A New Representation of the Gamma Functions

In this article, we present a new representation of gamma function as a series of complex delta functions. We establish the convergence of this representation in the sense of distributions. It turns out that the gamma function can be defined over a space of complex test functions of slow growth denoted by Z. Some properties of the gamma function are discussed by using the properties of the delta function. The plan of this Chapter is as follows. After giving a necessary and brief introduction to the gamma function in Section 2, we discuss the distributional representation of gamma function in Section 3. The convergence of the series along with its properties is also discussed here. Some properties of gamma function w.r.t Fourier transformation are discussed in Section 4.

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The problem of giving s! a useful meaning when s is any complex number was solved by Euler (1707-1783), who defined what is now called the gamma function,
Γ(s):=∫_0^∞▒〖e^(-t) t^(s-1) dt〗  (s=σ+iτ,σ=R(s)>0). (1.1)
He extended the domain of gamma function from natural numbers, , to complex numbers . This is an extensively studied classical function and can be considered as a basic special function. It can be written as a sum of two incomplete gamma functions
Γ(s):=∫_0^1▒〖e^(-t) t^(s-1) dt〗 +∫_1^∞▒〖e^(-t) t^(s-1) dt〗 =γ(s,x)+Γ(s,x). (1.2)
The function Γ(s) itself, and its incomplete versions γ(s,x) and Γ(s,x), are known to play important role in the study of the analytic solutions of a variety of problems in diverse areas of science and engineering (see, for example, [1] [2] and [3]; see also the recent papers [4] [5] [6] and [7]). By using the Fourier transform representation, a distributional representation of the gamma function
By following the methodology of [1], we introduce
Γ_ε (z)=∫_0^∞▒〖t^(z-1) e^(-t) 〗 dt; zϵC
and for ϵ=1, we get the original gamma function. Next by following N. N. Lebedeve we write
∫_0^∞▒〖t^(z-1) e^(-t) 〗 dt=∫_0^1▒〖t^(z-1) e^(-t) 〗 dt+∫_1^∞▒〖t^(z-1) e^(-t) 〗 dt=P(z)+Q(z)
where it can be easily shown that the first integral is an analytic function of z while the second integral ∫_1^∞▒〖t^(z-1) e^(-t) 〗 dt is an entire function for zϵC;0<ϵ<1.
Now we replace the e^(-t) by its power series such as

P(z)=∑_(n=0)^∞▒(-1)^n/n! ∫_0^1▒〖t^n t〗^(z-1) dt=∑_(n=0)^∞▒(-1)^n/n! 1/(z+n)
where as usual it is permissible to reverse the order of integration and summation due to the uniform convergence of the integral. It shows that Γ(z) is an entire function with simple pole.

A new representation
Theorem 1. Gamma functions have the following representation.
█(Γ(z)=2π∑_(n=0)^∞▒(-)^n/n! δ(θ–i(ν+n)).)
Proof. Let us first replace t=e^x and z=ν+iθ in Equation (3) then we get:
█(Γ(z)=∫_(-∞)^∞▒〖e^(x(ν+iθ)) exp⁡(-e^x ) 〗 dx, ) (16)
Now writing the series form of the exponential function:
█(exp⁡(-e^x )=∑_(n=0)^∞▒(-e^x )^n/n!) (17)
And then collecting and expanding the exponential terms:
█(Γ(z)=∫_(-∞)^∞▒〖e^(x(ν+iθ)) ∑_(n=0)^∞▒(-e^x )^n/n!〗 dx, ) (18)
We get:
Γ(z)=∑_(n=0)^∞▒(-)^n/n! ∫_(-∞)^∞▒e^ixθ e^(ν+n)x dx. (19)
We can interchange the operations of integration and summation because of uniform convergence of the integral. By means of Equation (11), we find:
∫_(-∞)^∞▒e^iθx e^(ν+n)x dx=F [e^(ν+n)x;θ]=2πδ(θ–i(ν+n)) (20)
These Equalities (19–20) combined together produces the result (15) as mentioned above.
Remark 1. One can develop equivalent outcomes for supplementary related functions, in the form of following corollaries.

was obtained in [8]. This led to new integral identities involving the gamma function.

