Mellin-Barnes Integrals

Appendix A Mellin-Barnes Integrals

In analogy to integer-order calculus, fractional calculus also presents classes of functions that can be called special. In solving a fractional differential equation with constant coefficients, the classical Mittag-Leffler function and its particular cases emerge naturally. The most general special function appearing in problems of fractional calculus is the Fox’s H-function, which is defined in terms of the Mellin-Barnes integral. Here we present Mellin-Barnes integrals associated with Fox’s H-function and some particulares cases, among them, the classical hypergeometric function.

In this appendix we will present the called Mellin-Barnes integrals [1, 2], because of their importance in the inversion of the Laplace and Mellin transforms. The par- ticularity of these integrals lies in the fact that they are integral in the complex plane and, therefore, the integration contour plays a fundamental rule [3]. First, before presenting the deﬁnition of the Mellin-Barnes integral, it should be pointed out that Pincherle [4], in 1888, therefore, before Mellin [2] and Barnes [1], obtained the following formula

m (x − ρi ) = (t) = i 1 ext dx m−1 γ (x − σi ) i=1 whose convergence was proved using an asymptotic formula for the gamma function. We note the similarity of this expression with the inverse Laplace transform, since by defining the quotient involving the gamma functions, by f (x, ρi , σi ) for example, we have exactly the integral, in the complex plane, that recovers the function through the inverse Laplace transform. Thus, this integral can be considered as the first example in the literature of what is now known as the Mellin-Barnes integral [5]. In order to make explicit the calculations, we will discuss, through examples, involving the Mittag-Leffler function with three parameters and the classical hyper-

© Springer Nature Switzerland AG 2019 309 E. Capelas de Oliveira, Solved Exercises in Fractional Calculus, Studies in Systems, Decision and Control 240, https://doi.org/10.1007/978-3-030-20524-9 310 Appendix A: Mellin-Barnes Integrals geometric function, how to present the function as an integral in the complex plane and vice versa, that is, from the integral in complex plane, to recover the particular function.

Definition A.1 (Mellin-Barnes integral)Letγ be a contour in the complex plane starting at c − i∞ and ending at c + i∞, with Re(s) = c > 0. It is called Mellin- Barnes integral to any integral in the complex plane whose integrand contemplates at least one gamma function, given by 1 I (z) = f (s) z−s ds (A.1) 2πi γ where the density function, f (s), in general, solution of a differential equation with polynomial coefficients, is given by a quotient of products of gamma functions de- pending on parameters. The choice of the contour γ, as we have already mentioned, plays a crucial rule, and will be explained through examples, involving a Mittag- Leffler function with three parameters, the hypergeometric function as well as, the confluent hypergeometric function, because with these functions we can discuss the case of the called ‘empty product’, expressing these functions as a Meijer’s G- function or a Fox’s H-function.

Moreover, we will direct this appendix to be able to use it, when appropriate, in the inversion of integrals, in particular, involving the Meijer and Fox functions. Therefore, in this sense, we will consider the density function in the following form

A(s)B(s) f (s) = C(s)D(s) with A(s), B(s), C(s) and D(s) are products of gamma functions, as in Sect. 2.4.3.

Example A.1 (Mittag-Lefﬂer function with three parameters) Let α ∈ R+; β, ρ ∈ C, β = 0 and Re(ρ)>0. Show that the integral 1 (s)(ρ − s) ≡ (−z)−s ds 2πi L (β − αs) for a convenient integration contour, L, in the complex plane, is a less than adequate gamma function, a Mittag-Lefﬂer function with three parameters. We consider |arg(z)| < π and the contour L,asinFig.A.1, starting at c − i∞ going to c + i∞, being 0 < c < Re(ρ), separating all the poles s =−k, with k = 0, 1, 2,... to the left and all the poles s = ρ + n, with n = 0, 1, 2,... to the right (do not contribute to the residue theorem) relatively to the line Re(s) = c. To compute the integral in the complex plane, we close the integration contour so that only the poles at s =−k, with k = 0, 1, 2,...contribute, and we use the residue theorem, that is, the sum of the residues at the poles s = 0, −1, −2,..., from where follows Appendix A: Mellin-Barnes Integrals 311

