Two forms of the integral representations of the Mittag-Leffler function

The integral representation of the two-parameter Mittag-Leffler function $E_{\rho,\mu}(z)$ is considered in the paper that expresses its value in terms of the contour integral. For this integral representation, the transition is made from integration over a complex variable to integration over real variables. It is shown that as a result of such a transition, the integral representation of the function $E_{\rho,\mu}(z)$ has two forms: the representation ``A'' and ``B''. Each of these representations has its advantages and drawbacks. In the paper, the corresponding theorems are formulated and proved, and the advantages and disadvantages of each of the obtained representations are discussed.


Introduction
The Mittag-Leffler function is an entire function defined by a power series This function was introduced by Mittag-Leffler in a number of works published from 1902 to 1905 in connection with his development of a method of summing divergent series. For more detailed information on the content of these works and on the history of the introduction of the Mittag-Leffler function, we refer the reader to the book [1] (see chapter 2 in [1]). The function itself E ρ (z) was introduced in the work [2]. In the paper [3] the integral representation for this function was obtained that expresses its value in terms of the contour integral. An important generalization of the Mittag-Leffler function E ρ,µ (z) = ∞ k=0 z k Γ(µ + k/ρ) , ρ > 0, µ ∈ C, z ∈ C was obtained by A. Wiman in 1905 [4,5] and was developed in the works [6][7][8]. The expression (1) is customary called the two-parameter Mittag-Leffler function. As we can see, the two-parameter Mittag-Leffler function E ρ,µ (z) is connected with the classical Mittag-Leffler function E ρ (z) with a simple relation E ρ,1 (z) = E ρ (z). In this paper integral representations of the function E ρ,µ (z) will be obtained and investigated. The integral representation of the Mittag-Leffler function is important from the point of view of its practical use, as well as for studying the asymptotic properties and zeros of this function. The integral representation expressed through the contour integral is used for these purposes. Several such representations are known for the Mittag-Leffler function. One of the earliest integral representations of the function E ρ,µ (z) was given in the book [9] (see §18.1, formula (20)). Further development of the issue of the integral representation of the Mittag-Leffler function and the study of its asymptotic properties was carried out in the works of M.M. Djrbashian. In the work [10] the integral representation was obtained that expressed the Mittag-Leffler function through the contour integral. Later it was included in his monograph [11] (see chapter 2, §2, lemma 3.2.1). Using this integral representation, asymptotic formulas and the distribution of the zeros of the Mittag-Leffler function were obtained. Further, the results of the work [10,11] were used in the books of [1,12], as well as in the works of [13][14][15][16][17] to develop the methods and algorithms of calculating the Mittag-Leffler function. However, despite a wide use of the integral representation for the Mittag-Leffler function that was obtained in the works of [10,11], it turned out that there was a mistake in that representation. This fact was pointed out in the work of [18]. In this regard, there is an issue of obtaining the correct integral representation for the Mittag-Leffler function.
One of the possible solutions to this issue was given in the works of [18,19]. In these works, the representation of the Mittag-Leffler function was obtained that expresses its value through the contour integral. As it was noted earlier, this representation is used to study the asymptotic properties of the Mittag-Leffler function. However, for practical use and for calculating the value of the function, it is convenient to have integral representations expressing the function in terms of the integrals of real variables. This paper is devoted to obtaining such integral representations for the Mittag-Leffler function. The starting point of the solution to this problem is the integral representation of the function E ρ,µ (z) obtained in the work [18]. Running a little ahead, we will say that the transition from integration over a complex variable to integration over real variables leads to the appearance of two forms of the integral representation of the Mittag-Leffler function. The first form will be abbreviated as the representation "A", the second as the representation "B". The representation "A" is a direct consequence of the integral representation obtained in the work [18]. It is obtained as a result of the transition from the contour integral to integration over real variables. To obtain the representation "B" in addition to performing such a transition, it is necessary to carry out a terminal transition ε → 0. This limits the possible values of the parameter µ in which this integral representation is true. As a result, both the representation "A" and the representation "B" have its advantages and drawbacks which will be discussed in detail in the paper.

Integral representation "A"
The purpose of this paper is to obtain integral representations for the function E ρ,µ (z) expressing this function in terms of integrals over real variables. It is convenient to use the representations of such a kind in practical problems as well as for calculating the values of the Mittag-Leffler function.
