Numerical study and optimization of GO/ZnO based perovskite solar cell using SCAPS

Norsakinah Johrin; Fuei Pien Chee; Syafiqa Nasir; Pak Yan Moh; Norsakinah Johrin; Fuei Pien Chee; Syafiqa Nasir; Pak Yan Moh

doi:10.3934/energy.2023034

AIMS Energy

2023, Volume 11, Issue 4: 683-693. doi: 10.3934/energy.2023034

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Research article

Numerical study and optimization of GO/ZnO based perovskite solar cell using SCAPS

1.
Physics with Electronic Programme, Faculty of Science and Natural Resources, Universiti Malaysia Sabah
2.
Industrial Chemistry Programme, Faculty of Science and Natural Resources, Universiti Malaysia Sabah
3.
Water Research Unit, Faculty of Science and Natural Resources, Universiti Malaysia Sabah

Received: 20 April 2023 Revised: 19 June 2023 Accepted: 09 July 2023 Published: 28 August 2023

This paper focuses on the numerical study of hybrid organic-inorganic perovskite solar cells. It investigates the incorporation of a graphene oxide (GO) thin layer to enhance solar cell efficiency. The study demonstrates that the GO layer improves interaction with the absorber layer and enhances hole transportation, resulting in reduced recombination and diffusion losses at the absorber and hole transport layer (HTL) interface. The increased energy level of the Lower Unoccupied Molecular Orbital (LUMO) in GO acts as an excellent electron-blocking layer, thereby improving the V_OC. The objective is to explore different structures of perovskite solar cells to enhance their performance. The simulated solar cell comprises a GO/FASnI₃/TiO₂/ZnO/ITO sandwich structure, with FASnI₃ and ZnO thicknesses adjusted to improve conversion efficiency. The impact of thickness on device performance, specifically the absorber and electron transport layers, is investigated. The fill factor (FF) changes as the absorber and electron transport layers (ETL) increase. The FF is an important parameter that determines PSC performance since it measures how effectively power is transferred from the cell to an external circuit. The optimized solar cell achieves a short-circuit current density (J_SC) of 27.27 mA/cm², an open-circuit voltage (V_OC) of 2.76 V, a fill factor (FF) of 27.05% and the highest power conversion efficiency (PCE) of 20.39% with 400 nm of FASnI₃ and 300 nm of ZnO. These findings suggest promising directions for the development of more effective GO-based perovskite solar cells.

Keywords:

Citation: Norsakinah Johrin, Fuei Pien Chee, Syafiqa Nasir, Pak Yan Moh. Numerical study and optimization of GO/ZnO based perovskite solar cell using SCAPS[J]. AIMS Energy, 2023, 11(4): 683-693. doi: 10.3934/energy.2023034

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Abstract

1. Introduction

We will derive integrals as indicated in the abstract in terms of special functions. Some special cases of these integrals have been reported in Gradshteyn and Ryzhik ^[5]. In 1867 David Bierens de Haan derived hyperbolic integrals of the form

$\begin{equation} \int_{0}^{\infty}\left((\log (a)-i x)^k+(\log (a)+i x)^k\right) \log (\cos (\alpha ) \text{sech}(x)+1)dx \end{equation}$

(1.1)

In our case the constants in the formulas are general complex numbers subject to the restrictions given below. The derivations follow the method used by us in ^[6]. The generalized Cauchy's integral formula is given by

$\begin{equation} \frac{y^k}{k!} = \frac{1}{2\pi i}\int_{C}\frac{e^{wy}}{w^{k+1}}dw. \end{equation}$

(1.2)

2. Definite integral of the contour integral

We use the method in ^[6]. Here the contour is in the upper left quadrant with $\Re(w) < 0$ and going round the origin with zero radius. Using a generalization of Cauchy's integral formula we first replace $y$ by $ix+\log(a)$ for the first equation and then $y$ by $-ix+\log(a)$ to get the second equation. Then we add these two equations, followed by multiplying both sides by $\frac{1}{2} \log (\cos (\alpha) \text{sech}(x)+1)$ to get

$\begin{multline} \frac{\left((\log (a)-i x)^k+(\log (a)+i x)^k\right) \log (\cos (\alpha ) \text{sech}(x)+1)}{2 k!}\\ = \frac{1}{2\pi i}\int_{C}a^w w^{-k-1} \cos (w x) \log (\cos (\alpha ) \text{sech}(x)+1)dw \end{multline}$