Corollary 6. Gamma functions have the following representation:
█(Γ(z)=2π∑_(n=0)^∞▒(-1)^n/n! δ(z+n).)
Proof. The result follows simply by replacing k=1 in Equation (23).
Corollary 7. Gamma functions have the following representation:
█(Γ(z)=2π∑_(n,r=0)^∞▒(-1)^n/n! n^r/r! δ^((r) ) (z).) (30)
Proof. The result follows simply by replacing k=1 in Equation (25).
Note that we have acquired a series in terms of Dirac delta functions that are only significant in the sense of distributions once defined as an inner product with some appropriate function. For a simple observation, multiply Equation (29) with 1/Γ(z) in a conventional way.
█(1=(2π∑_(n=0)^∞▒(-1)^n/n! δ(z+n) )/Γ(z) .) (31)
Note that, at z=-n, the singularities of delta function are canceled by the zeros of the reciprocal gamma function in the involved product lim┬(z→-n) δ(z+n)/Γ(z) =lim┬(z→-n)⁡〖(1/(z+n))/(1/(z+n))〗=lim┬(z→-n)⁡〖(z+n)/(z+n)〗=1. Therefore, by using the basic definition of delta function:
█(δ(t)={█(∞ (t=0)@0 (t≠0),# )┤ ) (32)
In Equation (31), we obtain:
█(1={■(2π/e;[email protected];&z∈C∖\{-n})┤ ) (33)
That is an apparent inconsistency.
For the moment, if we think through the inner product
█(〈Γ(z),1/Γ(z) 〉=2π∑_(n=0)^∞▒(-1)^n/n! 〈δ(z+n),1/Γ(z) 〉 ) (34)
We get
█(∫_zϵC▒〖 1 〗 dz=2π∑_(n=0)^∞▒(-1)^n/n!Γ(-n) ) (35)
Since 1/Γ(-n) =0,n=0,1,2,… so we get from Equation (35)
█(∫_zϵC▒〖 1 〗 dz=0) (36)
█(∫_zϵC▒〖 1 〗 dz=∫_(-∞)^(+∞)▒1 dz=0) (37)
█(⟹∞=0) (38)
And again, an apparent inconsistency. So, we should be precise in choosing a space of functions to consider the convergence of this series representation.
3.2. Convergence and Applications of New Representation
The representation of thegamma functions Γ(z) is achieved as an infinite sum of delta functions that is definite if it converges in the distributional sense. Hence, it is interesting to prove that a series of delta functions are distributions (generalized functions) on the space Z as illustrated in the following theorem.
Theorem 2. Show that gamma function is defined as a distribution for space Z.
Proof. Consider the following combination for arbitrary〖 Λ_1 (z),Λ〗_2 (z)ϵZ and 〖c_1,c〗_2 ϵC.
〈Γ(z),c_1 Λ_1 (z)+c_2 Λ_2 (z)〉=〈2π∑_(n=0)^∞▒〖(-1)^n/n! δ(z+n),c_1 Λ_1 (z)+c_2 Λ_2 (z)〗〉. (39)