Fig. A.1 Contour L for Mittag-Lefﬂer with three parameters

∞ (s)(ρ − s) = lim (s + k) (−z)−s s→−k (β − αs) k=0 that, using the fact that the limit of the product is equal to the product of the limits,

∞ (k + ρ) = (−z)k lim (s + k)(s)· (αk + β) s→−k k=0

Note that we have an indetermination of type 0 ·∞. In order to raise this indetermination, we will calculate the remaining limit separately,

(s + k − 1) ···s(s) (s + k + 1) lim (s + k)(s) = lim (s + k) = lim s→−k s→−k (s + k − 1) ···s s→−k (s + k − 1) ···s

( ) (− )k = 1 = 1 = 1 · (−1)(−2) ···(−k) (−1)k (1 · 2 ···k) k!

This result is the residue of the gamma function at the poles s =−k with k = 0, 1, 2,... Another way to obtain this result is through the called Mittag-Lefﬂer expansion, according to Exercises (2) and (3) from Chap. 2. 312 Appendix A: Mellin-Barnes Integrals

Thus, returning with this result in the expression for , we obtain

∞ ∞ (−1)k (ρ + k) (ρ + k) zk = (−z)k = · k! (β + αk) (β + αk) k! k=0 k=0

Using the deﬁnition of the Pochhammer symbol, we can write

∞ (ρ) k = (ρ) k z (αk + β) k! k=0 which can be identiﬁed with a Mittag-Lefﬂer function with three parameters, that is,

ρ = (ρ)Eα,β(z) from which it follows that the integral is the product of a gamma function by the Mittag-Lefﬂer function with three parameters ( )(ρ − ) 1 s s −s ρ (−z) ds = (ρ)Eα,β(z) (A.2) 2πi L (β − αs) which is the desired result

Example A.2 (Particular cases) Since the Mittag-Leffler function with three parameters contains as particular cases Mittag-Leffler functions with two and one parameter, we will express such functions in terms of an integral in the complex plane. Using the previous result, taking ρ = 1inEq.(A.2) we obtain ( )( − ) 1 s 1 s −s (−z) ds = Eα,β(z) 2πi L (β − αs) an integral representation for the Mittag-Leffler function with two parameters. Also, taking ρ = 1 = β in Eq. (A.2) we obtain ( )( − ) 1 s 1 s −s (−z) ds = Eα(z) 2πi L (1 − αs) an integral representation for the classical Mittag-Leffler function. Finally, for ρ = 1 = β = α in Eq. (A.2) we obtain 1 (s)(−z)−s ds = ez 2πi L an integral representation for the exponential function. The last three expressions can be interpreted as inverse Mellin transforms Appendix A: Mellin-Barnes Integrals 313

Example A.3 (Mittag-Lefﬂer function as a Fox’s H-function) Here we will address the inverse problem, that is, we have the integral in the complex plane and we want to determine the function, that is, to calculate the integral explicitly.

Let us show that this function can not be written in terms of a Meijer function, which, as particular cases, contains all classical hypergeometric functions. Just to mention, the particular Mittag-Leffler function that can be written in terms of a confluent hypergeometric function, yes, can be written as a Meijer function. In order to write the Mittag-Leffler function with three parameters, let us first identify its representation in terms of a Mellin-Barnes integral, Eq. (A.2), with the Mellin-Barnes integral defining Fox’s H-function, that is,

m n (b j + β j s) (1 − a j − α j s) 1 i=1 j=1 (−z)s ds · π q p 2 i L (1 − b j − β j s) (a j + α j s) i=m+1 j=n+1