(2.1)

where the logarithmic function is defined in Eq (4.1.2) in ^[2]. We then take the definite integral over $x \in [0, \infty)$ of both sides to get

$\begin{equation} \int_{0}^{\infty}\frac{\left((\log (a)-i x)^k+(\log (a)+i x)^k\right) \log (\cos (\alpha ) \text{sech}(x)+1)}{2 k!}dx \\ = \frac{1}{2\pi i}\int_{0}^{\infty}\int_{C}a^w w^{-k-1} \cos (w x) \log (\cos (\alpha ) \text{sech}(x)+1)dwdx \\ = \frac{1}{2\pi i}\int_{C}\int_{0}^{\infty}a^w w^{-k-1} \cos (w x) \log (\cos (\alpha ) \text{sech}(x)+1)dxdw \\ = \frac{1}{2\pi i}\int_{C}\pi a^w w^{-k-2} \cosh \left(\frac{\pi w}{2}\right) \text{csch}(\pi w)dw -\frac{1}{2\pi i}\int_{C}\pi a^w w^{-k-2} \text{csch}(\pi w) \cosh (\alpha w)dw \end{equation}$

(2.2)

from Eq (1.7.7.120) in ^[1] and the integral is valid for $\alpha$ , $a$ , and $k$ complex and $|\Re(\alpha)| < \pi$ .

3. Infinite sums of the contour integral

In this section we will again use the generalized Cauchy's integral formula to derive equivalent contour integrals. First we replace $y$ by $y-\pi/2$ for the first equation and $y$ by $y+\pi/2$ for second then add these two equations to get

$\begin{equation} \frac{\left(y-\frac{\pi }{2}\right)^k+\left(y+\frac{\pi }{2}\right)^k}{k!} = \frac{1}{2\pi i}\int_{C}2 w^{-k-1} e^{w y} \cosh \left(\frac{\pi w}{2}\right)dw \end{equation}$

(3.1)

Next we replace $y$ by $\log(a)+\pi(2p+1)$ then we take the infinite sum over $p\in[0, \infty)$ to get

$\begin{equation} \sum\limits_{p = 0}^{\infty}\frac{2 \pi \left(\left(\log (a)+\pi (2 p+1)-\frac{\pi }{2}\right)^k+\left(\log (a)+\pi (2 p+1)+\frac{\pi }{2}\right)^k\;\;\;\right)}{k!} \\ = \frac{1}{2\pi i}\sum\limits_{p = 0}^{\infty}\int_{C}4 \pi w^{-k-1} \cosh \left(\frac{\pi w}{2}\right)e^{w (\log (a)+\pi (2 p+1))}\;\;dw \\ = \frac{1}{2\pi i}\int_{C}\sum\limits_{p = 0}^{\infty}4 \pi w^{-k-1} \cosh \left(\frac{\pi w}{2}\right)e^{w (\log (a)+\pi (2 p+1))}\;\;dw \end{equation}$

(3.2)

where $\Re(w) < 0$ according to (1.232.3) in ^[5]. Then we simplify the left-hand side to get the Hurwitz zeta function

$\begin{equation} -\frac{2^{k+1} \pi ^{k+2} }{(k+1)!}\left(\zeta \left(-k-1,\frac{2 \log (a)+\pi }{4 \pi }\right)+\zeta \left(-k-1,\frac{2\log (a)+3\pi}{4 \pi }\right)\right) \\ = \frac{1}{2\pi i}\int_{C}\pi a^w w^{-k-2} \cosh\left(\frac{\pi w}{2}\right) \text{csch}(\pi w)dw \end{equation}$

(3.3)

Then following the procedure of (3.1) and (3.2) we replace $y$ by $y+\alpha$ and $y-\alpha$ to get the second equation for the contour integral given by

$\begin{equation} \frac{(y-\alpha )^k+(\alpha +y)^k}{k!} = \frac{1}{2\pi i}\int_{C}2 w^{-k-1} e^{w y} \cosh (\alpha w)dw \end{equation}$

(3.4)

next we replace $y$ by $\log(a)+\pi(2p+1)$ and take the infinite sum over $p\in[0, \infty)$ to get

