Bidirectional relaying with energy harvesting capable relay: outage analysis for Nakagami-m fading

Abstract

This paper analyzes outage probability of bidirectional relaying (BDR) where two power-unconstrained single-antenna sources communicate with each other under assistance of a self-powered half-duplex single-antenna relay capable of energy harvesting and amplify-and-forward implementation. The relay harvests radio energy from both sources to power its relaying operation with the power splitting method. For outage analysis of the BDR for Nakagami-m fading, an exact formula is first proposed in closed-form. Through this formula, influences of important specifications (time switching ratio, power splitting ratio, energy conversion efficiency, fading severity, target transmission rate, transmit power of each source, distances from sources to relay) on the outage probability are then evaluated. Finally, Monte-Carlo simulations are generated to corroborate the proposed formula.

Appendices

Appendix A: Proof of Lemma 1

This appendix proves (28). By inserting ${\Phi _1} = \frac{{\texttt {a}_2xy}}{{x + \texttt {b}}}$ in (26) into ${\Upsilon _1}$ in (21), one can rewrite ${\Upsilon _1}$ as

$$\begin{aligned} {\Upsilon _1}= & {} \Pr \left\{ {\frac{{\texttt {a}_2xy}}{{x + \texttt {b}}}< {\tau _1}} \right\} = \Pr \left\{ {y < \frac{{{\tau _1}}}{\texttt {a}_2}\frac{{x + \texttt {b}}}{x}} \right\} \nonumber \\= & {} {\Xi _x}\left\{ {{F_y}\left( {\left. {\frac{{{\tau _1}}}{\texttt {a}_2}\frac{{x + \texttt {b}}}{x}} \right| x} \right) } \right\} . \end{aligned}$$

(42)

Using (3) for $F_y(y)$, one can further simplify (42) as

$$\begin{aligned} {\Upsilon _1}= & {} {\Xi _x}\left\{ {1 - {e^{ - \frac{{{\tau _1}}}{{{\Omega _2}{} \texttt {a}_2}}\frac{{x + \texttt {b}}}{x}}}\sum \limits _{k = 0}^{{\alpha _2} - 1} {\frac{{{{\left( {\frac{{{\tau _1}}}{\texttt {a}_2}\frac{{x + \texttt {b}}}{x}} \right) }^k}}}{{k!\Omega _2^k}}} } \right\} \nonumber \\= & {} 1 - \sum \limits _{k = 0}^{{\alpha _2} - 1} {\frac{{{e^{ - \frac{{{\tau _1}}}{{{\Omega _2}{} \texttt {a}_2}}}}}}{{k!\Omega _2^k}}} {\left( {\frac{{{\tau _1}}}{\texttt {a}_2}} \right) ^k}{\Xi _x}\left\{ {{e^{ - \frac{{{\tau _1}{} \texttt {b}}}{{{\Omega _2}{} \texttt {a}_2}}\frac{1}{x}}}{{\left( {\frac{{x + \texttt {b}}}{x}} \right) }^k}} \right\} .\nonumber \\ \end{aligned}$$

(43)

Applying the definition of the expectation to rewrite (43) as

$$\begin{aligned} {\Upsilon _1} = 1 - \sum \limits _{k = 0}^{{\alpha _2} - 1} {\frac{{{e^{ - \frac{{{\tau _1}}}{{{\Omega _2}{} \texttt {a}_2}}}}}}{{k!\Omega _2^k}}} {\left( {\frac{{{\tau _1}}}{\texttt {a}_2}} \right) ^k}\int \limits _0^\infty {{e^{ - \frac{{{\tau _1}{} \texttt {b}}}{{{\Omega _2}{} \texttt {a}_2}}\frac{1}{x}}}{{\left( {\frac{{x + \texttt {b}}}{x}} \right) }^k}{f_x}\left( x \right) dx}. \end{aligned}$$

(44)

