Defining Incomplete Gamma Type Function with Negative Arguments and Polygamma functions ψ (n) ( m)

Transcription

1 1 Defining Incomplete Gamma Type Function with Negative Arguments and Polygamma functions ψ (n) (m) arxiv: v1 [math.ca] 27 Jun 214 Emin Özc. ağ and İnci Ege Abstract. In this paper the incomplete gamma function γ(α, x) and its derivative is considered for negative values of α and the incomplete gamma type function γ (α,x ) is introduced. Further the polygamma functions ψ (n) (x) are defined for negative integers via the neutrix setting. AMS Mathematics Subject Classification (21): 33B15, 33B2, 33D5 Key words and phrases: Gamma function, incomplete gamma function, Digamma Function, Polygamma Function, neutrix, neutrix limit. 1. Introduction The incomplete gamma function of real variable and its complement are defined via integrals and γ(α,x) Γ(α,x) x u α e u du (1) u α e u du (2) respectively, for α > and x see [1, 8]. Note that the definition of γ(α,x) does not valid for negative α, but it can be extended by the identity γ(α,x) Γ(α)Γ(α,x). The incomplete gamma function does not exists for negative integers α or zero [13]. However, Fisher et. al. [4] defined γ(,x) by γ(,x) u (e u )dulnx. The problem of evaluating incomplete gamma function is subject of some earlier articles. The general problem in which α and x are complex was considered by Winitzki in [15], but no method is found to be satisfactory in cases where Re[x] <. The incomplete gamma function with negative arguments are difficult to compute, see [6]. In [14] Thompson gave the algorithm for accurately computing the incomplete gamma function γ(α,x) in the cases where α n1/2,n Z and x <.

2 2 2. Incomplete Gamma Function A series expansion of incomplete gamma function can be obtained by replacing the exponential in (1) with its Maclaurin series, the result is that γ(α,x) k () k k!(αk) xαk. (3) A single integration by parts in (1) yields the recurrence relation γ(α1,x) αγ(α,x)x α e x (4) and thus the incomplete gamma function can be defined for α < and α,2,... By repeatedly applying recurrence relation, we obtain γ(α,x) x α e x k x k (α) k 1 (α) m γ(αm,x) where (.) k is Pochhamer s symbol [1]. The last term on the right-hand side disappears in the limit m. By regularization we have γ(α,x) u α [e u (u) i ] du () i x αi (αi). (5) for m < α < m1 and x >. It follows from the definition of gamma function that lim γ(α,x) Γ(α) x for α,,2,..., see [7, 11]. InthefollowingweletNbetheneutrix[2,3,4]havingdomainN { : < < } and range N the real numbers, with negligible functions finite linear sums of the functions λ ln r, ln r (λ <, r Z ) (6) and all functions of which converge to zero in the normal sense as tends to zero. If f() is a real (or complex) valued function defined on N and if it is possible to find a constant c such that f()c is in N, then c is called the neutrix limit of f() as and we write f() c. Note that if a function f() tends to c in the normal sense as tends to zero, it converges to c in the neutrix sense. On using equation (5), the incomplete gamma function γ(α, x) was also defined by γ(α,x) u α e u du (7)

3 3 for all α R and x >, and it was shown that lim x γ(m,x) Γ(m) for m N, see [4]. Further, the r-th derivative of γ(α, x) was similarly defined by γ (r) (α,x) u α ln r ue u du (8) for all α and r,1,2,..., provided that the neutrix limit exists, see [11]. By using the neutrix N given in equation(6), the interesting formula was obtained in [11] for m N. γ(m,x) 1 ()m γ(m1,x) 1 m mm! m ex x m It can be easily seen that equation (3) is also valid for α < and α,2,... In fact it follows from the definition that γ(α,x) k [ u α e u () k du k! k () k [ x αk αk] k!(αk) k () k k!(αk) xαk. Now assume that α Z, then writing ] u αk du u e u du () k k, k m k, k m k! u mk du ()m m! () k k!(k m) xkm ()m lnxo() m! u du where O() is negligible function, and taking the neutrix limit we obtain γ(m,x) k, k m () k k!(k m) xkm ()m lnx m! for x > and m N. If m we particularly have γ(,x) k1 (x) k kk! lnx