$\begin{equation} \sum\limits_{p = 0}^{\infty}\frac{(-\alpha +\log (a)+\pi (2 p+1))^k+(\alpha +\log (a)+\pi (2 p+1))^k}{k!} \\ = \frac{1}{2\pi i}\sum\limits_{p = 0}^{\infty}\int_{C}2 w^{-k-1} \cosh (\alpha w) e^{w (\log (a)+\pi (2 p+1))}\;\;dw \\ = \frac{1}{2\pi i}\int_{C}\sum\limits_{p = 0}^{\infty}2 w^{-k-1} \cosh (\alpha w) e^{w (\log (a)+\pi (2 p+1))}\;\;dw \end{equation}$

(3.5)

Then we simplify to get

$\begin{equation} \frac{2^{k+1} \pi ^{k+2} }{(k+1)!}\left(\zeta \left(-k-1,\frac{-\alpha +\log (a)+\pi }{2 \pi }\right)+\zeta \left(-k-1,\frac{\alpha +\log (a)+\pi }{2 \pi }\right)\right) \\ = \frac{1}{2\pi i}\int_{C}\pi \left(-a^w\right)w^{-k-2} \text{csch}(\pi w) \cosh (\alpha w)dw \end{equation}$

(3.6)

4. Definite integral in terms of the Hurwitz zeta function

Since the right-hand sides of Eqs (2.2), (3.3) and (3.5) are equivalent we can equate the left-hand sides to get

$\begin{equation} \int_{0}^{\infty}\left((\log (a)-i x)^k+(\log (a)+i x)^k\right) \log (\cos (\alpha ) \text{sech}(x)+1)dx \\ = 2 \left(\frac{2^{k+1} \pi ^{k+2} \left(\zeta \left(-k-1,\frac{-\alpha +\log (a)+\pi }{2 \pi }\right)+\zeta\left(-k-1,\frac{\alpha +\log (a)+\pi }{2 \pi }\right)\;\;\right)}{k+1}\;\;\;\;\right) \\ -2\left(\frac{2^{k+1} \pi ^{k+2} \left(\zeta \left(-k-1,\frac{2 \log (a)+\pi }{4 \pi }\right)+\zeta \left(-k-1,\frac{2\log (a)+3\pi}{4 \pi}\right)\;\;\right)}{k+1}\;\;\;\;\right) \end{equation}$

(4.1)

from (9.521) in ^[5] where $\zeta(z, q)$ is the Hurwitz zeta function. Note the left-hand side of Eq (4.1) converges for all finite $k$ . The integral in Eq (4.1) can be used as an alternative method to evaluating the Hurwitz zeta function. The Hurwitz zeta function has a series representation given by

$\begin{equation} \zeta(z,q) = \sum\limits_{n = 0}^{\infty}\frac{1}{(q+n)^z} \end{equation}$

(4.2)

where $\Re(z) > 1, q \neq 0, -1, ..$ and is continued analytically by (9.541.1) in ^[5] where $z = 1$ is the only singular point.

5. Derivation of special cases

In this section we have evaluated integrals and extended the range of the parameters over which the integrals are valid. The aim of this section is to derive a few integrals in ^[5] in terms of the Lerch function. We also present errata for one of the integrals and faster converging closed form solutions.

5.1. Derivation of entry 3.531.7 in ^[5]

Using Eq (4.1) and taking the first partial derivative with respect to $\alpha$ and setting $a = 1$ and simplifying the left-hand side we get

$\begin{equation} \int_{0}^{\infty}\frac{x^k}{\cos (\alpha )+\cosh (x)}dx = 2^k \pi ^{k+1} \csc (\alpha ) \sec \left(\frac{\pi k}{2}\right) \left(\zeta \left(-k,\frac{\pi -\alpha }{2 \pi }\right)-\zeta \left(-k,\frac{\alpha +\pi }{2 \pi }\right)\right) \end{equation}$

(5.1)

from Eq (7.102) in ^[3].

