{swt}}{P_m} - {2^{ - {C_0}}}\left( {{{\bar{B}}_{sdt}}{P_m} + {2^{{C_0}}} - 1} \right) }}\frac{1}{{{\lambda _{sr}}}}{e^{ - \frac{x}{{{\lambda _{sr}}}}}}dx \\&\quad + \int \limits _{\frac{{{I_m}}}{{{P_m}}}}^\infty {{{\bar{B}}_{sdt}}\frac{{{I_m}}}{x}{e^{ - \frac{{{2^{{C_0}}} - 1}}{{{\lambda _{sd}}\frac{{{I_m}}}{x}}}}}{2^{ - {C_0}}}\left( {{A_{tw}} + \frac{{{{\bar{B}}_{swt}}\frac{{{I_m}}}{x}}}{{{{\bar{B}}_{swt}}\frac{{{I_m}}}{x} - {2^{ - {C_0}}}\left( {{{\bar{B}}_{sdt}}\frac{{{I_m}}}{x} + {2^{{C_0}}} - 1} \right) }}} \right) } \\&\quad \times \frac{{{\mathcal{V}}\left( {\frac{{{2^{{C_0}}}}}{{{\lambda _{sd}}\frac{{{I_m}}}{x}}} + \frac{1}{{{\lambda _{sw}}\frac{{{I_m}}}{x}}},{2^{ - {C_0}}}\left( {{{\bar{B}}_{sdt}}\frac{{{I_m}}}{x} + {2^{{C_0}}} - 1} \right) } \right) }}{{{{\bar{B}}_{swt}}\frac{{{I_m}}}{x} - {2^{ - {C_0}}}\left( {{{\bar{B}}_{sdt}}\frac{{{I_m}}}{x} + {2^{{C_0}}} - 1} \right) }}\frac{1}{{{\lambda _{sr}}}}{e^{ - \frac{x}{{{\lambda _{sr}}}}}}dx\\ &=\left( {{A_{tw}} + \frac{{{{\bar{B}}_{swt}}{P_m}}}{{{{\bar{B}}_{swt}}{P_m} - {2^{ - {C_0}}}\left( {{{\bar{B}}_{sdt}}{P_m} + {2^{{C_0}}} - 1} \right) }}} \right) \frac{{{{\bar{B}}_{sdt}}{P_m}{e^{ - \frac{{{2^{{C_0}}} - 1}}{{{\lambda _{sd}}{P_m}}}}}{2^{ - {C_0}}}}}{{{{\bar{B}}_{swt}}{P_m} - {2^{ - {C_0}}}\left( {{{\bar{B}}_{sdt}}{P_m} + {2^{{C_0}}} - 1} \right) }} \\&\quad \times \left( {1 - {e^{ - \frac{{{I_m}}}{{{P_m}{\lambda _{sr}}}}}}} \right) {\mathcal{V}}\left( {\frac{{{2^{{C_0}}}}}{{{\lambda _{sd}}{P_m}}} + \frac{1}{{{\lambda _{sw}}{P_m}}},{2^{ - {C_0}}}\left( {{{\bar{B}}_{sdt}}{P_m} + {2^{{C_0}}} - 1} \right) } \right) \\&\quad + \frac{{{{\bar{B}}_{sdt}}{I_m}{A_{tw}}}}{{\left( {1 - {2^{{C_0}}}} \right) {\lambda _{sr}}}}\int \limits _{\frac{{{I_m}}}{{{P_m}}}}^\infty \frac{{{e^{ - \left( {\frac{{{2^{{C_0}}} - 1}}{{{\lambda _{sd}}{I_m}}} + \frac{1}{{{\lambda _{sr}}}}} \right) x}}}}{{x + \frac{{{{\bar{B}}_{swt}}{I_m} - {2^{ - {C_0}}}{{\bar{B}}_{sdt}}{I_m}}}{{{2^{ - {C_0}}} - 1}}}}\\&\quad \times {\mathcal{V}} \left( {\left[ {\frac{{{2^{{C_0}}}}}{{{\lambda _{sd}}}} + \frac{1}{{{\lambda _{sw}}}}} \right] \frac{x}{{{I_m}}},{2^{ - {C_0}}}\left( {\frac{{{{\bar{B}}_{sdt}}{I_m}}}{x} + {2^{{C_0}}} - 1} \right) } \right) dx \\&\quad + \frac{{{{\bar{B}}_{sdt}}{{\bar{B}}_{swt}}}}{{{\lambda _{sr}}{2^{{C_0}}}}}{\left( {\frac{{{I_m}}}{{{2^{ - {C_0}}} - 1}}} \right) ^2} \\&\quad\times \int \limits _{\frac{{{I_m}}}{{{P_m}}}}^\infty \frac{{{e^{ - \left( {\frac{{{2^{{C_0}}} - 1}}{{{\lambda _{sd}}{I_m}}} + \frac{1}{{{\lambda _{sr}}}}} \right) x}}}}{{{{\left( {x + \frac{{{{\bar{B}}_{swt}}{I_m} - {2^{ - {C_0}}}{{\bar{B}}_{sdt}}{I_m}}}{{{2^{ - {C_0}}} - 1}}} \right) }^2}}}\\&\quad {\mathcal{V}}\left( {\left[ {\frac{{{2^{{C_0}}}}}{{{\lambda _{sd}}}} + \frac{1}{{{\lambda _{sw}}}}} \right] \frac{x}{{{I_m}}},{2^{ - {C_0}}}\left[ {\frac{{{{\bar{B}}_{sdt}}{I_m}}}{x} + {2^{{C_0}}} - 1} \right] } \right) dx \end{aligned} \end{aligned}$$
(97)
Two integrands in the last equality of (97) are solved in closed-form with the aids of (37) and (49), respectively, reducing (97) to (60) and completing the proof.
Appendix 5: Proof of Theorem 3
Inserting \(B_{sdt}\) in (13), \(A_{sd}\) in (12), \(B_{swt}\) in (20), D in (28) and \(A_{sw}\) in (19) into (33), one can rewrite (33) as
$$\begin{aligned} \begin{aligned} {{\mathcal{Q}}_3} &={\mathrm{{E}}_{{{\left| {{h_{sr}}} \right| }^2}}}\left\{ {\frac{{{{\bar{B}}_{sdt}}{P_s}{e^{ - \frac{{{2^{{C_0}}} - 1}}{{{\lambda _{sd}}{P_s}}}}}{2^{ - {C_0}}}}}{{{{\bar{B}}_{swt}}{P_s} - {2^{ - {C_0}}}\left( {{{\bar{B}}_{sdt}}{P_s} + {2^{{C_0}}} - 1} \right) }}\left[ {{A_{tw}} + \frac{{{{\bar{B}}_{swt}}{P_s}}}{{{{\bar{B}}_{swt}}{P_s} - {2^{ - {C_0}}}\left( {{{\bar{B}}_{sdt}}{P_s} + {2^{{C_0}}} - 1} \right) }}} \right. } \right. \\&\quad - \left. {\left. {{{\bar{B}}_{swt}}{P_s}\left( {\frac{{{2^{{C_0}}}}}{{{\lambda _{sd}}{P_s}}} + \frac{1}{{{\lambda _{sw}}{P_s}}}} \right) } \right] {\mathcal{V}}\left( {\frac{{{2^{{C_0}}}}}{{{\lambda _{sd}}{P_s}}} + \frac{1}{{{\lambda _{sw}}{P_s}}},{{\bar{B}}_{swt}}{P_s}} \right) } \right\} \end{aligned} \end{aligned}$$
(98)
Recalling \({P_s} = \min \left( {\frac{{{I_m}}}{{{{\left| {{h_{sr}}} \right| }^2}}},{P_m}} \right) \) to further simplify (98) as