The proof of this theorem can be found in the work [18]. The form of the contour of integration γ ζ on the complex plane ζ is given in Fig. 1. The cut of the complex plane goes along the positive part of a real axis. The contour of integration consists of the The contour of integration γ ζ . The region that corresponds to the condition (2) is shaded in grey half-line S 1 , the arc of the circle C ǫ radius 1 + ǫ and the half-line S 2 . The contour γ ζ is traversed in a positive direction. The parameters δ 1ρ and δ 2ρ have the meaning of inclination angles of the half-lines S 1 and S 2 in relation to the contour axis γ ζ . In Fig. 1 the contour axis γ ζ coincides with a real axis. The values of the angles δ 1ρ and δ 2ρ are measured from the negative part of the real axis. The reference directions of these angles in Fig. 1 are shown by arrows.
We will obtain the integral representations that of interest to us by going from integration over the complex variable ζ to integration over variables r and ϕ interconnected by the relation ζ = re iϕ . As a result, the following theorem is true for the function E ρ,µ (z).
We consider now the integral I Cǫ . Getting rid of the complexity in the denominator we get where Using (16) and (17) in (15) we obtain the representation (7). It is important to pay attention that in the process of proving no additional limitations on the values of parameters ρ, µ, δ 1ρ , δ 2ρ and the argument z were imposed and it means that the ranges of admissible values pass from theorem 1 without change. Thus, the representation (7) is valid for any real ρ > 1/2, any real δ 1ρ , δ 2ρ satisfying the conditions (5), any complex µ and any complex z satisfying the condition (6).
The theorem is proved.
The proved theorem formulates an integral representation for the Mittag-Leffler function that expresses this function in terms of the sum of improper and definite integrals. To be definite, we will call this integral representation of the Mittag-Leffler function the representation "A". As we can see, the representation "A" is a direct consequence of the representation (3). It is obtained by passing from the contour integral to integrals over the real variable. Moreover, the improper integral in (7) corresponds to the sum of integrals along the half-lines S 1 and S 2 of the contour γ ζ and the definite integral corresponds to the integral along the arc of a circle C ǫ . It should be noted that this integral is taken along the arc of a circle of radius 1 + ǫ, where ǫ > 0. The representation (7) is valid for arbitrary values ρ > 1/2, any µ and any δ 1ρ and δ 2ρ that satisfy the condition (5).
Now substituting this result in (25) we obtain (19). Since in the proof process no additional restrictions on the values of parameters ρ, µ and on the value arg z were imposed, then the conditions for these parameters go from theorem 2 without change. Thus, we come to the conditions of the corollary.
2) We consider the case δ 1ρ = δ 2ρ = π/ρ. As we can see, the case considered is a particular case of the previous one. It follows from (5) that this case can be implemented if ρ 1. For the range of values arg z from (23) we get − π 2ρ + π < arg z < π 2ρ + π.
Now we consider the representation (18). In the case under consideration it will be written in the form We denote K ρ,µ (r, z) ≡ K ρ,µ r, π ρ , z .
The corollary is proved.
From the proved corollary it follows that if the parameter values δ 1ρ and δ 2ρ coincide, then in this the kernel function K ρ,µ (r, ϕ 1 , ϕ 2 , z) is significantly simplified. Recall that the parameters δ 1ρ and δ 2ρ are inclination angles of half-lines S 1 and S 2 in the contour γ ζ relative to the axis of this contour (see Fig. 1). Since in the theorem 1 the axis of the contour γ ζ coincides with the real axis, then the selection of δ 1ρ = δ 2ρ means that half-lines S 1 and S 2 run symmetrically in relation to the real axis. The kernel function K ρ,µ (r, ϕ 1 , ϕ 2 , z) takes the simplest form in the case δ 1ρ = δ 2ρ = π/ρ. Further, it is necessary for us to know the position of singular points of the integrand of the representation (3). This issue was studied in the work [20]. For completeness of the statement here we give the result obtained in the work [20] and formulate it in the form of a lemma. Lemma 1. For any real ρ > 1/2 and any complex values of the parameter µ = µ R +iµ I the integrand of the representation (3) The proof of this lemma can be found in the work [20]. We will make the following remark to corollary 1. Remark 1. In corollary 1 the special case under consideration δ 1ρ = δ 2ρ = π/ρ. We will assume that ρ = 1 and study the behavior of the formula (20) in this case. As a result, we obtain Using (21) for K 1,µ (z) we obtain From here it is clear, if µ = n, where n = 0, ±1, ±2, ±3, . . . , then sin(π(1 − µ)) = 0. Consequently, Thus, with integer values of µ the first summand in (28) becomes zero and to calculate the value of E 1,n (z) it remains to calculate the second integral. To calculate this integral, numerical methods can be used. However, in this case, this integral can be calculated analytically using the residue theory.