5.2. Derivation of entry 4.371.2 in ^[5]

Using Eq (5.1) and taking the first partial derivative with respect to $k$ and setting $k = 0$ and simplifying the left-hand side we get

$\begin{equation} \int_{0}^{\infty}\frac{\log (x)}{\cos (\alpha )+\cosh (x)}dx = \csc (\alpha ) \left(\alpha \log (2 \pi )-\pi \log (-\alpha -\pi )+\pi \log (\alpha -\pi )-\pi \text{log$\Gamma $}\left(-\frac{\alpha +\pi }{2 \pi }\right) \\ +\pi \text{log$\Gamma $}\left(-\frac{\pi-\alpha }{2\pi }\right)\right)) \\ = \csc (\alpha ) \left(\alpha \log (2 \pi )+\pi \log \left(\frac{\Gamma \left(\frac{\alpha +\pi }{2 \pi }\right)}{\Gamma \left(\frac{\pi -\alpha }{2 \pi }\right)}\right)\right) \end{equation}$

(5.2)

from (7.105) in ^[3].

5.3. Derivation of entry 4.371.1 in ^[5]

Using Eq (5.2) and setting $\alpha = \pi/2$ and simplifying we get

$\begin{equation} \int_{0}^{\infty}\log (x) \text{sech}(x)dx = \pi \log \left(\frac{\sqrt{2 \pi } \Gamma \left(\frac{3}{4}\right)}{\Gamma \left(\frac{1}{4}\right)}\right) \end{equation}$

(5.3)

5.4. Derivation of entry 4.371.3 in ^[5]

Using Eq (5.2) and taking the first derivative with respect to $\alpha$ and setting $\alpha = \pi/2$ and simplifying we get

$\begin{equation} \int_{0}^{\infty}\log (x) \text{sech}^2(x)dx = \log \left(\frac{\pi }{4}\right)-\gamma \end{equation}$

(5.4)

where $\gamma$ is Euler's constant.

5.5. Derivation of entry 4.376.1 in ^[5]

Using Eq (5.1) and taking the first partial derivative with respect to $k$ then setting $k = -1/2$ and $\alpha = \pi/2$ and simplifying we get

$\begin{equation} \int_{0}^{\infty}\frac{\log (x) \text{sech}(x)}{\sqrt{x}}dx = \frac{1}{2} \sqrt{\pi } \left(-2 \zeta'\left(\frac{1}{2},\frac{1}{4}\right)+2 \zeta'\left(\frac{1}{2},\frac{3}{4}\right)+\left(\zeta \left(\frac{1}{2},\frac{3}{4}\right)-\zeta \left(\frac{1}{2},\frac{1}{4}\right)\right) \left(\pi +\log \left(\frac{1}{4 \pi ^2}\right)\right)\right) \end{equation}$

(5.5)

The expression in ^[4] is correct but converges much slower than Eq (5.5).

5.6. Derivation of Eq (8) Table 257 in ^[4]

Using Eq (5.1) and taking the first partial derivative with respect to $\alpha$ we get

$\begin{equation} \int_{0}^{\infty}\frac{x^k}{(\cos (\alpha )+\cosh (x))^2}\;\;dx = -2^{k-1} \pi ^k \csc ^2(\alpha ) \sec \left(\frac{\pi k}{2}\right) \left(k \left(\zeta \left(1-k,\frac{\pi -\alpha }{2 \pi }\right)+\zeta \left(1-k,\frac{\alpha +\pi }{2 \pi }\right)\right)\right) \\ -2^{k-1} \pi ^k \csc ^2(\alpha ) \sec \left(\frac{\pi k}{2}\right) \left(2 \pi \cot (\alpha ) \left(\zeta \left(-k,\frac{\pi -\alpha }{2 \pi }\right)-\zeta \left(-k,\frac{\alpha +\pi }{2 \pi }\right)\right)\right) \end{equation}$

(5.6)

from (7.102) in ^[3]. Next we use L'Hopital's rule and take the limit as $\alpha \to 0$ to get

$\begin{equation} \int_{0}^{\infty}\frac{x^k}{(\cosh (x)+1)^2}\;\;dx = -\frac{1}{3} 2^{1-k} \left(\left(2^k-8\right) \zeta (k-2)-\left(2^k-2\right) \zeta (k)\right) \Gamma (k+1) \end{equation}$

(5.7)