4 4 and thus it follows by the calculation of Mathematica that γ(, 1/2).13699, γ(, 3/4) , γ(, 1) Similar equation for γ (r) (α,x) can be obtained using the following lemma. Lemma 2.1 u ln r udu lnr1 x r1 r u α ln r udu for all α,x > and r N. Proof. Straight forward. k () k r!x α1 ln rk x (α1) k1 (r k)! ()r r!x α1 (α1) r1 Now let α < and α,,2,... Then it follows from lemma 2.1 that and similarly γ (r) (α,x) k r k γ (r) (m,x) k, k m for r 1,2,... and m N. r u α ln r ue u du () k k! u αk ln r udu () ik r! k!(r i)!(αk) i1xαk ln ri x k u ln r ue u du () kr r! k!(αk) r1xαk () ik r! k!(ri)!(mk) i1xkm ln ri x k, k m () kr r! k!(mk) r1xkm ()m (r 1)m! lnr1 x It was indicated in [3] that equation (1) could be replaced by the equation γ(α,x) u α e u du (9)

5 5 and this equation was used to define γ(α,x) for all x and α >, the integral again diverging for α. The locally summable function γ (α,x ) for α > was then defined by { γ (α,x ) u α e u du, x,, x > If α >, then the recurrence relation u α e u du. (1) γ (α1,x ) αγ (α,x )x α e x. (11) holds. So we can use equation (11) to extend the definition of γ (α,x ) to negative non-integer values of α. More generally it can be easily proved that γ (α,x ) u α [e u if m < α < m1 for m 1,2,... (u) i Now if m < α < m1,m N and x <, we have and thus u α e u du u α [e u (u) i ] x αi du (αi). (12) ] du γ (α,x ) u α e u du [ x αi αi ] (αi) on using equation (12). This suggests that we define γ (m,x ) by γ (m,x ) u e u du (13) for x < and m N. If x <, then we simply write thus we have u e u du u e u du u (e u )duln x ln γ (,x ). u (e u )dulnx

6 6 Next, writing similarly u e u du u [e u for x < and m N, then it follows that γ (m,x ) u [e u u e u du m (u) i m (u) i ] du [ x im im ] (mi) ] du 1 m! [ ] ln x ln x im (mi) 1 m! lnx. (14) Equation (14) can be regarded as the definition of incomplete gamma function γ (m,x ) for negative integers and also written in the form γ (m,x ) u [e u m (u) i ] du u e u 1 du (mi) Replacing e x by its Maclaurin series yields in equation (13) γ (m,x ) If m we particular have and by Mathematica k, k m γ (,x ) Differentiating equation (12) we get (15) x km (mk)k! 1 m! lnx (m N). (16) k1 x k kk! lnx. (17) γ (,/3) γ (,/2) γ (,3/4) γ (,) γ (α,x ) [ u α ln u e u (u) i (αi)x αi lnx x αi (αi) 2 ] du

7 7 for m < α < m1 and x <. On the other hand, arranging the following integral and taking the neutrix limit u α ln u e u du [ u α ln u e u (u) i ] du 1 x u αi ln u du [ u α ln u e u (u) i ] du 1 [ ] x αi (αi) lnx αi ln 1 [ x αi (αi) 2 ]. αi Thus γ (α,x ) u α ln u e u du for α,,2,... and x <. More generally it can be shown that γ (r) (α,x ) u α ln r u e u du (18) for α,,2,...,r,1,2,... and x <. This suggests the following definition. Definition 2.2 The r-th derivative of incomplete gamma function γ (r) (m,x ) is defined by γ (r) (m,x ) u ln r u e u du (19) for r,m,1,2,... and x < provided that the neutrix limit exists. Equation(18) will then define γ (r) (α,x ) for all α and r,1,2,... To prove the neutrix limit above exists we need the following lemma. Lemma 2.3 and ln r u du r () i r! (ri)! ulnri u () r r!u (2) r u s ln r r! u du (r i)! si u s ln ri u r!s r u s (21)

8 8 for r,s 1,2,... Proof. Equation (2) follows by induction and equation (21) follows on using equation (2) and making the substitution w u s. Theorem 2.4 The functions γ (r) (,x ) and γ (r) (m,x ) exist for x < and γ (r) (,x ) u ln r u e u du for r,1,2,... and u ln r u [e u ]du (22) γ (r) (m,x ) u ln r u e u du [ u ln r u e u m (u) i ] du r!(mi) r (23) for r,m 1,2,... Proof. We have u ln r u e u du u ln r u e u du u ln r u e u du u ln r u [e u ]du u ln r u [e u ]du lnr1 r 1. u ln r u du for r,1,2,... It follows that u ln r u e u du and so equation (22) follows. Now let us consider u ln r u e u du u ln r u e u du u ln r u e u du [ u ln r u e u m () i u ln r u [e u ]du. m (u) i ] du u mi ln r u du

