The incomplete gamma functions. Notes by G.J.O. Jameson. These notes incorporate the Math. Gazette article [Jam1], with some extra material.

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1 The incomplete gamma functions Notes by G.J.O. Jameson These notes incorporate the Math. Gazette article [Jam], with some etra material. Definitions and elementary properties functions: Recall the integral definition of the gamma function: Γ(a) t a e t dt for a >. By splitting this integral at a point, we obtain the two incomplete gamma γ(a, ) Γ(a, ) t a e t dt, () t a e t dt. (2) Γ(a, ) is sometimes called the complementary incomplete gamma function. These functions were first investigated by Prym in 877, and Γ(a, ) has also been called Prym s function. There are not many books that give these functions much space. Massive compilations of results about them can be seen stated without proof in [Erd, chapter 9] and [Olv, chapter 8]. Here we offer a small selection of these results, with proofs and some discussion of contet. Clearly, Γ(a, ) Γ(a) and γ(a, ) + Γ(a, ) Γ(a) (3) for all and a >. Also, Γ(, ) e and γ(, ) e. For >, the integral (2) converges for all real a, so we regard it as defining Γ(a, ) for all such a. In particular, Γ(, ) is the eponential integral t e t dt. This case has a number of features of its own: it is discussed in companion notes [Jam2]. The integral () only converges for a >, but in fact the definition of γ(a, ) can be etended to negative a, as we see later. Note on comple a and. The definition of γ(a, ) makes sense for comple a with Re a >, and the definition of Γ(a, ) for all comple a. Also, with due interpretation of the integrals, one can replace by a comple variable z. However, in these notes we confine ourselves to the case where a and are real. Some of the results, which interested readers will be able to recognise, apply without change when a is comple. Of course, γ(a, ) and Γ(a, ) can be considered both as functions of (for fied a) and as functions of a (for fied ). Our emphasis will be firmly on them as functions of

2 . First, some simple facts. Since the integrand is non-negative, so are γ(a, ) and Γ(a, ). For fied a, γ(a, ) is an increasing function of, with lim γ(a, ) Γ(a), and Γ(a, ) is a decreasing function of with lim Γ(a, ) (this applies also for a ). By the fundamental theorem of calculus, we have If a >, this is largest when a. d d γ(a, ) d d Γ(a, ) a e. (4) We record some inequalities that follow very easily from the integrals () and (2), and shed some light on their nature for small and large. Firstly, since e e t for t, we have t a e t a e t t a for such t. Now ta dt a /a, so a e a γ(a, ) a a. (5) Hence γ(a, ) a /a as +. (Recall that the notation f() g() as means f()/g() as.) Secondly, if a, then t a a for t, hence Γ(a, ) a e t dt a e, (6) and clearly the opposite holds for a. For a <, another inequality, comparable to the left-hand side of (5), and stronger than (6) for small, is Γ(a, ) e t a a dt e a. (7) We remark that (5) gives γ(a, a) a a e a, while (6) gives Γ(a, a) a a e a. Hence we have shown, with minimal effort, that Γ(a) 2a a e a for a, an elementary inequality of the same type as Stirling s formula. Further inequalities and estimations for Γ(a, ) will be presented later. We mention some equivalent forms given by simple substitutions. substitution ct u gives t a e ct dt c ( u c For c >, the ) a e u c du γ(a, c), (8) ca and similarly ta e ct dt c a Γ(a, c). Note the case : γ(a, c) c a ta e ct dt. The substitution t + u gives Γ(a, ) e ( + u) a e u du. (9) 2