In fact, we return to the integral representation formulated in theorem 1. Recall that we consider the case ρ = 1. Using the notation (27) the representation (3) takes the form where the contour of integration γ ζ , defined by (4), is written in the form We represent the complex parameter µ in the form µ = µ R +iµ I and make use of lemma 1. According to this lemma, the function Φ 1,µ (ζ, z) at values µ I = 0 and µ R = 1 − m 1 , where m 1 = 0, 1, 2, 3, . . . has one singular point ζ = 1 which is a pole of the first order. The point ζ = 0, in this case, is the regular point. In case, if µ I = 0 and µ R = 1+ m 2 , where m 2 = 1, 2, 3, . . . the function Φ 1,µ (ζ, z) has two singular points: the point ζ = 1 is a pole of the first order and the point ζ = 0 is a pole of the order m 2 . As we can see, in both cases the point ζ = 0 is not a branch point. As a result, in these two cases, the function Φ 1,µ (ζ, z) is the entire function of a complex variable ζ. From here it follows that when µ I = 0, and µ R = n, where n = 0, ±1, ±2, ±3, . . . the arc of the circle C ǫ that enters the contour (31) is the closed circle of radius 1 + ǫ. The half-lines S 1 and S 2 pass along the positive part of a real axis in mutually opposite directions. With all other values of the parameter µ (when µ I = 0 or µ R = n), according to lemma 1, the point ζ = 0 is a branch point of the function Φ 1,µ (ζ, z). In this case, the circle C ǫ of the contour (31) will not close up and half-lines S 1 and S 2 will go along the upper and lower banks of the cut of the complex plane which runs along the positive part of a real axis.
It is clear from here that the result (29) is a consequence of lemma 1 In fact, in the case when the parameter µ takes integer real values, the first and second items of lemma 1 turn out to be true. As we have already pointed out, in this case the arc of the circle C ǫ of the contour (31) is a closed circle and the half-lines S 1 and S 2 run along the positive part of a real axis in mutually opposite directions. Consequently, the sum of the integrals along the half-lines S 1 and S 2 will be equal to zero. Next, it is necessary to recall that the improper integral in the expression (28) just corresponds to the sum of the integrals along the half-lines S 1 and S 2 . Therefore, with integer real values µ it should be equal to zero which has been obtained. A definite integral in (28) corresponds to integration along the closed circle. Therefore, one can use the theory of residues to calculate it.
The calculation of the integral in (30) using the theory of residues with integer real values of the parameter µ was carried out in the work [20]. For completeness of the statement, we will give the results obtained in this paper and formulate them in the form of a corollary to lemma 1.

Integral representation "B"
The integral representation "A" consists of the sum of two integrals. As it has been found earlier, the improper integral in (7) corresponds to the sum of integrals along the half-lines S 1 and S 2 of the contour γ ζ in the representation (3), a definite integral corresponds to the integral along the arc of the circle C ǫ . As a result, in analytical studies of the function E ρ,µ (z), as well as in the solution of problems where it is encountered, one should conduct studies of these two integrals. This causes certain difficulties. As it will be shown below in the representation (7) one can get rid of an integral on the arc of the circle C ǫ and write the integral representation for the function E ρ,µ (z) in the form of an improper integral. This representation will be much easier to use. However, as a result of such a transition, some restrictions are imposed on the parameter values µ. The integral representation of the Mittag-Leffler function represented in the following theorem will be called the representation "B".
Proof. The starting point of the proof is theorem 1 and the integral representation (3) which is defined in it. In view of the notation (13), the representation (3) will be written in the form The problem consists in calculating this contour integral. We consider an auxiliary integral where the contour Γ (see Fig. 2) consists of the arc of the circle C ǫ of radius 1 + ǫ with the center at the origin of coordinates, the segment Γ 2 , the arc of the circle C ε the radius ε with the center at the origin of coordinates and the segment Γ 1 , which are defined in the following way: The contour is traversed in the positive direction. The cut of the complex plane ζ goes along the positive part of a real axis.