Then we take the first partial derivative with respect to $k$ to get

$\begin{equation} \int_{0}^{\infty}\frac{x^k \log (x)}{(\cosh (x)+1)^2}\;\;dx = \frac{1}{3} 2^{4-k} k \Gamma (k) \zeta '(k-2)-\frac{2}{3} k \Gamma (k) \zeta '(k-2) -\frac{1}{3} 2^{2-k} k \Gamma (k) \zeta '(k) \\ +\frac{2}{3} k \Gamma (k)\zeta '(k)-\frac{1}{3} 2^{1-k} k \log (256) \zeta (k-2) \Gamma (k)+\frac{1}{3} 2^{1-k} k \log (4) \zeta (k) \Gamma (k) \\ +\frac{1}{3} 2^{4-k} k \zeta (k-2) \Gamma (k) \psi ^{(0)}(k+1) -\frac{2}{3} k\zeta (k-2) \Gamma (k) \psi ^{(0)}(k+1) \\ -\frac{1}{3} 2^{2-k} k \zeta (k) \Gamma (k) \psi ^{(0)}(k+1) +\frac{2}{3} k \zeta (k) \Gamma (k) \psi ^{(0)}(k+1) \end{equation}$

(5.8)

Finally we set $k = 0$ to get

$\begin{equation} \int_{0}^{\infty}\frac{\log (x)}{(\cosh (x)+1)^2}\;\;dx = \frac{1}{3} \left(14 \zeta '(-2)-\gamma +\log \left(\frac{\pi }{2}\right)\right) \end{equation}$

(5.9)

The integral listed in ^[4] appears with an error in the integrand.

5.7. Derivation of entry 3.522.6 in ^[5]

Using Eq (4.1) we first take the limit as $k\to-1$ by applying L'Hopital's rule and using (7.105) in ^[3] and simplifying the right-hand side we get

$\int_0^\infty {\frac{{\log (\cos (\alpha ){\rm{sech}}(x) + 1)}}{{{a^2} + {x^2}}}\;\;} dx = \frac{\pi }{a}\log \left( {\frac{{\sqrt \pi {2^{\frac{1}{2} - \frac{a}{\pi }}}\;\;\Gamma \left( {\frac{a}{\pi } + \frac{1}{2}} \right)}}{{\Gamma \left( {\frac{{a - \alpha + \pi }}{{2\pi }}\;\;} \right)\Gamma \left( {\frac{{a + \alpha + \pi }}{{2\pi }}} \right)}}} \right)$

(5.10)

Next we take the first partial derivative with respect to $\alpha$ and set $\alpha = 0$ to get

$\int_0^\infty {\frac{{{\rm{sech}}(x)}}{{{a^2} + {x^2}}}} \;dx = - \frac{{{\psi ^{(0)}}\left( {\frac{{2a + \pi }}{{4\pi }}\;\;} \right) - {\psi ^{(0)}}\left( {\frac{{2a + 3\pi \;\;}}{{4\pi }}} \right)}}{{2a}}$

(5.11)

from Eq (8.360.1) in ^[5] where $\Re(a) > 0$ , next we replace $x$ with $bx$ to get

$\begin{equation} \int_{0}^{\infty}\frac{\text{sech}(b x)}{a^2+b^2 x^2}dx = -\frac{\psi ^{(0)}\left(\frac{2 a+\pi }{4 \pi }\right)-\psi ^{(0)}\left(\frac{2a+3\pi}{4 \pi }\right)}{2 a b} \end{equation}$

(5.12)

Next we set $a = b = \pi$ to get

$\begin{equation} \int_{0}^{\infty}\frac{\text{sech}(\pi x)}{x^2+1}dx = \frac{1}{2} \left(\psi ^{(0)}\left(\frac{5}{4}\right)-\psi ^{(0)}\left(\frac{3}{4}\right)\right) = 2-\frac{\pi }{2} \end{equation}$

(5.13)

from Eq (8.363.8) in ^[5].

5.8. Derivation of entry 3.522.8 in ^[5]

Using Eq (5.12) and setting $a = b = \pi/2$ we get

$\begin{equation} \int_{0}^{\infty}\frac{\text{sech}\left(\frac{\pi x}{2}\right)}{x^2+1}dx = \frac{1}{2} \left(-\gamma -\psi ^{(0)}\left(\frac{1}{2}\right)\right) = \log (2) \end{equation}$

(5.14)

from Eq (8.363.8) in ^[5].