3 The substitution u t n gives e tn dt n n e u u n du n γ( n, n ). We include here some brief remarks about γ(a, ) and Γ(a, ) as functions of a (which, however, will not be used below). Now t a decreases with a for < t <, and increases with a for t >. Consequently, γ(a, ) decreases with a for fied, and Γ(a, ) increases with a for fied. Recall, however, that Γ(a, ) Γ(a) decreases for < a < a, where a.466, and increases for a > a. By (5) and (6), γ(a, ) and Γ(a, ) tend to infinity as a for fied >. Also, for any >, γ(a, ) Γ(a) < γ(a, ) γ(a, a) ea a a a, hence γ(a, )/Γ(a) and Γ(a, )/Γ(a) as a. In the same way as for the gamma function, one can show that γ(a, ) and Γ(a, ) are conve (even log-conve) functions of a, and that they are differentiable, with (for eample) d γ(a, ) da t a e t log t dt. Integration by parts; two basic identities; evaluation for positive integer a The most basic property of the gamma function is the identity Γ(a + ) aγ(a). We now show how this identity decomposes into two companion ones for the incomplete gamma functions. This is achieved by a very simple integration by parts. Clarity and simplicity are gained by stating the basic result for general integrals of the same type. Given a function f(t) on (, ), write I f () if these integrals eist. f(t)e t dt, J f () f(t)e t dt If I f () and I f () eist and f(), then for >, [ ] I f () f(t)e t + f (t)e t dt f()e + I f (). () Similarly, if J f () and J f () eist and f()e as, then J f () f()e + J f (). () If f(t) t a, then I f () γ(a +, ) and I f () aγ(a, ), and similarly for J f and J f. Also, f() if a >, and lim f()e for any a, so we conclude: 3

4 THEOREM. For > and a >, γ(a +, ) aγ(a, ) a e. (2) For > and all a, Γ(a +, ) aγ(a, ) + a e. (3) Added together, (2) and (3), with (3), reproduce the identity Γ(a + ) aγ(a). The process in () and () can be repeated by application to f and higher derivatives. In the case of (), the statement is: THEOREM 2. Suppose that J f (r)() eists for r k and that f (r) ()e as for r k. Then J f () e [f() + f () + f (k ) ()] + J f (k)(). (4) With f() a, the condition lim f (r) ()e is, of course, satisfied for all r. The corresponding development of () is trickier to apply, because it requires the condition f (r) () for successive r, which will fail for some r. One can write out eplicitly what (4) says for Γ(a, ) in general: we return to this later. However, for positive integers a, it simplifies and delivers at once a closed epression for Γ(a, ). For more pleasant notation, we will state this with a n + : note that Γ(n +, ) tn e t dt. Of course, once we have evaluated Γ(n +, ), the value of γ(n +, ) is given by γ(n +, ) n! Γ(n +, ). If f() n, then f (n+) (), so the epression in (4) terminates, and we have Γ(n +, ) J f () e [f() + f () + f (n) ()]. (5) The bracketed sum simply comprises successive derivatives until they become zero. With no further effort, we can write down the first few cases: Γ(2, ) e ( + ), Γ(3, ) e ( ), Γ(4, ) e ( ). We now give an epression for the general case. For this purpose, write e n () n r r r!, the eponential series truncated after n + terms. 4

5 THEOREM 3. For integers n and, Γ(n +, ) n!e n ()e. (6) Proof. For f() n, we have f (k) () n(n )... (n k + ) n k Now applying (5) and substituting r for n k, we obtain Γ(n +, ) n!e n k n k (n k)! n!e n! (n k)! n k. n r r r!. This displays Γ(n+, ) in a rather pleasing way as a fraction of Γ(n+) n!, and shows that Γ(n +, )/n! as n for any fied. It also shows that Γ(n +, ) n e as, supplementing (6). While this derivation was, surely, attractive enough, it is instructive to see a second, equally efficient proof. Alternative proof of (6). By (9), the binomial epansion and the fact that Γ(n r+) (n r)!, we have Γ(n +, ) e ( + u) n e u du n ( n e ) r u n r e u du r r n ( ) n e r Γ(n r + ) r r e n r n! r! r. Note. Write I n for γ(n +, ) tn e t dt. It has been known for students (including my students) to be set the eercise of evaluating, say, I 3 by repeated application of the recurrence relation implied by (2), starting with I. After at least as much work as either of the methods just described, the student arrives (barring accidents) at the answer I 3 6 6e, but this approach does little to reveal the pattern, and the formula for general n. I am indebted to the Math. Gazette referee for pointing out that Theorem 3 leads to the following neat proof of the irrationality of e. By (6), Γ(n +, ) B n e, where 5