Using (2) for ${f_x}\left( x \right)$ to simplify (44) as

$$\begin{aligned} {\Upsilon _1}= & {} 1 - \sum \limits _{k = 0}^{{\alpha _2} - 1} {\frac{{{e^{ - \frac{{{\tau _1}}}{{{\Omega _2}{} \texttt {a}_2}}}}}}{{k!\Omega _2^k}}} {\left( {\frac{{{\tau _1}}}{\texttt {a}_2}} \right) ^k}\nonumber \\&\times \int \limits _0^\infty {{e^{ - \frac{{{\tau _1}{} \texttt {b}}}{{{\Omega _2}{} \texttt {a}_2}}\frac{1}{x}}}{{\left( {\frac{{x + \texttt {b}}}{x}} \right) }^k}\frac{{{x^{{\alpha _1} - 1}}}}{{\Gamma \left( {{\alpha _1}} \right) \Omega _1^{{\alpha _1}}}}{e^{ - \frac{x}{{{\Omega _{_1}}}}}}dx} \nonumber \\= & {} 1 - \sum \limits _{k = 0}^{{\alpha _2} - 1} {\frac{{{e^{ - \frac{{{\tau _1}}}{{{\Omega _2}{} \texttt {a}_2}}}}}}{{k!\Omega _2^k\Gamma \left( {{\alpha _1}} \right) \Omega _1^{{\alpha _1}}}}} {\left( {\frac{{{\tau _1}}}{\texttt {a}_2}} \right) ^k}\nonumber \\&\times \int \limits _0^\infty {{e^{ - \frac{{{\tau _1}{} \texttt {b}}}{{{\Omega _2}{} \texttt {a}_2}}\frac{1}{x} - \frac{x}{{{\Omega _{_1}}}}}}{{\left( {x + \texttt {b}} \right) }^k}{x^{{\alpha _1} - k - 1}}dx}. \end{aligned}$$

(45)

Applying the binominal expansion to ${\left( {x + \texttt {b}} \right) ^k}$ to further rewrite (45) as

$$\begin{aligned} {\Upsilon _1}= & {} 1 - \sum \limits _{k = 0}^{{\alpha _2} - 1} {\frac{{{e^{ - \frac{{{\tau _1}}}{{{\Omega _2}{} \texttt {a}_2}}}}}}{{k!\Omega _2^k\Gamma \left( {{\alpha _1}} \right) \Omega _1^{{\alpha _1}}}}} {\left( {\frac{{{\tau _1}}}{\texttt {a}_2}} \right) ^k}\nonumber \\&\times \int \limits _0^\infty {{e^{ - \frac{{{\tau _1}{} \texttt {b}}}{{{\Omega _2}{} \texttt {a}_2}}\frac{1}{x} - \frac{x}{{{\Omega _{_1}}}}}}\left[ {\sum \limits _{u = 0}^k {C_k^u{x^u}{\texttt {b}^{k - u}}} } \right] {x^{{\alpha _1} - k - 1}}dx} \nonumber \\= & {} 1 - \sum \limits _{k = 0}^{{\alpha _2} - 1} {\sum \limits _{u = 0}^k {\frac{{C_k^u{\texttt {b}^{k - u}}{e^{ - \frac{{{\tau _1}}}{{{\Omega _2}{} \texttt {a}_2}}}}}}{{k!\Omega _2^k\Gamma \left( {{\alpha _1}} \right) \Omega _1^{{\alpha _1}}}}{{\left( {\frac{{{\tau _1}}}{\texttt {a}_2}} \right) }^k}} }\nonumber \\&\underbrace{\int \limits _0^\infty {{e^{ - \frac{{{\tau _1}{} \texttt {b}}}{{{\Omega _2}{} \texttt {a}_2}}\frac{1}{x} - \frac{x}{{{\Omega _{_1}}}}}}{x^{{\alpha _1} - k + u - 1}}dx} }_\mathcal {D}. \end{aligned}$$

(46)

By denoting

$$\begin{aligned} \Psi \left( {l,m,j} \right) = \int \limits _0^\infty {{x^{l - 1}}{e^{ - \frac{m}{x} - jx}}dx}, \end{aligned}$$

(47)

it is straightforwardly seen that $\mathcal {D}$ can be expressed as $\mathcal {D} = \Psi \left( {{\alpha _1} - k + u,\frac{{{\tau _1}{} \texttt {b}}}{{{\Omega _2}{} \texttt {a}_2}},\frac{1}{{{\Omega _{_1}}}}} \right)$, resulting in the exact agreement between (28) and (46). This completes the proof of Lemma 1 if (47) coincides (29). With the aid of [56, Eq. (3.471.9)], one can easily reduce (47) to (29).