As it follows from lemma 1 depending on a value of the parameter µ the integrand (38) has one or two poles. If ℜµ 2, then there are two poles in the points ζ = 0 and ζ = 1. If ℜµ 1, then there is one pole in the point ζ = 1. In addition, at non-integer values ρ and µ the point ζ = 0 is the branch point. In the case when 1 2 < ρ 1 and δ 1ρ = π the segment Γ 1 will pass through a singular point ζ = 1 (Fig. 2). The similar situation will be at 1 2 < ρ 1 and δ 2ρ = π. In this case the segment Γ 2 will pass through the point ζ = 1. These two cases will be considered separately. The case 1 2 < ρ 1, δ 1ρ = δ 2ρ = π also requires a separate consideration. In this case each of the segments Γ 1 and Γ 2 passes through a singular point ζ = 1. In view of the foregoing, it is necessary to consider four cases: Case 1. We consider the first case at the beginning. Directly calculating the integral I and substituting the variable of integration ζ = re iϕ we obtain Now we let ε → 0 in this expression and we will study the behavior of I Cε , I Γ 1 and I Γ 2 . We consider the integral I Cε at the beginning. For this integral the relation is true We consider the integrand. To get rid of the complexity in the denominator we multiply and divide the integrand by εe −iϕ − 1. We also represent z = |z|e i arg z and µ = µ R + iµ I , and for a power function we will use the representation ξ a = exp{a ln ξ}, ξ > 0. As a result, we obtain For the numerator we have the estimate From this it follows that lim Now we consider the behavior of the integral I Γ 1 at ε → 0. We have the estimate Consequently, it is necessary to study the behavior of the integrand at r → 0. Similar to the previous case we get rid of the complexity in the denominator and represent z = |z|e i arg z , µ = µ R + µ I and denote ϕ 1 = −δ 1ρ − π. For a power function we will make use of the representation ξ a = exp{a ln ξ}.
To parametrize the arc of the circle C ′ 1ε 1 we consider the mapping u = ζ − 1 of the complex plane ζ on the complex plane u. This mapping is a conformal mapping and is a left shift of the entire complex plane ζ by the value 1. As a result of such a shift the point ζ = 1 in the plane ζ is mapped onto the point u = 0 of the plane u. Thus, the circle C ′ 1ε 1 with the center in the point ζ = 1 and the radius ε 1 < 1 of the plane ζ will be mapped on the circle C u 0ε 1 with the center in the point u = 0 and the radius ε 1 on the plane u. Thus, to parametrize the arc of the circle C ′ 1ε 1 in the plane ζ it is enough to parametrize the arc of the circle C u 0ε 1 in the complex plane u and then to map the complex plane u on to the plane ζ with the help of an inverse conformal mapping ζ = u + 1.
Here it is necessary to note that arctan(x) is a multivalued function. The principal branch of this function takes values in the interval [−π/2, π/2]. However, as one can see from the definition of the auxiliary contour Γ ′ , the center of the arc of the circle C ′ 1ε 1 lies in the point ζ ′ = 1 in which arg ζ ′ = −2π. Therefore, it is necessary to choose the required branch in arctan(x) in such a way that the mapping (49) could map an arc of a circle C u 0ε 1 of the plane u on to an arc of a circle C ′ 1ε 1 of the plane ζ with the center in the point ζ ′ = 1 and arg ζ ′ = −2π. As a result, the mapping (49) will take the form |ζ| = τ 2 + 2τ cos ψ + 1, arg ζ = arctan τ sin ψ τ cos ψ + 1 + kπ.
For the arc of the circle C ′ 1ε 1 we obtain k = −2. Here it should be pointed out that these formulas produce the mapping of the circle C u 0ε 1 on to the circle C ′ 1ε 1 only in the case τ 1. If τ > 1, then these formulas will not be true.
Next, we consider an auxiliary integral By calculating this integral we have One should pay attention to the fact that inside the region limited by the contour Γ ′ the integrand φ ρ,µ (ζ, z)/(ζ − 1) of the integral I ′ is an analytical function of the complex variable ζ. Consequently, according to the Cauchy integral theorem From here we get that Thus, it is possible to substitute integration on the arc of the circle C ǫ by integration on the remaining parts of the contour Γ ′ . We consider each integral in the right part of this expression. The integral I Cε has already been considered by us earlier in this lemma. Using (42) we obtain that in the case considered lim ε→0 I Cε = 0, µ R < 1 + 1/ρ.