In order to identify, we start with the numerator, that is, the integrand numerator in Eq. (A.2), has two gamma functions, where we consider m = 1 = n while in the denominator we have only one gamma function, then we consider q = 2 and p = 1, the smallest that does not contribute, that is, it will generate an ‘empty product’, since the initial index begins with j = n + 1 and, like n = 1, the beginning will be j = 2, so it is equal to unity. Let’s explain these products individually, identifying them as follows:

1 (b j + β j s) = (b1 + β1s) numerator, j=1 1 (1 − a j − α j s) = (1 − a1 − α1s) numerator, j=1 2 (1 − b j − β j s) = (1 − b2 − β2s) denominator, j=2 1 (a j + α j s) = 1(not contributes) denominator. j=2

Using the integrand in Eq. (A.2) and identifying with the products, ( )(ρ − ) ( + β )( − − α ) s s = b1 1s 1 a1 1s (β − αs) (1 − b2 − β2s) 314 Appendix A: Mellin-Barnes Integrals we have a1 = 1 − ρ, α1 = 1, b1 = 0, β1 = 1, b2 = 1 − β, β2 = α and z →−z. Iden- tifying the symbols for the Fox’s H-function, we have ( , α ) ( − ρ, ) 1,1 − a1 1 = 1,1 − 1 1 H1,2 z H1,2 z (b1, β1), (b2, β2) (0, 1), (1 − β, α) which, ﬁnally, allows us to write

ρ , (1 − ρ, 1) (ρ)E (z) = H 1 1 −z α,β 1,2 (0, 1), (1 − β, α) which is the desired result Example A.4 (A particular case. Meijer’s G-function) From the above expression it is clear that Fox’s H-function can not be reduced to a Meijer’s function, except if α = ρ (·) 1, a particular Mittag-Lefﬂer function E1,β , that is,

ρ , 1 − ρ (ρ)E (z) = G1 1 −z 1,β 1,2 0, 1 − β · 1,1 · with G1,2 · is a Meijer’s G-function. Since this particular Mittag-Lefﬂer function with three parameters is related to a conﬂuent hypergeometric function, through the expression

(β) ρ ( ) = (ρ; β; ) E1,β z 1 F1 z we can write this conﬂuent hypergeometric function in terms of a Meijer’s G- function,

, 1 − ρ F (ρ; β; x) = G1 1 −z 1 1 1,2 0, 1 − β which is the desired result Example A.5 (Hypergeometric function) Here, we star with the calculation of the Mellin transform, denoted by M , of the classical hypergeometric function 2 F1(a, b; c;−x), that is

∞ s−1 ≡ M [2 F1(a, b; c;−x)]= x 2 F1(a, b; c;−x) dx · 0

By introducing the integral representation for the hypergeometric function, given by Eq. (2.10) and changing the integration orders, we can write (c) 1 ∞ xs−1 = tb−1( − t)c−b−1 x t· 1 a d d (b)(c − b) 0 0 (1 + xt) Appendix A: Mellin-Barnes Integrals 315

To calculate the integral in the variable x, we introduce the change of variable xt = ξ from where we obtain, only for the integral in x ∞ xs−1 ∞ ξ s−1 x = t−s ξ· a d a d 0 (1 + xt) 0 (1 + ξ)

In order to calculate the integral in the variable ξ, we introduce another change of variable, η, given by ξ = η + ξη, from where it follows to the integral, already simplifying ∞ ξ s−1 1 t−s ξ = t−s ηs−1( − η)−s+a−1 η· a d 1 d 0 (1 + ξ) 0

The remaining integral is nothing more than a beta function, according to Eq. (2.2), then ∞ xs−1 (s)(a − s) x = t−s B(s, a − s) = t−s a d 0 (1 + xt) (a) where, in the last step, we use the relation between the gamma and beta functions. Returning in the expression to we can write (c) (s)(a − s) 1 = tb−s−1(1 − t)c−b−1 dt (b)(c − b) (a) 0 which, also, is another beta function. Proceeding as in the previous one and expressing the result in terms of gamma functions, we obtain, already simplifying ( ) ( )( − )( − ) = c s a s b s · (a)(b) (c − s)