Appendix B: Proof of $\Upsilon _3$

This appendix proves (31). Using the explicit forms of ${\Phi _1} = \frac{{\texttt {a}_2xy}}{{x + \texttt {b}}}$ and ${\Phi _2} = \frac{{\texttt {a}_1xy}}{{y + \texttt {b}}}$ in (26) and (27), respectively, one can rewrite ${\Upsilon _3}$ in (21) as

$$\begin{aligned} {\Upsilon _3}= & {} \Pr \left\{ {\left\{ {\frac{{\texttt {a}_2xy}}{{x + \texttt {b}}}< {\tau _1}} \right\} \cap \,\,\left\{ {\,\frac{{\texttt {a}_1xy}}{{y + \texttt {b}}}< \,{\tau _2}} \right\} } \right\} \nonumber \\= & {} \Pr \left\{ {\left\{ {y< \frac{{{\tau _1}(x + \texttt {b})}}{{\texttt {a}_2x}}} \right\} \cap \,\,\left\{ {x < \frac{{{\tau _2}(y + \texttt {b})}}{{\texttt {a}_1y}}} \right\} } \right\} \nonumber \\= & {} \int \limits _0^{{x_0}} {\int \limits _{\frac{{{y_0}}}{{{x_0}}}x}^{\frac{{{\tau _1}(x + \texttt {b})}}{{\texttt {a}_2x}}} {{f_{x,y}}(x,y)dydx} }\nonumber \\&+ \int \limits _0^{{y_0}} {\int \limits _{\frac{{{x_0}}}{{{y_0}}}y}^{\frac{{{\tau _2}(y + \texttt {b})}}{{\texttt {a}_1y}}} {{f_{x,y}}(x,y)dxdy} }, \end{aligned}$$

(48)

where $f_{x,y}\left( x,y\right)$ is the joint PDF of x and y; ($x_0, y_0$) is the intersection point of two curves, $y = \frac{{{\tau _1}(x + \texttt {b})}}{{\texttt {a}_2x}}$ and $x = \frac{{{\tau _2}(y + \texttt {b})}}{{\texttt {a}_1y}}$, which can be expressed as

$$\begin{aligned} {x_0}= & {} \frac{{{\mu } + \sqrt{{\mu }^2 + 4\texttt {a}_1\texttt {b}{\tau }_1^2{\tau _2}} }}{{2\texttt {a}_1{\tau _1}}}, \end{aligned}$$

(49)

$$\begin{aligned} {y_0}= & {} \frac{{{\tau _2}{} \texttt {b}}}{{\texttt {a}_1{x_0} - {\tau _2}}}, \end{aligned}$$

(50)

with

$$\begin{aligned} {\mu } = {\tau _1}{\tau _2} - \texttt {a}_1\texttt {b}{\tau _1} + \texttt {a}_2\texttt {b}{\tau _2}. \end{aligned}$$

(51)

Because x is statistically independent of y, one can decompose $f_{x,y}\left( x,y\right)$ as $f_{x,y}\left( x,y\right) =f_{x}(x)f_{y}(y)$. Therefore, (48) can be rewritten as

$$\begin{aligned} {\Upsilon _3}= & {} \underbrace{\int \limits _0^{{x_0}} {\left[ {\int \limits _{\frac{{{y_0}}}{{{x_0}}}x}^{\frac{{{\tau _1}(x + \texttt {b})}}{{\texttt {a}_2x}}} {{f_y}(y)dy} } \right] {f_x}(x)dx} }_{{H_1}}\nonumber \\&+ \underbrace{\int \limits _0^{{y_0}} {\left[ {\int \limits _{\frac{{{x_0}}}{{{y_0}}}y}^{\frac{{{\tau _2}(y + \texttt {b})}}{{\texttt {a}_1y}}} {{f_x}(x)dx} } \right] } {f_y}(y)dy}_{{H_2}}. \end{aligned}$$