9 9 Thus u ln r u e u du r j [ u ln r u e u () i r! (mi) j mi ln rj (r j)! m (u) i ] du () i (mi) r r! [ ] () mi mi ) lnr1 m!(r 1). γ (r) (m,x ) u ln r u e u du r!(mi) r. for r,m 1,2,... Equation (23) follows. u ln r u e u du [ u ln r u e u m (u) i ] du 3. Polygamma Functions ψ (n) (m) The polygamma function is defined by It may be represented as which holds for x >, and It satisfies the recurrence relation ψ (n) (x) dn dn1 dxnψ(x) lnγ(x) (x > ). (24) dxn1 ψ (n) (x) () n1 t n e xt 1e t ψ (n) t x ln n t (x) dt. (25) 1t ψ (n) (x1) ψ (n) (x) ()n n! x n1 (26) see [8]. This is used to define the polygamma function for negative non-integer values of x. Thus if m < x < m1, m 1,2,..., then ψ (n) t x () n n! (x) 1t lnn tdt (xk) n1. (27) k

10 1 Kölbig gave the formulae for theintegral tλ (1t) ν ln m tdtfor integer and halfinteger values of λ and ν in [1]. As the integral representation of the polygamma function is similar to the integral mentioned above, we prove the existence of the integral in equation (25) for all values of x by using the neutrix limit. Now we let N be a neutrix having domain the open interval {ǫ : < < 1 2 } with the same negligible functions as in equation (6). We first of all need the following lemma. Lemma 3.1. The neutrix limits as tends to zero of the functions /2 t x ln n tln r (1t)dt, exists for n,r,1,2,... and all x. 1/2 (1t) x ln n tln r (1t)dt Proof. Suppose first of all that n r. Then /2 { 2 x t x x1, x, dt x1 ln2ln, x /2 and so t x dt exists for all x. Now suppose that r and that integer n and all x. Then /2 t x ln n1 tdt and it follows by induction that all x. Finally we note that we can write /2 t x ln n tdt exists for some nonnegative 2 x ln n1 2 x1 ln n1 n1 1/2 t x ln n tdt, x, x1 x1 () n ln n2 2ln n2 /2 ǫ ln r (1t) m2, x t x ln n tdt exists for n,1,2,... and α ir t i for r 1,2,..., the expansion being valid for t < 1. Choosing a positive integer k such that xk >, we have /2 t x ln n tln r (1t)dt k /2 α in t xi ln n tdt i1 i1 /2 α in t xi ln n tdt. ik It follows from what we have just proved that k /2 α in t xi ln n tdt i1

11 11 exists and further /2 α in ik t xi ln n tdt lim /2 α in t xi ln n tdt ik /2 α in t xi ln n tdt, proving that the neutrix limit of /2 t x ln n tln r (1t)dt exists for n,r,1,2,... and all x. Making the substitution 1t u in 1/2 ik (1t) x ln n tln r tdt, it follows that 1/2 (1t)x ln n tln r tdt also exits for n,r,1,2,... and all x. Lemma 3.2. The neutrix limit as of the integral t ln n tdt exists for m,n 1,2,... and Proof. Integrating by parts, we have and so t ln n tdt n! mn1. (28) t lntdt m m lnm t dt t lntdt 1 n 2 proving equation (28) for n 1 and m 1,2,... Now assume that equation (28) holds for some m and n 1,2,... Then t m2 ln n tdt (m1) ln n n t m2 ln n tdt m1 (m1) ln n n (n)! m1 (m1) n and it follows that t m2 ln n n! tdt (m1) n1 proving equation (28) for m1 and n 1,2,... Using the regularization and the neutrix limit, we prove the following theorem. Theorem 3.3. The function ψ (n) (x) exists for n,1,2,..., and all x.

12 12 Proof. Choose positive integer r such that x > r. Then we can write t x 1t lnn tdt /2 [ r 1 t x ln n t 1t () i t ]dt i r /2 () i t xi ln n tdt 1/2 t x 1t lnn tdt. and We have /2 t x [ r 1 lim 1t lnn t 1t () i t ]dt i /2 t x [ r 1 1t lnn t 1t () i t ]dt i lim 1/2 t x 1t lnn tdt 1/2 t x 1t lnn tdt the integrals being convergent. Further, from the Lemma 3.1 we see that the neutrix limit of the function r /2 () i t xi ln n tdt exists and implying that t x 1t lnn tdt exists. This proves the existence of the function ψ (n) (x) for n,1,2,..., and all x. Before giving our main theorem, we note that ψ (n) t x (x) 1t lnn tdt since the integral is convergent in the neighborhood of the point t 1. Theorem 3.4. The function ψ (n) (m) exists and ψ (n) (m) m i1 n! i n1 ()n1 n!ζ(n1) (29) for n 1,2,... and m,1,2,..., where ζ(n) denotes zeta function.