Returning with the inverse we obtain the Mellin-Barnes integral ( )( ) i∞ ( )( − )( − ) a b 1 s a s b s −s 2 F1(a, b; c; z) = (−z) ds (A.3) (c) 2πi −i∞ (c − s) which is also an integral representation for the classical hypergeometric function. In order to show, from the previous integral representation, the inverse process, that is, from the integral representation, to obtain the hypergeometric function, we proceed as in the case of the Mittag-Lefﬂer function with three parameters. We explain only the steps, and then write this function as a particular case of the Meijer’s G-function. Moreover, just to remember, we can also write it in terms of a Fox’s H-function, but, as we have already mentioned, Meijer’s G-function is a particular case of Fox’s H-function, so, we write only in terms of Meijer’s G- function. 316 Appendix A: Mellin-Barnes Integrals

Then we return the integral given by Eq. (A.3) and outline the steps, in analogy to the case of the Mittag-Lefﬂer function with three parameters. First the contour, where we admit arg(−z)<π and separating the poles from the (a − s) and (b − s) those of function (s), as well as considering (−z)−s with its principal value. Then we consider the poles of (s) such that −s = k = 0, 1, 2,..., so we must calculate the limit ∞ (−s + a)(−s + b) lim (s + k) (s) (−z)−s · s→−k (−s + c) k=0

Even more, analogous to the previous one, we have (−1)k lim (k + s)(s) = s→−k k! from where follows the result for the integral (a)(b) F (a, b; c; z), (c) 2 1 which is the desired result Example A.6 (Hypergeometric function as a Meijer’s G-function) Consider the following integral in the complex plane 1 i∞ (−s)(a + s)(b + s) = (−z)s ds · (A.4) 2πi −i∞ (c + s)

We recall that, as in all gamma functions in the integrand of the previous one, the variable s is not multiplied by any coefﬁcient, we can directly identify with the Meijer’s G-function, being the contour as described in the previous example. Then we must compare the integrants, given in with that of the Meijer’s G-function, given by m n (bi − s) · (1 − a j + s) i=1 j=1 G m,n(s) = · p,q q p (1 − b j + s) · (a j − s) i=m+1 j=n+1

Then, since in the denominator we only have a gamma function, we take m = 1 and q = 2 and p < n + 1 so that the product is ‘empty’, identiﬁed with the unit. Let n = 2 (we still have two more gamma functions in the numerator) and p = 2. Let’s explain these products individually, identifying them as follows:

1 (b j − s) = (b1 − s) numerator, i=1 Appendix A: Mellin-Barnes Integrals 317

2 (1 − a j + s) = (1 − a1 + s)(1 − a2 + s) numerator, j=1 2 (1 − b j + s) = (1 − b2 + s) denominator, i=2 2 (a j − s) = 1(not contributes) denominator. j=3

Using the integrand in Eq. (A.4) and identifying with the products, (− )( + )( + ) ( − )( − + )( − + ) s a s b s = b1 s 1 a1 s 1 a2 s (c + s) (1 − b2 + s) we have a1 = 1 − a, a2 = 1 − b, b1 = 0 and b2 = 1 − c. Identifying with the symbols for the Meijer’s G-function, we have , − , − 1,2 − a1 a2 = 1,2 − 1 a 1 b G2,2 z G2,2 z b1, b2 0, 1 − c which ﬁnally allows us to write ( )( ) a b 1,2 1 − a, 1 − b 2 F1(a, b; c; z) = G , −z (c) 2 2 0, 1 − c which is the desired result Example A.7 (Hypergeometric function. A particular case) Knowing that the relation is worth 1 1 1 (1 + x2)−1/2 = F , ; ;−x2 2 1 2 2 2 express this function in terms of the Meijer’s G-function. Using the previous result, we can write, already simplifying