Thus, for the integral I ′ AB we get Now consider the integral I ′ CD . Representing the complex number ζ in the form ζ = re iϕ and using (51) we get Comparing this expression with (55) we notice that the integrand of the integral obtained is similar to the integrand of the integral (55). Therefore, using (56) we get that the integrand can be represented in the form (57). Thus, for the integral I ′ CD we obtain Similarly, using (51) for the integral I ′ EF we obtain Using now (56) and (57) the integral I ′ EF takes the form In the sum (52) it remains to consider the integral I ′ . This integral is taken along the arc of the circle C ′ 1ε 1 with the center in the point ζ = 1 (at that arg ζ = −2π) and radius ε 1 . The contour is traversed in the direction from the point D to the point E (see Fig. 3). As it was shown earlier the arc of the circle C ′ 1ε 1 in the plane ζ can be given in the form (see (51)) As one can see, the arc traverse in the direction from the point D to the point E corresponds to a change of the parameter ψ from −π to −2π. Thus, in the case under consideration the contour C ′ 1ε 1 on the complex plane ζ can be represented in the form where −π ψ −2π, 0 < ε 1 < 1 and r(τ, ψ) = τ 2 + 2τ cos ψ + 1, ϕ(τ, ψ, k) = arctan τ sin ψ τ cos ψ+1 + kπ Representing now in the integral I ′ the complex number ζ in the form (62) we find One should pay attention that traversing along the arc of the circle C ′ 1ε 1 in the direction from the point D to the point E (see Fig. 3) corresponds to the negative traversing direction along the contour. Here the value of the parameter ψ = −π, corresponds to the point D and the value of the parameter ψ = −2π to the point E. That is why in the integral derived one must transpose the limits of integration.
As a result the integral (64) takes the form Now we get back to the expression (53) and let ε → 0 in this expression. It is necessary to point out that the integrals I Cǫ , I ′ C ′ 1ε 1 and I ′ EF do not depend on ε and, consequently, they will not change with such a passage to the limit. In view of this, we have Taking into account (54) and using the expressions (58), (59), (60) and (67) we get

From this we obtain
Now we get back to the Mittag-Leffler function. Assuming that δ 1ρ = π the contour of integration γ ζ in (37) takes the form By calculating (37) directly and representing a complex number ζ in the form ζ = re iϕ we obtain using the notation (56) this expression can be written in the form where the form of the functions K ′ ρ,µ (r, δ, z) is defined by (57). Now we will make use of the representation here (68) for the integral I Cǫ . As a result, we obtain where ℜµ < 1 + 1/ρ. It remains to consider how to change the condition (2) in this case. Since, in the case considered 1 2 < ρ 1, δ 1ρ = π, π/(2ρ) < δ 2ρ π, then the condition (2) takes the form π 2ρ − δ 2ρ + π < arg z < − π 2ρ + 2π. Thus, we have obtained the statement of the lemma for the second case.
As one can see, only the arc of the circle C ′′ 1ε 1 remains non-parametric. For its parametrization we will fulfill the procedure as in the previous case. Considering the conformal mapping u = ζ − 1 and representing the complex number u in the form u = τ e iψ we get that the arc of the circle C ′′ 1ε 1 can be represented in the form (50), where the parameter ψ varies within the limits from −π to 0. Note that the value of the parameter ψ = −π corresponds to the point C and the value ψ = 0 to the point B. It should be pointed out that the center of a circle lies in the point ζ ′′ = 1 with arg ζ ′′ = 0. Consequently, in the formulas (50) one should choose the principal branch arctan(x), i.e. to take k = 0. As a result, an arc of a circle C ′′ 1ε 1 can be represented in the form arg ζ = arctan τ sin ψ τ cos ψ+1 + kπ, |ζ| = τ 2 + 2τ cos ψ + 1, 0 ψ −π, τ = ε 1 , k = 0.
(69) Fig. 4. The auxiliary contour of integration Γ ′′ In view of the foregoing, the contour Γ ′′ is written in the form The contour is traversed in a positive direction.
We consider now an auxiliary integral In the similar way how it was done in the previous case, we calculate this integral. Using the definition of the contour Γ ′′ we have It is clear from this expression that inside the region limited by the contour Γ ′′ the integrand of the integral I ′′ is an analytical and continuous function of a variable ζ. Consequently, according to the Cauchy theorem I Cǫ + I ′′ AB + I ′′ From this we obtain that We consider now each of the integrals in the right part separately.