(52)

By using (2) for ${f_x}(x)$ and ${f_y}(y)$, one rewrites $H_1$ in a more compact form as

$$\begin{aligned} {H_1}= & {} \int \limits _0^{{x_0}} {\left[ {\int \limits _{\frac{{{y_0}}}{{{x_0}}}x}^{\frac{{{\tau _1}\left( {x + \texttt {b}} \right) }}{{\texttt {a}_2x}}} {\frac{{{y^{{\alpha _2} - 1}}}}{{\Gamma \left( {{\alpha _2}} \right) \Omega _2^{{\alpha _2}}}}{e^{ - \frac{y}{{{\Omega _2}}}}}dy\,} } \right] } \frac{{{x^{{\alpha _1} - 1}}}}{{\Gamma \left( {{\alpha _1}} \right) \Omega _1^{{\alpha _1}}}}{e^{ - \frac{x}{{{\Omega _{_1}}}}}}dx\nonumber \\= & {} \frac{\Omega _1^{{-\alpha _1}}\Omega _2^{{-\alpha _2}}}{{\Gamma \left( {{\alpha _1}} \right) \Gamma \left( {{\alpha _2}} \right) }}\int \limits _0^{{x_0}} {{x^{{\alpha _1} - 1}}{e^{ - \frac{x}{{{\Omega _1}}}}}\left[ {\int \limits _{\frac{{{y_0}}}{{{x_0}}}x}^{\frac{{{\tau _1}(x + \texttt {b})}}{{\texttt {a}_2x}}} {{y^{{\alpha _2} - 1}}{e^{ - \frac{y}{{{\Omega _2}}}}}dy\,} } \right] dx} \nonumber \\= & {} \frac{{\Omega _1^{ - {\alpha _1}}\Omega _2^{ - {\alpha _2}}}}{{\Gamma \left( {{\alpha _1}} \right) \Gamma \left( {{\alpha _2}} \right) }}\int \limits _0^{{x_0}} {x^{{\alpha _1} - 1}}{e^{ - \frac{x}{{{\Omega _1}}}}}\nonumber \\&\times \left[ {\underbrace{\int \limits _{\frac{{{y_0}}}{{{x_0}}}x}^\infty {{y^{{\alpha _2} - 1}}{e^{ - \frac{y}{{{\Omega _2}}}}}dy\,} }_{{{\bar{H}}_{11}}} - \underbrace{\int \limits _{\frac{{{\tau _1}(x + \texttt {b})}}{{\texttt {a}_2x}}}^\infty {{y^{{\alpha _2} - 1}}{e^{ - \frac{y}{{{\Omega _2}}}}}dy\,} }_{{{\bar{H}}_{12}}}} \right] dx\nonumber \\= & {} \frac{{{H_{11}} - {H_{12}}}}{{\Gamma \left( {{\alpha _1}} \right) \Gamma \left( {{\alpha _2}} \right) \Omega _1^{{\alpha _1}}\Omega _2^{{\alpha _2}}}}. \end{aligned}$$

(53)

where

$$\begin{aligned} {H_{11}}= & {} \int \limits _0^{{x_0}} {{x^{{\alpha _1} - 1}}{e^{ - \frac{x}{{{\Omega _1}}}}}{{\bar{H}}_{11}}} dx, \end{aligned}$$

(54)

$$\begin{aligned} {H_{12}}= & {} \int \limits _0^{{x_0}} {{x^{{\alpha _1} - 1}}{e^{ - \frac{x}{{{\Omega _1}}}}}{{\bar{H}}_{12}}} dx. \end{aligned}$$

(55)