5.9. Derivation of entry 3.522.10 in ^[5]

Using Eq (5.12) and setting $a = b = \pi/4$ we get

$\begin{equation} \int_{0}^{\infty}\frac{\text{sech}\left(\frac{\pi x}{4}\right)}{x^2+1}dx = \frac{1}{2} \left(\psi ^{(0)}\left(\frac{7}{8}\right)-\psi ^{(0)}\left(\frac{3}{8}\right)\right) = \frac{\pi -2 \coth ^{-1}\left(\sqrt{2}\right)}{\sqrt{2}} \end{equation}$

(5.15)

from Eq (8.363.8) in ^[5].

5.10. Derivation of a definite integral in terms of the trigamma function

Using Eq (5.10) and taking the second partial derivative with respect to $\alpha$ we get

$\begin{equation} \int_{0}^{\infty}\frac{\text{sech}^2(x)}{a^2+x^2}dx = \frac{\psi ^{(1)}\left(\frac{a}{\pi }+\frac{1}{2}\right)}{\pi a} \end{equation}$

(5.16)

from Eq (8.363.8) in ^[5] where $\Re(a) > 0$ .

6. Discussion

In this paper we were able to present errata and express our closed form solutions in terms of special functions and fundamental constants such $\pi$ , Euler's constant and $\log(2)$ . The use of the trigamma function is quite often necessary in statistical problems involving beta or gamma distributions. This work provides both an accurate and extended range for the solutions of the integrals derived.

7. Conclusion

We have presented a novel method for deriving some interesting definite integrals by Bierens de Haan using contour integration. The results presented were numerically verified for both real and imaginary and complex values of the parameters in the integrals using Mathematica by Wolfram.

Acknowledgments

This research is supported by Natural Sciences and Engineering Research Council of Canada NSERC Canada under Grant 504070.

Conflict of interest

The authors declare there are no conflicts of interest.

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AIMS Energy

Numerical study and optimization of GO/ZnO based perovskite solar cell using SCAPS

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Abstract

1. Introduction

2. Definite integral of the contour integral

3. Infinite sums of the contour integral

4. Definite integral in terms of the Hurwitz zeta function

5. Derivation of special cases

5.1. Derivation of entry 3.531.7 in ^[5]

5.2. Derivation of entry 4.371.2 in ^[5]

5.3. Derivation of entry 4.371.1 in ^[5]

5.4. Derivation of entry 4.371.3 in ^[5]

5.5. Derivation of entry 4.376.1 in ^[5]

5.6. Derivation of Eq (8) Table 257 in ^[4]

5.7. Derivation of entry 3.522.6 in ^[5]

5.8. Derivation of entry 3.522.8 in ^[5]

5.9. Derivation of entry 3.522.10 in ^[5]

5.10. Derivation of a definite integral in terms of the trigamma function

6. Discussion

7. Conclusion

Acknowledgments

Conflict of interest

References

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AIMS Energy

Numerical study and optimization of GO/ZnO based perovskite solar cell using SCAPS

Related Papers:

Abstract

1. Introduction

2. Definite integral of the contour integral

3. Infinite sums of the contour integral

4. Definite integral in terms of the Hurwitz zeta function

5. Derivation of special cases

5.1. Derivation of entry 3.531.7 in [5]

5.2. Derivation of entry 4.371.2 in [5]

5.3. Derivation of entry 4.371.1 in [5]

5.4. Derivation of entry 4.371.3 in [5]

5.5. Derivation of entry 4.376.1 in [5]

5.6. Derivation of Eq (8) Table 257 in [4]

5.7. Derivation of entry 3.522.6 in [5]

5.8. Derivation of entry 3.522.8 in [5]

5.9. Derivation of entry 3.522.10 in [5]

5.10. Derivation of a definite integral in terms of the trigamma function

6. Discussion

7. Conclusion

Acknowledgments

Conflict of interest

References

Reader Comments

通讯作者: 陈斌, bchen63@163.com

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5.1. Derivation of entry 3.531.7 in ^[5]

5.2. Derivation of entry 4.371.2 in ^[5]

5.3. Derivation of entry 4.371.1 in ^[5]

5.4. Derivation of entry 4.371.3 in ^[5]

5.5. Derivation of entry 4.376.1 in ^[5]

5.6. Derivation of Eq (8) Table 257 in ^[4]

5.7. Derivation of entry 3.522.6 in ^[5]

5.8. Derivation of entry 3.522.8 in ^[5]

5.9. Derivation of entry 3.522.10 in ^[5]