− / 1 , 1/2, 1/2 (1 + x2) 1 2 = √ G1 2 x2 · π 2,2 0, 1/2

This expression can be simpliﬁed. To this end, we use Exercise (30) from Chap.2, then − / 1 , 1/2 (1 + x2) 1 2 = √ G1 2 x2 π 2,2 0 which is the desired result. 318 Appendix A: Mellin-Barnes Integrals

References

1. Barnes, E.W.: A new development in the theory of the hypergeometric functions. Proc. Lond. Math. Soc. 6, 141–177 (1908) 2. Mellin, H.: Om deﬁnita integraler. Acta Societatis Scientiarum Fennicae 20, 1–39 (1895) 3. Costa, F.S., Vaz Jr., J., Capelas de Oliveira, E., Figueiredo Camargo, R.: Mellin-Barnes integrals and the Fox function. Tend. Mat. Apl. Comput. 12, 157–169 (2011) 4. Pincherle, S.: On generalized hypergeometric functions. Atti R. Accad. Lincei Rend. Cl. Sci. Fis. Natur. 4, 694–700 and 792–799 (1888). (in Italy) 5. Mainardi, F., Pagnini, G.: Salvatore Pincherle: the pioneer of the Mellin-Barnes integral. J. Comput. Appl. Math. 153, 331–342 (2003) Index

A Euler-Mascheroni constant, 165 Abel Euler relation, 188, 252 equation, 187, 234, 273 Exponential integral, 289 integral equation, 234

F B Fick law, 248 Bessel function, 32, 49, 64, 189, 208, 242 Fourier transform, 117, 128, 139, 151, 205, modified, 59, 242 279 second kind, 243, 245 composition, 131 zero order, 135, 159, 165, 184, 218 convolution, 118, 131, 142, 153 Beta function, 20, 45, 189, 247, 288 derivative, 119 Bromwich contour, 123, 167, 250 displacement, 131 modified, 155, 276 duality, 131 inverse, 118, 140, 153, 280 C properties, 118, 162 Caputo fractional derivative, 177, 182, 196, scale, 131 237, 239, 257, 269, 277, 289, 293 translation, 131 Christofell-Darboux formula, 74 Fox function, 28, 54, 109, 111, 227, 298, 310, 313 Fractional D Caputo derivative, 264, 269, 277 Diffusion equation, 78, 228, 281 derivative, 169, 259, 260 anomalous, 228 differential equation, 277, 279, 296 fractional, 281, 298 diffusion equation, 227, 281 standard, 78 Hadamard derivative, 176 Dirac delta function, 151, 278, 281 harmonic oscillator, 237 Dirichlet Hilfer derivative, 263 formula, 246 integral, 259, 260 problem, 134, 153 mass-spring, 239 partial differential equation, 279 Riemann-Liouville derivative, 263 E Function Elliptic integral admissible, 121 first kind, 225 exponential order, 120 second kind, 225 limited, 120 © Springer Nature Switzerland AG 2019 319 E. Capelas de Oliveira, Solved Exercises in Fractional Calculus, Studies in Systems, Decision and Control 240, https://doi.org/10.1007/978-3-030-20524-9 320 Index

