Security protocols analysis including various time parameters

Sabina Szymoniak; Sabina Szymoniak

doi:10.3934/mbe.2021061

Mathematical Biosciences and Engineering

2021, Volume 18, Issue 2: 1136-1153. doi: 10.3934/mbe.2021061

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Research article Special Issues

Security protocols analysis including various time parameters

Sabina Szymoniak ^,

Department of Computer Science, Czestochowa University of Technology, Dabrowskiego 69, 42-201 Czestochowa, Poland

Received: 28 September 2020 Accepted: 17 December 2020 Published: 11 January 2021

Communication is a key element of everyone's life. Nowadays, we most often use internet connections for communication. We use the network to communicate with our family, make purchases, we operate home appliances, and handle various matters related to public administration. Moreover, the Internet enables us to participate virtually in various official meetings or conferences. There is a risk of losing our sensitive data, as well as their use for malicious purposes. In this situation, each connection should be secured with an appropriate security protocol. These protocols also need to be regularly checked for vulnerability to attacks. In this article, we consider a tool that allows to analyze different executions of security protocols as well as simulate their operation. This research enables the gathering of knowledge about the behaviour of the security protocol, taking into account various time parameters (time of sending the message, time of generation confidential information, encryption and decryption times, delay in the network, lifetime) and various aspects of computer networks. The conducted research and analysis of the obtained results enable the evaluation of the protocol security and its vulnerability to attacks. Our approach also assumes the possibility of setting time limits that should be met during communication and operation of the security protocol. We show obtained results on Andrew and Needham Schroeder Public Key protocols examples.

Keywords:

Citation: Sabina Szymoniak. Security protocols analysis including various time parameters[J]. Mathematical Biosciences and Engineering, 2021, 18(2): 1136-1153. doi: 10.3934/mbe.2021061

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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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沈阳化工大学材料科学与工程学院沈阳 110142

Mathematical Biosciences and Engineering

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Mathematical Biosciences and Engineering

Security protocols analysis including various time parameters

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

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通讯作者: 陈斌, bchen63@163.com

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Mathematical Biosciences and Engineering

Security protocols analysis including various time parameters

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]