13 13 Proof. From Theorem 3.3, we have ψ (n) (m) t ln n t dt 1t [m1 i1 t i (1t) ] ln n tdt. (3) We first of all evaluate the neutrix limit of integral ti ln n tdt for i 1,2,... and n 1,2,... It follows from Lemma 3.2 that For i 1, we have Next t i ln n n! tdt (i > 1). (31) (i) n1 ln n t 1t dt t ln n tdt O(). k ln n t 1t dt () n n! (k 1) n1 k t k ln n tdt () n n!ζ(n1), (32) where t k ln n tdt ()n n! (k 1) n1. It now follows from equations (3), (31) and (32) that ψ (n) (m) m1 implying equation (29). m i1 i1 t ln n t dt 1t t i ln n tdt n! i n1 ()n1 n!ζ(n1) Note that the digamma function ψ(x) can be defined by ln n t (1t) dt. 1t x ψ(x) γ dt (33) 1t

14 14 for all x. Also note that using Lemma 3.1 we have 1t 1t dt and it follows from equation (33) that m1 i1 t i dt [ln1ln] m1 i2 [1 i1 ] i1 1t ψ(m) γ dt γ 1t γ φ(m) (34) which was also obtained in [5] and [9]. Equation (34) can be regarded as the definition of the digamma function for negative integers. m i1 i References [1] M. Abramowitz and I. Stegun, Hand of Mathematical Functions, Dover Publications, Newyork, [2] J.G. van der Corput, Introduction to the neutrix calculus, J. Analyse Math., 7(1959), [3] B. Fisher, On defining the incomplete gamma function γ(m,x ), Integral Trans. Spec. Funct., 15(6)(24), [4] B. Fisher, B. Jolevska-Tuneska and A. Kılıc.man, On defining the incomplete gamma function, Integral Trans. Spec. Funct., 14(4)(23), [5] Fisher, B. and Kuribayashi, Y., Some results on the Gamma function, J. Fac. Ed. Tottori Univ. Mat. Sci., 3(2)(1988) [6] W. Gautschi, the incoplete gamma function since Tricomi. Tricomi s Ideas Contemporary Applied Mathematics. Number 147 in Atti dei Convegni Lincei. Roma : Accedemia nazionale dei Lincel, Roma, Italy, 1998, [7] I. M. Gel fand and G. E. Shilov, Generalized Functions, Vol. I., Academic Press, Newyork/London, (1964). [8] I. S. Gradshhteyn, and I. M. Ryzhik, Tables of integrals, Series, and Products, Academic Press, San Diego, 2.

15 15 [9] Jolevska-Tuneska, B. and Jolevski, I., Some results on the digamma function, Appl. Math. Inf. Sci., 7(213) [1] Kölbig, K. S., On the integral xν (1 x) λ ln m xdx, J. Comput. Appl. Math., 18(1987) [11] E. Özc.ağ, İ. Ege, H. Gürc.ay and B. Jolevska-Tuneska, Some remarks on the incomplete gamma function, in: Kenan Taş et al. (Eds.), Mathematical Methods in Engineering, Springer, Dordrecht, 27, pp [12] Özc.ağ, E., Ege, İ., Gürc.ay, H. and Jolevska-Tuneska, B., On partial derivatives of the incomplete beta function, Appl. Math. Lett., 21(28) [13] I. Thompson, A note on real zeros of the incomplete gamma function, Integral Trans. Spec. Funct., 23(6)(212), [14] I. Thompson, Algorithm 926: Incomplete Gamma Function with Negative Arguments, ACM Trans. Math. Softw. 39(2)(213), 9 pp. [15] S. Winitzki, computing the incomplete gamma function to arbitrary precision, In Proceedings of International Conference on Computational Science and Its Applications (ICCSA 3). Lecturer Notes in Computer Science, Vol (23), Emin Özc. ağ Department of Mathematics Hacettepe University, Beytepe, Ankara, Turkey ozcag1@hacettepe.edu.tr İnci Ege Department of Mathemacis Adnan Menderes University, Aydın, Turkey iege@adu.edu.tr