$$\begin{aligned} \begin{aligned} {{\mathcal{Q}}_3} &=\int \limits _0^{\frac{{{I_m}}}{{{P_m}}}} {\frac{{{{\bar{B}}_{sdt}}{P_m}{e^{ - \frac{{{2^{{C_0}}} - 1}}{{{\lambda _{sd}}{P_m}}}}}{2^{ - {C_0}}}}}{{{{\bar{B}}_{swt}}{P_m} - {2^{ - {C_0}}}\left( {{{\bar{B}}_{sdt}}{P_m} + {2^{{C_0}}} - 1} \right) }} } \\& \quad \times \left[ {{A_{tw}} + \frac{{{{\bar{B}}_{swt}}{P_m}}}{{{{\bar{B}}_{swt}}{P_m} - {2^{ - {C_0}}}\left( {{{\bar{B}}_{sdt}}{P_m} + {2^{{C_0}}} - 1} \right) }} - {{\bar{B}}_{swt}}{P_m}\left( {\frac{{{2^{{C_0}}}}}{{{\lambda _{sd}}{P_m}}} + \frac{1}{{{\lambda _{sw}}{P_m}}}} \right) } \right] \\&\quad \times {\mathcal{V}}\left( {\frac{{{2^{{C_0}}}}}{{{\lambda _{sd}}{P_m}}} + \frac{1}{{{\lambda _{sw}}{P_m}}},{{\bar{B}}_{swt}}{P_m}} \right) \frac{1}{{{\lambda _{sr}}}}{e^{ - \frac{x}{{{\lambda _{sr}}}}}}dx \\&\quad + \int \limits _{\frac{{{I_m}}}{{{P_m}}}}^\infty {\frac{{{{\bar{B}}_{sdt}}\frac{{{I_m}}}{x}{e^{ - \frac{{{2^{{C_0}}} - 1}}{{{\lambda _{sd}}\frac{{{I_m}}}{x}}}}}{2^{ - {C_0}}}}}{{{{\bar{B}}_{swt}}\frac{{{I_m}}}{x} - {2^{ - {C_0}}}\left( {{{\bar{B}}_{sdt}}\frac{{{I_m}}}{x} + {2^{{C_0}}} - 1} \right) }} } \\&\quad \times \left[ {{A_{tw}} + \frac{{{{\bar{B}}_{swt}}\frac{{{I_m}}}{x}}}{{{{\bar{B}}_{swt}}\frac{{{I_m}}}{x} - {2^{ - {C_0}}}\left( {{{\bar{B}}_{sdt}}\frac{{{I_m}}}{x} + {2^{{C_0}}} - 1} \right) }} - {{\bar{B}}_{swt}}\frac{{{I_m}}}{x}\left( {\frac{{{2^{{C_0}}}}}{{{\lambda _{sd}}\frac{{{I_m}}}{x}}} + \frac{1}{{{\lambda _{sw}}\frac{{{I_m}}}{x}}}} \right) } \right] \\&\quad \times {\mathcal{V}}\left( {\frac{{{2^{{C_0}}}}}{{{\lambda _{sd}}\frac{{{I_m}}}{x}}} + \frac{1}{{{\lambda _{sw}}\frac{{{I_m}}}{x}}},{{\bar{B}}_{swt}}\frac{{{I_m}}}{x}} \right) \frac{1}{{{\lambda _{sr}}}}{e^{ - \frac{x}{{{\lambda _{sr}}}}}}dx, \end{aligned} \end{aligned}$$
(99)
which can be rewritten in a compact form as