⟹〈Γ(z),c_1 Λ_1 (z)+c_2 Λ_2 (z)〉=c_1 〈Γ(z),Λ_1 (z)〉+c_2 〈Γ(z),Λ_2 (z)〉. (40)
Next, consider any sequence Z⊃{Λ_μ }_(μ=1)^∞⟶0 and the fact that δ(z) is continuous over Z we can assert that {〈δ(z+n),Λ_μ 〉 }_(μ=1)^∞→0.
⇒{〈Γ(z)),Λ_μ (z)〉 }_(μ=1)^∞=2π∑_(n=0)^∞▒〖(-1)^n/n!{〈δ(z+n),Λ_μ (z)〉 }_(μ=1)^∞ 〗 (41)
converges to zero. Hence, k-gamma function is a generalized function over Z in view of the convergence of series representation (15) as follows:
█(〈Γ(z),Λ(z)〉=2π∑_(n=0)^∞▒(-1)^n/n! 〈δ(z+n),Λ(z)〉; (∀Λ(z)ϵΖ) )
█(=2π∑_(n=0)^∞▒(-1)^n/n! Λ(-n),) (42)
Where we have used the shifting property of delta functions:
〈δ(z+n),Λ(z)〉=Λ(-n), (43)
Which being the elements of space Z are slowly increasing (bounded by a polynomial) test functions and note that sum over the coefficients is:
█(sum over the coefficients=∑_(n=0)^∞▒(-1)^n/[email protected] = exp⁡(-1).) (44)
This sum is finite and well defined. So from Equation (42), it is evident that 〈Γ(z),Λ(z)〉 being a product of a rapidly decreasing and slowly increasing function is convergent, for ∀Λ(z)ϵΖ. It can also be verified in view of well-known Abel theorem to check the behavior of series. It indicates the same statement for other specific cases as presented in Equations (21)–(29).
In this investigation, the convergence is shown for the functions of slow growth; nevertheless, it can be perceived that the series converges for a wider space of functions. Therefore, for ∀Λ(z)ϵΖ the investigated sum is finite, but if this sum is finite, then Λ(z) may be an element of some wider domain where delta function is defined.
Subsequently, by means of this acquired representation of Γ(z) we establish some integral identities and validate these identities by making use of standard Fourier transform in the ensuing section. Initially, we thought through a simple example of particular test functions
Λ(z)=τ^zξ (ξ>0;z∈C)
Remark 3. The gamma function has a classical representation [30, p. 2],
█(Γ(z)=2π∑_(n=0)^∞▒(-1)^n/n! 1/(z+n),)
Which demonstrates that gamma function has poles at negative integers z=-n;n=0,1,2,3,…. and the sum over the residues of these poles is ∑_(n=0)^∞▒(-1)^n/n. Now if we compare this representation with (29), then the result can be interpreted in the same way.