With the aid of [56, Eq. (3.351.2)], ${\bar{H}_{11}}$ is expressed in closed-form as

$$\begin{aligned} {{\bar{H}}_{11}} = \int \limits _{\frac{{{y_0}}}{{{x_0}}}x}^\infty {{y^{{\alpha _2} - 1}}{e^{ - \frac{y}{{{\Omega _2}}}}}dy} = {e^{ - \frac{{{y_0}}}{{{x_0}{\Omega _2}}}x}}\sum \limits _{k = 0}^{{\alpha _2} - 1} {\frac{{\left( {{\alpha _2} - 1} \right) !}}{{k!\Omega _2^{k - {\alpha _2}}}}} {\left( {\frac{{{y_0}}}{{{x_0}}}} \right) ^k}{x^k}. \end{aligned}$$

(56)

Inserting (56) into (54), one obtains

$$\begin{aligned} {H_{11}}= & {} \int \limits _0^{{x_0}} {{x^{{\alpha _1} - 1}}{e^{ - \frac{x}{{{\Omega _1}}}}}{e^{ - \frac{{{y_0}}}{{{x_0}{\Omega _2}}}x}}\sum \limits _{k = 0}^{{\alpha _2} - 1} {\frac{{\left( {{\alpha _2} - 1} \right) !}}{{k!\Omega _2^{k - {\alpha _2}}}}} {{\left( {\frac{{{y_0}}}{{{x_0}}}} \right) }^k}{x^k}} dx \nonumber \\= & {} \sum \limits _{k = 0}^{{\alpha _2} - 1} {\frac{{\left( {{\alpha _2} - 1} \right) !}}{{k!\Omega _2^{k - {\alpha _2}}}}} {\left( {\frac{{{y_0}}}{{{x_0}}}} \right) ^k}\,\int \limits _0^{{x_0}} {{x^{k + {\alpha _1} - 1}}{e^{ - \left( {\frac{1}{{{\Omega _1}}} + \frac{{{y_0}}}{{{x_0}{\Omega _2}}}} \right) x}}dx}.\nonumber \\ \end{aligned}$$

(57)

By using [56, Eq. (3.381.1)], one can represent the last integral in (57) in terms of the lower incomplete gamma function $\gamma \left( \cdot ,\cdot \right)$. Therefore, (57) is rewritten in closed-form as

$$\begin{aligned} {H_{11}}= & {} \sum \limits _{k = 0}^{{\alpha _2} - 1} {\frac{{\left( {{\alpha _2} - 1} \right) !}}{{k!\Omega _2^{k - {\alpha _2}}}}} {\left( {\frac{{{y_0}}}{{{x_0}}}} \right) ^k}{\left( {\frac{1}{{{\Omega _1}}} + \frac{{{y_0}}}{{{x_0}{\Omega _2}}}} \right) ^{ - \left( {k + {\alpha _1}} \right) }}\nonumber \\&\times \,\gamma \left( {k + {\alpha _1},\left[ {\frac{1}{{{\Omega _1}}} + \frac{{{y_0}}}{{{x_0}{\Omega _2}}}} \right] {x_0}} \right) . \end{aligned}$$

(58)

Now, we process $H_{12}$. First, by imitating the derivation of ${\bar{H}_{11}}$, one can compute ${\bar{H}_{12}}$ as

$$\begin{aligned} {{\bar{H}}_{12}}= & {} \int \limits _{\frac{{{\tau _1}\left( {x + \texttt {b}} \right) }}{{\texttt {a}_2x}}}^\infty {{y^{{\alpha _2} - 1}}{e^{ - \frac{y}{{{\Omega _2}}}}}dy}\nonumber \\= & {} {e^{ - \,\frac{{{\tau _1}(x + \texttt {b})}}{{{\Omega _2}{} \texttt {a}_2x}}}}\sum \limits _{k = 0}^{{\alpha _2} - 1} {\frac{{\left( {{\alpha _2} - 1} \right) !}}{{k!\Omega _2^{k - {\alpha _2}}}}{{\left( {\frac{{{\tau _1}\left[ {x + \texttt {b}} \right] }}{{\texttt {a}_2x}}} \right) }^k}}. \end{aligned}$$

(59)