$$\begin{aligned} \begin{aligned} {{\mathcal{Q}}_3} &=\frac{{{{\bar{B}}_{sdt}}{P_m}{e^{ - \frac{{{2^{{C_0}}} - 1}}{{{\lambda _{sd}}{P_m}}}}}{2^{ - {C_0}}}}}{{{{\bar{B}}_{swt}}{P_m} - {2^{ - {C_0}}}\left( {{{\bar{B}}_{sdt}}{P_m} + {2^{{C_0}}} - 1} \right) }} \\&\times \left( {1 - {e^{ - \frac{{{I_m}}}{{{P_m}{\lambda _{sr}}}}}}} \right) {\mathcal{V}}\left( {\frac{{{2^{{C_0}}}}}{{{\lambda _{sd}}{P_m}}} + \frac{1}{{{\lambda _{sw}}{P_m}}},{{\bar{B}}_{swt}}{P_m}} \right) \\&\times \left[ {{A_{tw}} + \frac{{{{\bar{B}}_{swt}}{P_m}}}{{{{\bar{B}}_{swt}}{P_m} - {2^{ - {C_0}}}\left( {{{\bar{B}}_{sdt}}{P_m} + {2^{{C_0}}} - 1} \right) }} - {{\bar{B}}_{swt}}\left( {\frac{{{2^{{C_0}}}}}{{{\lambda _{sd}}}} + \frac{1}{{{\lambda _{sw}}}}} \right) } \right] \\&+ \frac{{{{\bar{B}}_{sdt}}{I_m}{2^{ - {C_0}}}}}{{\left( {{2^{ - {C_0}}} - 1} \right) {\lambda _{sr}}}}\left[ {{A_{tw}} - {{\bar{B}}_{swt}}\left( {\frac{{{2^{{C_0}}}}}{{{\lambda _{sd}}}} + \frac{1}{{{\lambda _{sw}}}}} \right) } \right] \\&\times \int \limits _{\frac{{{I_m}}}{{{P_m}}}}^\infty {\frac{{{e^{ - \left( {\frac{{{2^{{C_0}}} - 1}}{{{\lambda _{sd}}{I_m}}} + \frac{1}{{{\lambda _{sr}}}}} \right) x}}}}{{x + \frac{{{{\bar{B}}_{swt}}{I_m} - {2^{ - {C_0}}}{{\bar{B}}_{sdt}}{I_m}}}{{{2^{ - {C_0}}} - 1}}}}{\mathcal{V}}\left( {\left[ {\frac{{{2^{{C_0}}}}}{{{\lambda _{sd}}}} + \frac{1}{{{\lambda _{sw}}}}} \right] \frac{x}{{{I_m}}},\frac{{{{\bar{B}}_{swt}}{I_m}}}{x}} \right) dx} \\&+ \frac{{{2^{{C_0}}}{{\bar{B}}_{swt}}{{\bar{B}}_{sdt}}}}{{{\lambda _{sr}}}}{\left( {\frac{{{I_m}}}{{1 - {2^{{C_0}}}}}} \right) ^2} \\&\times \int \limits _{\frac{{{I_m}}}{{{P_m}}}}^\infty {\frac{{{e^{ - \left( {\frac{{{2^{{C_0}}} - 1}}{{{\lambda _{sd}}{I_m}}} + \frac{1}{{{\lambda _{sr}}}}} \right) x}}}}{{{{\left( {x + \frac{{{{\bar{B}}_{swt}}{I_m} - {2^{ - {C_0}}}{{\bar{B}}_{sdt}}{I_m}}}{{{2^{ - {C_0}}} - 1}}} \right) }^2}}}{\mathcal{V}}\left( {\left[ {\frac{{{2^{{C_0}}}}}{{{\lambda _{sd}}}} + \frac{1}{{{\lambda _{sw}}}}} \right] \frac{x}{{{I_m}}},\frac{{{{\bar{B}}_{swt}}{I_m}}}{x}} \right) dx} \end{aligned} \end{aligned}$$
(100)
Two integrands in the last equality of (100) are solved in closed-form with the aids of (61) and (62), respectively, reducing (100) to (63) and completing the proof.