Now consider the integral I ′′ AB . Representing ζ in the form ζ = re iϕ we find From this it is clear that this integral is similar to the integral (55). That is why, one can use the study results of this integral. Using (57) we obtain In the same way, using (70), (56) and (57) for the integrals I ′′ CD and I ′′ EF we find It remains to consider the integral I C ′′ 1ε 1 . This integral is taken along the arc of the C ′′ 1ε 1 with the center in the point ζ ′′ = 1, where arg ζ ′′ = 0. This means that the contour is traversed from the point B of the complex plane ζ in which arg ζ B = 0 to the point C in which arg ζ C = 0. In other words, the starting and ending points of this contour have the same value of the argument. As it was shown above, the arc of a circle C ′′ 1ε 1 can be represented in the form (69). Traversing this arc in the direction from the point B to the point C corresponds to a change of the parameter ψ from 0 to −π. Thus, the points of the arc of the circle C ′′ 1ε 1 can be represented in the form where r(τ, ψ) and ϕ(τ, ψ, k) have the form (63). Using now the representation (75), in the integral I C ′′ 1ε 1 we obtain φ ρ,µ r(τ, ψ)e iϕ(τ,ψ,0) , z iτ e iψ r(τ, ψ)e iϕ(τ,ψ,0) − 1 dψ τ =ε 1 = −iε 1 0 −π φ ρ,µ r(ε 1 , ψ)e iϕ(ε 1 ,ψ,0) , z e iψ r(τ, ψ)e iϕ(τ,ψ,0) − 1 dψ.
Comparing this expression with (64) it is clear that these two integrals are similar. Therefore, using (65) and (66) for the integral I ′′ We return to the expression (71) and let ε → 0 in this expression. Note that the integrals I Cǫ , I ′′ AB , do not depend on ε and, consequently, with such a passage to the limit they will not change. As a result we obtain Using here the expressions (72), (73), (74), (76) we obtain 1+ǫ ε K ′ ρ,µ (r, π, z)dr, ℜµ < 1 + 1/ρ.
The contour is traversed in a positive direction. As a result, the contour Γ ′′′ is written in the form arg ζ = arctan τ sin ψ τ cos ψ+1 + kπ, |ζ| = τ 2 + 2τ cos ψ + 1, −π ψ −2π, τ = ε 1 , k = −2, Next, we consider an auxiliary integral where φ ρ,µ (ζ, z) is defined by (13). By calculating this integral directly we obtain We immediately note that inside the region bounded by the contour Γ ′′′ , the function φ ρ,µ (ζ, z)/(ζ − 1) is an analytical function of the complex variable ζ. Consequently, according to the Cauchy integral theorem, From this it directly follows that Next, we calculate each integral on the right-hand side of this expression. However, these integrals were already calculated by us earlier. The integral I ′′′ AB corresponds to the integral I ′′ AB . Therefore, using (73) we obtain The integral I ′′′ CD corresponds to the integral I ′′ CD . Using (74) we obtain The integral I ′′′ The integral I Cε was considered by us when dealing with case 1. Using (42) for the case considered we obtain lim ε→0 I Cε = 0, ℜµ < 1 + 1/ρ.
The integral I ′′′ EF corresponds to the integral I ′ CD . Using (59), we obtain The integral I ′′′ GH corresponds to the integral I ′ EF . Therefore, using (60) we obtain The integral I ′′′ . Using (67) we obtain We get back to the expression (78) and let ε → 0 in it. Note that the integrals I ′′′ AB , I ′′′ , I ′′′ GH do not depend on ε. Therefore, with such a passage to the limit, they do not change. Taking into account the above expressions for the integrals, the sum (78) takes the form Performing the passage to the limit in this expression ε → 0 we obtain Now we consider the Mittag-Leffler (37). Taking into account that in the studied case δ 1ρ = δ 2ρ = π we get where the contour γ ζ takes the form and the condition (2) is written in the form π 2ρ < arg z < − π 2ρ + 2π. Directly calculating this integral and representing the complex number ζ in the form ζ = re iϕ we have , z re i(−π−π) − 1 e i(−π−π) dr + I Cǫ + ∞ 1+ǫ φ ρ,µ re i(π−π) , z re i(π−π) − 1 e i(π−π) dr. Now using here (56) we obtain that the expression may be represented in the form Now here we make use of the representation (79) for the integral I Cǫ . As a result, we obtain In the expression derived there is the difference of K ′ ρ,µ (r, π, z) − K ′ ρ,µ (r, −π, z). We transform this difference. For this we use the representation (57) for the function K ′ ρ,µ (r, δ, z). As a result, we obtain exp (zr) ρ e −iπρ cos(ρδ) (zre −iπ ) ρ(1−µ) r 2 + 2r cos δ + 1 e iη(r,δ,z) r + e iδ − e iη(r,−δ,z) r + e −iδ .