3. Distributional Representation of the Gamma Function and Some Properties
Theorem 1. Gamma function has a series representation
Proof. Consider Equation (1.2)
Now, by making use of the basic property of delta function given in Equation (2.3) and taking , we obtain
which leads to (3.1).
Theorem 2. Prove that (3.1) converges over the space of complex test functions denoted by .
Proof. The series can act over test functions as
⟨Γ(s),ϕ(s)⟩=2π∑_(n=0)^∞▒〖(-1)^n/n! ϕ(-n) 〗. (3.4)
We can note that
∑_(m=0)^∞▒(-1)^n/n!=1/e (3.5)
and ϕ(-n) is slowly increasing sequence of test functions. Therefore, the series (3.4) is convergent by using famous Abel convergent test or by ([13], Proposition 1, p. 46).
Remark 1. Note that ϕ∈Z implies that the series is convergent. Conversely if the series is convergent then ϕ may or may not belong to Z, therefore the condition is necessary and not sufficient.
Remark 2. Note that (3.1) generalizes the domain of (1.2) from real τ to complex s and then to ϕ(s).
Remark 3. Any sequence of complex delta functions {δ(s+n)}_(n=0)^∞, when multiplied and divided by "2π" /e gives us e/"2π" times gamma function
(2π∑_(n=0)^∞▒(-1)^n/n!)/(2π∑_(n=0)^∞▒(-1)^n/n!) δ(s+n)=e/"2π" Γ(s). (3.6)
Theorem 3. Prove that gamma function is a continuous linear functional over Z.
Proof. If ϕ_1 (s) and ϕ_2 (s) are test functions in Z and α and β are two complex numbers, then
⟨Γ(s),αϕ_1 (s)+βϕ_2 (s)⟩=⟨2π∑_(n=0)^∞ (-1)^n/n! δ(s+n),αϕ_1 (s)+βϕ_2 (s)⟩. (3.7)
Since complex delta is linear over , this implies that gamma function is also linear over . It gives us
⟨Γ(s),αϕ_1 (s)+βϕ_2 (s)⟩=α⟨Γ(s),ϕ_1 (s)⟩+β⟨Γ(s),ϕ_2 (s)⟩. (3.8)
If Γ(s) is known to be linear, definition of continuity will be wane to as “the numerical sequence {⟨Γ(s),ϕ_v (s)⟩}_(v=1)^∞ converges to zero, whenever the sequence converges to zero.” Let {ϕ_μ }_(μ=1)^∞ be a sequence of test functions in Z con- verges to zero. Then by the continuity of delta functional, the sequence
{⟨δ(s+n),ϕ_μ ⟩}_(μ=1)^∞ converges to zero, which implies that
{⟨Γ(s),ϕ_μ ⟩}_(μ=1)^∞=2π∑_(n=0)^∞ (-1)^n/n! {⟨δ(s+n),ϕ_μ ⟩}_(μ=1)^∞ (3.9)
converges to zero. For linear functionals continuity at any point is equivalent to continuity at every point. It proves that gamma function (3.1) is a continuous linear functional over .
Theorem 4. Prove that following properties hold for gamma function as a distribution.
1) Addition
⟨Γ(s),ϕ_1 (s)+ϕ_2 (s)⟩=⟨Γ(s),ϕ_1 (s)⟩+⟨Γ(s),ϕ_2 (s)⟩, (3.10)
2) Multiplication
⟨αΓ(s),ϕ(s)⟩=⟨Γ(s),αϕ(s)⟩, (3.11)
be any complex or real number.
3) Shifting or Translation
⟨Γ(s-τ),ϕ(s)⟩=⟨Γ(s),ϕ(s+τ)⟩. (3.12)
4) Multiplication of independent variable with a constant (positive or negative)
⟨Γ(as),ϕ(s)⟩=⟨Γ(s),1/a ϕ(s/a)⟩. (3.13)
5) Shifting including multiplication with independent variable
⟨Γ(as-τ),ϕ(s)⟩=⟨Γ(s),1/a ϕ(s/a+τ)⟩. (3.14)
6) The product of gamma function with a regular distribution ψ(s) is defined over Z.
7) The product ψ(s)Γ(s) is also continuous linear functional over Z.