Then, inserting (59) into (55) to yield

$$\begin{aligned} {H_{12}}= & {} \int \limits _0^{{x_0}} {{x^{{\alpha _1} - 1}}{e^{ - \frac{x}{{{\Omega _1}}}}}{e^{ - \frac{{{\tau _1}(x + \texttt {b})}}{{{\Omega _2}{} \texttt {a}_2x}}}} \sum \limits _{k = 0}^{{\alpha _2} - 1} {\frac{{\left( {{\alpha _2} - 1} \right) !}}{{k!\Omega _2^{k - {\alpha _2}}}}{{\left( {\frac{{{\tau _1}\left[ {x + \texttt {b}} \right] }}{{\texttt {a}_2x}}} \right) }^k}} dx }\nonumber \\= & {} \sum \limits _{k = 0}^{{\alpha _2} - 1} \frac{{\left( {{\alpha _2} - 1} \right) !}}{{k!\Omega _2^{k - {\alpha _2}}}}{{\left( {\frac{{{\tau _1}}}{\texttt {a}_2}} \right) }^k}{e^{ - \frac{{{\tau _1}{} \texttt {b}}}{{{\Omega _2}{} \texttt {a}_2}}}} \nonumber \\&\times \,\int \limits _0^{{x_0}} {{{\left( {\frac{{x + \texttt {b}}}{x}} \right) }^k}{x^{{\alpha _1} - 1}}{e^{ - \frac{x}{{{\Omega _1}}} - \frac{{{\tau _1}{} \texttt {b}}}{{{\Omega _2}{} \texttt {a}_2x}}}}} dx \nonumber \\= & {} \sum \limits _{k = 0}^{{\alpha _2} - 1} \frac{{\left( {{\alpha _2} - 1} \right) !}}{{k!\Omega _2^{k - {\alpha _2}}}}{{\left( {\frac{{{\tau _1}{} \texttt {b}}}{\texttt {a}_2}} \right) }^k}{e^{ - \frac{{{\tau _1}}}{{\texttt {a}_2{\Omega _2}}}}}\nonumber \\&\times \,\int \limits _0^{{x_0}} {{{\left( {\frac{1}{\texttt {b}} + \frac{1}{x}} \right) }^k}{x^{{\alpha _1} - 1}}{e^{ - \frac{x}{{{\Omega _1}}} - \frac{{{\tau _1}{} \texttt {b}}}{{\texttt {a}_2{\Omega _2}x}}}}} dx \nonumber \\= & {} \sum \limits _{k = 0}^{{\alpha _2} - 1} \frac{{\left( {{\alpha _2} - 1} \right) !}}{{k!\Omega _2^{k - {\alpha _2}}}}{{\left( {\frac{{{\tau _1}{} \texttt {b}}}{\texttt {a}_2}} \right) }^k}{e^{ - \frac{{{\tau _1}}}{{\texttt {a}_2{\Omega _2}}}}}\nonumber \\&\times \,\int \limits _0^{{x_0}} {\left[ {\sum \limits _{z = 0}^k {C_k^z{x^{ - z}}{{\left( {\frac{1}{\texttt {b}}} \right) }^{k - z}}} } \right] {x^{{\alpha _1} - 1}}{e^{ - \frac{x}{{{\Omega _1}}} - \frac{{{\tau _1}{} \texttt {b}}}{{\texttt {a}_2{\Omega _2}x}}}}dx} \nonumber \\= & {} \sum \limits _{k = 0}^{{\alpha _2} - 1} \frac{{\left( {{\alpha _2} - 1} \right) !}}{{k!\Omega _2^{k - {\alpha _2}}}}{{\left( {\frac{{{\tau _1}}}{\texttt {a}_2}} \right) }^k}{e^{ - \frac{{{\tau _1}}}{{\texttt {a}_2{\Omega _2}}}}} \sum \limits _{z = 0}^k {C_k^z{\texttt {b}^z}} \nonumber \\&\times \,\int \limits _0^{{x_0}} {{x^{{\alpha _1} - z - 1}}{e^{ - \frac{x}{{{\Omega _1}}} - \frac{{{\tau _1}{} \texttt {b}}}{{\texttt {a}_2{\Omega _2}x}}}}} dx . \end{aligned}$$

(60)