Here the function K ρ,µ (r, δ, z) was obtained by us earlier and has the form (19). As a result, the expression (81) takes the form The derived expression proves the lemma completely.
The theorem is proved.
The proved theorem shows that in the representation (3) at the values of parameters ρ, δ 1ρ , δ 2ρ satisfying the conditions (32) the integral over the arc of the circle C ǫ of the contour γ ζ can be replaced with integration with respect to the segments Γ 1 and Γ 2 of the contour (39). Such a transition makes it possible to replace the contour integral with one improper integral over the real variable of the complex-valued function, which simplifies the use and study of the Mittag-Leffler function. The integral representation of the Mittag-Leffler function obtained in this theorem will be called the integral representation "B".
Consider in detail the differences between two derived representations. The integral representation "A" (see (7)) of the Mittag-Leffler function consists of two summands. The first summand corresponds to the sum of integrals along the half-lines S 1 and S 2 and the second summand is the integral along the arc of a circle C ǫ . Such a representation is not convenient in analytical studies of the Mittag-Leffler function since one has to deal with two integrals. The representation "B" turns out to be more convenient and it consists of one improper integral (see (33)). However, this convenience is fraught with the appearance of constraints imposed on values of the parameters of the function E ρ,µ (z). As it was shown in theorem 3, the representation "B" is true only at values ℜµ < 1+1/ρ. If ℜµ 1+1/ρ, then it is necessary to use the representation "A" which is the main integral representation for the Mittag-Leffler function. It should also be noted that in the representation "B" with the parameter values 1/2 < ρ 1 and values δ 1ρ = π or δ 2ρ = π the segments Γ 1 or Γ 2 of the auxiliary contour Γ (see Fig. 2) pass through the pole ζ = 1. This leads to the need to deform the contour of integration so as to bypass the pole. As a result, in integral representations (34), (35), (36) there are summands describing the pole bypass along the arcs of the circle C ′ 1ε 1 and C ′′ 1ε 1 . In addition, one has to split the integrals along the half-lines Γ 1 + S 1 and Γ 2 + S 2 into parts to exclude the sections corresponding to the integrals along the arcs of the circle C ′ 1ε 1 and C ′′ 1ε 1 . All this leads to the appearance of additional terms in the corresponding integral representations. It should also be noted that at parameter values 1/2 < ρ 1, δ 1ρ = δ 2ρ = π in the representation (36) the integral 1−ε 1 0 K ρ,µ (r, π, z)dr corresponds to the sum of integrals along the segments CD and EF the contour of integration Γ ′′′ (see Fig. 5) the integral ∞ 1+ε 1 K ρ,µ (r, π, z)dr corresponds to the sum of integrals along AB + S 2 and GH + S 1 . As one can see the specified sections of the integration contour go along the positive part of a real axis. As it was shown in the lemma 1 at values of parameters 1/2 < ρ < 1 or ρ = 1 and ℑµ = 0 or ρ = 1, ℑµ = 0 and ℜµ is not an integer, the point ζ = 0 is a branch point of the integrand (37). Consequently, the indicated sections of the integration contour will go along different sides of the cut of the complex plane ζ. As a result, 1−ε 1 0 K ρ,µ (r, π, z)dr = 0 and ∞ 1+ε 1 K ρ,µ (r, π, z)dr = 0. In case, if ρ = 1, ℑµ = 0 and ℜµ is an integer, lemma 1 shows that in this case the point ζ = 0 is a regular point. Consequently, the segments CD and EF , as well as the half-lines AB + S 2 and GH + S 1 will go along one straight line of the complex plane, but in different directions, and arcs of the circles C ′ 1ε 1 and C ′′ 1ε 1 will close. As a result, we obtain 1−ε 1 0 K 1,µ (r, π, z)dr = 0 and ∞ 1+ε 1 K 1,µ (r, π, z)dr = 0, and the sum of integrals 0 −π P ′ 1,µ (ε 1 , ψ, 0, z)dψ + −π −2π P ′ 1,µ (ε 1 , ψ, −2, z)dψ corresponds to the integral over the closed contour and will be equal to the residue in the point ζ = 1. Thus, we come to the condition of Corollary 2.
The formulas of the integral representation obtained in theorem 3 are rather lengthy. They can be simplified and reduced to a simpler form. The representation "B" takes the simplest form in the case when δ 1ρ and δ 2ρ coincide, i.e. δ 1ρ = δ 2ρ = δ ρ . We will formulate the result obtained in the form of a corollary.
The first part of the corollary is proved.