Proof. Results (1-7) can easily be proved by using the properties of complex delta functions.
Theorem 5. Prove that distributional derivatives of gamma function also exist.
Proof. By using ⟨δ^((1) ) (s),ϕ(s)⟩=-ϕ^((1) ) (0), ([12], p.14, (12)), first derivative of (3.1) is
⟨Γ^((1) ) (s),ϕ(s)⟩=∑_(n=0)^∞ (-1)^n/n! (-1)^1 ϕ^1 (-n), (3.15)
which is convergent as discussed earlier. Similarly for kth order implies that
⟨Γ^k (s),ϕ(s)⟩=∑_(n=0)^∞ (-1)^n/n! (-1)^k ϕ^k (-n). (3.16)
Theorem 6. The recurrence relation for gamma function as a distribution holds true Γ(s+1)=sΓ(s) iff ϕ(s-1)=sϕ(s) where ϕ∈Z.
Proof. By using the Equation (3.10) and fixing α as a complex number”s” and using Equation (3.11), one can obtain the required result.
Theorem 7. The following result for the product of the gamma function holds true
Γ(a-s)Γ(s-b)=("2π" /e)^2 δ(a-b),∀ϕ(s)∈Z. (3.17)
Proof. Consider
=(2π∑_(n=0)^∞▒(-1)^n/n!)^2 ⟨δ(a-(s+n))δ((s+n)-b),ϕ(s)⟩ (3.18)
By making use of (2.4), we get
⟨Γ(a-s)Γ(s-b),ϕ(s)⟩=(2π∑_(n=0)^∞ (-1)^n/n!)^2 ⟨δ(a-b),ϕ(s)⟩ (3.19)
⟨Γ(a-s)Γ(s-b),ϕ(s)⟩=("2π" /e)^2 ⟨δ(a-b),ϕ(s)⟩. (3.20)
Hence proved.
4. Properties of Distributional Representation of Gamma Function w.r.t Fourier Transformation
Before stating the properties, we write the necessary notations used for this section. Here
F[Γ(s)]=Γ ̃(s); F[ϕ](s)=ϕ ̃(s). (4.1)
By using ([12], p. 184) gives
⟨Γ ̃(s),ϕ ̃(s)⟩=2π⟨Γ(t),ϕ(-t)⟩=2π⟨Γ(t),ϕ ̂(t)⟩, (4.3)
where ϕ ̂ denotes the transpose of test function ϕ.
here we define
Again, from parseval’s identity
⟨¯(Γ ̃(s) ),¯(ϕ ̃(s) )⟩=2π⟨Γ ̃(s),¯(¯(ϕ ̃(s) ))⟩=2π⟨Γ ̃(s),ϕ ̃(s)⟩ (4.6)
By applying double Fourier transform
⟨Γ ̃ ̃(s),ϕ(s)⟩=2π⟨Γ ̃(s),ϕ ̃(s)⟩=2π⟨Γ(t),ϕ(-t)⟩ (4.7)
i.e., double Fourier transform of a distribution is again a distribution. There will be a change of sign in range as one can see ([6], p. 189, problem(1)).
Theorem 8. By using the differentiation and integration rules of distribution prove that
Γ ̃^k (s)=[(-it)^k Γ(t)]. (4.8)
Proof. For ϕ is in ,
⟨Γ ̃^((1) ) (s),ϕ(s)⟩=⟨Γ ̃(s),ϕ^((1) ) (s)⟩
⟨Γ ̃^((1) ) (s),ϕ(s)⟩=⟨Γ(s),ϕ ̃^((1) ) (s)⟩ (4.9)
⟨Γ ̃^((1) ) (s),ϕ(s)⟩=⟨Γ(s),(-it)ϕ(s) e^(-ist) ⟩
⟨Γ ̃^((1) ) (s),ϕ(s)⟩=⟨(-it)Γ(s),ϕ ̃(s)⟩
⟨Γ ̃^((1) ) (s),ϕ(s)⟩=⟨(-it) Γ ̃(s),ϕ(s)⟩. (4.10)
On the same way differentiation w.r.t s upto k times, we get
⟨Γ ̃^k (s),ϕ(s)⟩=⟨(-it)^k Γ ̃(s),ϕ(s)⟩. (4.11)
It gives
Γ ̃^k (s)=(-it)^k Γ ̃(s) (4.12)
Γ ̃^k (s)=(-it)^k Γ(t). (4.13)

Theorem 9. If Γ(s) is in Z^' then its Fourier Γ ̃(s) is also in Z^'.
Proof. Since is closed under Fourier transformation therefore result follows.
Theorem 10. If Γ(s) is a distribution of complex space Z then its Fourier transformation Γ ̃(t) is an infinitely smooth distribution.
Proof. By following the lines of above theorem which shows that derivative of Fourier transform of Γ(s) exist upto kth order in distributional sense. So Γ ̃(t) is a distribution that is infinitely smooth

Cite this article

Asifa Tassaddiq, Hussain. A New Representation of the Gamma Functions, Encyclopedia, 2019, v1, Available online: https://encyclopedia.pub/297