G Jordan lemma, 104, 151 Gamma function, 19, 37, 98, 142, 228, 295, Juxtaposition of transforms, 279, 281 296 duplication formula, 39, 244, 245 incomplete, 20, 40, 187, 214 K integral representation, 19 Kronecker symbol, 63, 64 inverse, 19 Gaussian, 161, 269 Gegenbauer polynomials, 22 L Gel’fand-Shilov function, 18, 166, 204 Laguerre polynomials, 32, 76, 98, 108 Generalized geometric series, 90 Langevin Green function, 220, 229, 280, 281 equation, 267 Grünwald-Letnikov derivative, 171, 254, fractional equation, 267 261 Laplace transform, 119, 132, 151, 160, 181, 192, 232, 240, 250, 267, 275, 278, 281, 289, 293, 296 H bilateral, 119 Hadamard convergence region, 119 fractional derivative, 177, 190 convolution, 122, 166, 175, 238, 268, 295 integral, 176, 183, 190 derivative, 121 teorem, 265 existence, 121 Hadamard-Caputo derivative, 179 integral, 122 Hankel contour, 34, 57, 123, 250 inverse, 124, 134, 151, 192, 230, 232, Harmonic oscillator, 236, 239 240, 249, 250, 289, 293, 297 damped, 236, 239 Legendre simple, 236, 239 duplication formula, 30, 49, 104, 211, Heaviside function, 121, 165, 241, 256 244 Hermite polynomials, 64 functions, 184 Hilfer fractional derivative, 263 polynomials, 32, 47 Hypergeometric function, 21, 45, 211, 226, Leibniz rule, 186, 210, 265, 277 288, 314, 316 generalized, 293 conﬂuent, 23, 48, 63, 84, 142, 220, 285 integer order, 291 generalized, 24, 106 violation of the, 265 integral representation, 65, 315 Lerch theorem, 284 l’Hôpital rule, 88, 162, 285

I Integral M equation, 133, 237 Mainardi function, 72, 163 integer order, 174 Mass-spring sistem, 236, 239 derivative, 175 Maxwell model, 256 iterated, 258 Meijer function, 27, 54, 85, 227, 313, 316 non integer order, 174, 176 Mellin-Barnes integral, 36, 56, 111, 310, 313 derivative, 176 Mellin transform, 127, 133, 151, 161, 164, Integration order, 245, 272 183, 242, 298, 300, 314 Integrodifferential equation, 294 convolution, 130, 134, 164, 206 Inverse derivative, 129 Fourier transform, 118, 280 inverse, 128, 161, 245, 300 Laplace transform, 124, 280 particular function, 129 Mellin transform, 128, 242 properties, 129 Memory effect, 262 Miller-Ross function, 72, 108 J Mittag-Lefﬂer expansion, 38, 312 Jacobi function, 45 Mittag-Lefﬂer function, 70, 313 Index 321

classical, 30, 81, 102, 143, 154, 182, 231, Residue theorem, 104, 148, 243, 300, 310 236, 238, 250, 253, 269, 286, 294, 296 Riemann-Liouville duplication formula, 73, 109 derivative, 235, 264 ﬁve parameters, 110 fractional derivative, 177, 182, 273, 275, four parameters, 78, 102, 111 290 recurrence relation, 75 fractional integral, 200, 246, 290 six parameters, 100, 110 integral, 176, 183, 235 sums of, 285 Riemann zeta function, 134, 164 three parameters, 71, 76, 89, 97, 283, 310 RLC electric circuit, 241 two parameters, 71, 81, 132, 154, 232, Rodrigues formula, 31, 65, 184 238, 294, 296

S N Semigroup property, 202, 245, 246 Newton Simple law, 248 harmonic oscillator, 236 second law, 236 pendulum, 223 Nonconstant coefﬁcients, 275, 288, 289 Spectral function, 252, 253 Step function, 120

O Ordinary differential equation, 294, 297 T Tautochronous, 234 Theorem P ﬁnal value, 125 Parseval formula, 164 initial value, 125 Pochhammer symbol, 18, 37, 60, 75, 83, 225, Titchmarsh formula, 252 283 generalized, 76 Poisson integral, 187, 213 V Population growth, 235 Viscoelastic equation, 257 Volterra equation, 283 R integral equation, 166, 283 Rabotnov function, 72, 77, 108 Radioactive decay, 235 Reﬂection formula, 61, 66, 98, 173, 193, 245, W 261, 291 Wiman function, 70 Relaxation equation, 231 Wright function, 72, 232 radial, 226 generalized, 25