2) Consider the case δ ρ = π/ρ. Note immediately that the value of an angle δ ρ cannot exceed π. Consequently, this case is implemented only at values ρ 1. However, the value ρ = 1 should be excluded from consideration. In fact, at ρ = 1 the segments Γ 1 and Γ 2 of the auxiliary contour Γ (see Fig. 2) pass through the pole ζ = 1. Consequently, it is necessary to deform the contour Γ so as to bypass this pole. However, this has already been done by us in item 4 of theorem 3. By putting the value δ ρ = π/ρ in the representation (83) we obtain E ρ,µ (z) = ∞ 0 K ρ,µ (r, π/ρ, z)dr.
The corollary is proved.

Conclusion
It has been shown in the paper that when passing from the integral representation formulated in theorem 1 to integration over real variables, the integral representation of the Mittag-Leffler function can be written in two forms: the representation "A" and "B". The integral representation "A" was given in theorem 2 and the representation "B" in theorem 3. Each of these representations has its own advantages and drawbacks. The representation "A" is true for any complex µ and at values 1/2 < ρ 1 there is no need to bypass a pole in the point ζ = 1. This excludes the necessity to consider particular cases and greatly simplifies the presentation itself. These facts are the advantages of the representation "A". The disadvantages of this representation include the fact that it consists of the sum of two integrals: improper and definite. This leads to certain difficulties when working with this representation, since one has to study the behavior of these two integrals. The representation "B" is valid only for the parameter values µ satisfying the condition ℜµ < 1 + 1/ρ. Particular cases also arise at values 1/2 < ρ 1 in which one has to bypass a singular point ζ = 1. That is why, in the problems in which there is a necessity to investigate these particular cases this form of representation is not very convenient. These facts are the disadvantage of the representation "B". In all other cases the representation "B" consists of one improper integral and turns out to be more convenient in use than the representation "A". This is the advantage of the representation "B".
The forms of the representations "A" and "B" obtained in theorems 2 and 3 are given for the case of four parameters ρ, µ, δ 1ρ , δ 2ρ . The parameters ρ and µ are referred to the Mittag-Leffler function directly, and parameters δ 1ρ and δ 2ρ describe the integration contour. As a result, in the general case, the integral representations "A" and "B" of the Mittag-Leffler function turn out to be four-parameter. For definiteness, we call this four-parameter description parametrization 1. However, in parametrization 1 though the representations "A" and "B" of the function E ρ,µ (z) have a more general form but they are awkward enough. The representations "A" and "B" take a simpler form if δ 1ρ = δ 2ρ = δ ρ . In this case the integral representations "A" and "B" can be described in three parameters: ρ, µ, δ ρ . We call this parameterization parametrization 2. An even simpler view of the form "A" and "B" is adopted if δ 1ρ = δ 2ρ = π/ρ. In this case integral representations of the Mittag-Leffler function can be described in two parameters ρ and µ. We call this case of parametrization -parametrization 3. Thus, the integral representations of the Mittag-Leffler function in the form "A" and parameterizations 2 and 3 are given in Corollary 1, and the representation "B" in parametrizations 2 and 3 in Corollary 3.
Taking into account of the geometric meaning of the parameters δ 1ρ and δ 2ρ one can give the geometric interpretation of three introduced parametrizations of the representations "A" and "B". In fact, the parameters δ 1ρ and δ 2ρ describe an inclination angle of half-lines S 1 and S 2 in the contour γ ζ (see Fig. 1). Thus, in case if the half-lines S 1 and S 2 independently lie in the range of angles π 2ρ < δ 1ρ min π, π ρ , π 2ρ < δ 2ρ min π, π ρ , then we have parametrization 1. If these half-lines lie symmetrically in relation to the positive part of a real axis, then we obtain parametrizations 2 and 3. With this, parametrization 3 corresponds to the angle of inclination δ 1ρ = δ 2ρ = π/ρ.
In conclusion it should be pointed out that the representations for the function E ρ,µ (z), formulated in theorems 2 and 3 are valid for the vaues arg z satisfying the condition π 2ρ −δ 2ρ +π < arg z < − π 2ρ +δ 1ρ +π. This constraint appears as a result of the use of the proof of theorem 1 of the integral representation for the gamma function obtained in the work [18] (see Appendix in [18]). The presence of this constraint somewhat narrows the possibilities of using the obtained integral representations of the function E ρ,µ (z). Nevertheless, it is possible to get rid of this constraint and expand the range of admissible values arg z to the entire complex plane. However, this requires additional studies that are beyond the scope of this paper.