Plugging (58) and (60) in (53), one obtains

$$\begin{aligned} {H_1}= & {} \frac{{\Omega _1^{ - {\alpha _1}}\Omega _2^{ - {\alpha _2}}}}{{\Gamma \left( {{\alpha _1}} \right) \Gamma \left( {{\alpha _2}} \right) }}\left[ \sum \limits _{k = 0}^{{\alpha _2} - 1} {\frac{{\left( {{\alpha _2} - 1} \right) !}}{{k!\Omega _2^{k - {\alpha _2}}}}} {{\left( {\frac{{{y_0}}}{{{x_0}}}} \right) }^k}{\left( {\frac{1}{{{\Omega _1}}} + \frac{{{y_0}}}{{{x_0}{\Omega _2}}}} \right) ^{ - \left( {k + {\alpha _1}} \right) }}\right. \nonumber \\&\quad \times \,\gamma \left( {k + {\alpha _1},\left[ {\frac{1}{{{\Omega _1}}} + \frac{{{y_0}}}{{{x_0}{\Omega _2}}}} \right] {x_0}} \right) \nonumber \\&\quad -\,\sum \limits _{k = 0}^{{\alpha _2} - 1} \frac{{\left( {{\alpha _2} - 1} \right) !}}{{k!\Omega _2^{k - {\alpha _2}}}}{{\left( {\frac{{{\tau _1}}}{\texttt {a}_2}} \right) }^k}{e^{ - \frac{{{\tau _1}}}{{\texttt {a}_2{\Omega _2}}}}}\sum \limits _{z = 0}^k {C_k^z{\texttt {b}^z}}\nonumber \\&\quad \times \,\left. \int \limits _0^{{x_0}} {{x^{{\alpha _1} - z - 1}}{e^{ - \frac{x}{{{\Omega _1}}} - \frac{{{\tau _1}{} \texttt {b}}}{{\texttt {a}_2{\Omega _2}x}}}}} dx \right] \nonumber \\&= \frac{{\Omega _1^{ - {\alpha _1}}\Omega _2^{ - {\alpha _2}}}}{{\Gamma \left( {{\alpha _1}} \right) \Gamma \left( {{\alpha _2}} \right) }}\sum \limits _{k = 0}^{{\alpha _2} - 1} {\frac{{\left( {{\alpha _2} - 1} \right) !}}{{k!\Omega _2^{k - {\alpha _2}}}}} \left[ \phantom {\int \limits _0^{{x_0}} {{x^{{\alpha _1} - z - 1}}{e^{ - \frac{x}{{{\Omega _1}}} - \frac{{{\tau _1}{} \texttt {b}}}{{\texttt {a}_2{\Omega _2}x}}}}} dx} {{\left( {\frac{{{y_0}}}{{{x_0}}}} \right) }^k} \left( \frac{1}{{{\Omega _1}}} \right. \right. \nonumber \\&\quad \left. +\, \frac{{{y_0}}}{{{x_0}{\Omega _2}}} \right) ^{ - \left( {k + {\alpha _1}} \right) }\gamma \left( {k + {\alpha _1},\left[ {\frac{1}{{{\Omega _1}}} + \frac{{{y_0}}}{{{x_0}{\Omega _2}}}} \right] {x_0}} \right) \nonumber \\&\quad - \,{\left( {\frac{{{\tau _1}}}{\texttt {a}_2}} \right) ^k}{e^{ - \frac{{{\tau _1}}}{{\texttt {a}_2{\Omega _2}}}}}\left. {\sum \limits _{z = 0}^k {C_k^z{\texttt {b}^z}} \int \limits _0^{{x_0}} {{x^{{\alpha _1} - z - 1}}{e^{ - \frac{x}{{{\Omega _1}}} - \frac{{{\tau _1}{} \texttt {b}}}{{\texttt {a}_2{\Omega _2}x}}}}} dx} \right] . \end{aligned}$$

(61)

Given (34), one can write the last integral in (61) in terms of $U\left( {l,v,p,u} \right)$. Therefore, (61) exactly matches (32).

Following the same procedure as deriving (61), it is straightforwardly proven that $H_2$ in (52) coincides that in (33), completing the proof of (31).