6 B n n n! r, which is an integer. So γ(n +, ) A r! n B n e, where A n n!. Note that γ(n +, ) >. Suppose now that e p/q, where p and q are integers. Then pγ(n +, ) A n p B n q: for each n, this is positive and an integer, hence at least. But by (5), γ(n +, ), so pγ(n +, ) as n, a contradiction! n+ It is clear from (5) and (6) that functions like a γ(a, ) and e Γ(a, ) are particularly relevant. Using (2) and (3), we can derive very satisfactory epressions for the derivatives of these functions. By (4) and (2), we have d γ(a, ) aγ(a, ) + a e d a a+ a [ aγ(a, ) + a+ a e ] γ(a +, ). (7) a+ and similarly for Γ(a, ). So if f(a, ) a γ(a, ), then d f(a, ) f(a +, ). We can d deduce at once that the nth derivative is ( ) n f(a + n, ). By (4) and (3), we have d d [e Γ(a, )] e [Γ(a, ) a e ] e (a )Γ(a, ), (8) and similarly for γ(a, ). In particular, e Γ(a, ) increases with when a, and decreases when a <. Series epressions for γ(a, ) and etension to a < We now give an eplicit power-type series epression for γ(a, ). THEOREM 4. For a > and >, γ(a, ) a n ( ) n n n!(a + n) a ( a ) a !(a + 2). (9) Proof. This epression is obtained at once by termwise integration on [, ] of the series t a e t ( ) n t a+n. n! n Termwise integration is justified by uniform convergence of the power series for e t bounded intervals (after separating out the first term if a < ). on In principle, (9) enables us to calculate γ(a, ), although in practice the calculation is only pleasant for fairly small. 6

7 For a fied >, the series (9) converges for all a ecept and negative integers, so we take it as the definition of γ(a, ) for such a. We will show that the etended function still satisfies the basic identity (2). For the gamma function itself, the usual procedure is to etend the definition by the identity Γ(a + ) aγ(a): given Γ(a + ), this defines Γ(a). Meanwhile, Γ(a, ) is already defined for all a. We show that the two etensions are compatible, in the sense that the identity (3) still holds. PROPOSITION 5. For all a ecept and negative integers, and all >, γ(a +, ) aγ(a, ) a e. (2) If the definition of Γ(a) is etended in the way just stated, then γ(a, ) + Γ(a, ) Γ(a) for all such a. Proof. We have aγ(a, ) a ( ) n n a n! a + n n ( a ( ) n n n ) n! a + n n a e a n a e + a m a e + γ(a +, ). ( ) n n (n )!(a + n) ( ) m m+ m!(a + m + ) Now write γ(a, ) + Γ(a, ) F (a, ). We know that F (a, ) Γ(a) for a >. By (3) and (2), F (a +, ) af (a, ). Hence if we know that F (a +, ) Γ(a + ), it follows that F (a, ) Γ(a). By repeated backwards steps of length, it now follows that F (a, ) Γ(a) for all a ecept and negative integers. A satisfying application of Proposition 5 is that it gives an eplicit formula for the etended function Γ(a). For this purpose, we only need to take, obtaining: Γ(a) γ(a, ) + Γ(a, ) t a e t dt + n ( ) n n!(a + n). In fact, as an alternative to the procedure described above, one can adopt this as the definition of Γ(a) for such a. If this approach is chosen, then the conclusion from (2) is that the gamma function (etended in this way) still satisfies Γ(a + ) aγ(a). 7

8 Since the identity γ(a, ) + Γ(a, ) Γ(a) still applies, so does the identity d γ(a, ) d d Γ(a, ) d a e. In particular, γ(a, ) is an increasing function of. However, it may well be negative: indeed, (2) shows that it is certainly negative for < a <. Further note on Theorem 4. Write s n (t) n ( ) r r t r. It is easily proved by repeated r! integration that s 2n+ (t) e t s 2n (t) for all t >. Hence e t s n (t) t n+ /(n + )!, from which one can justify the termwise integration directly. Also, if we write s n (a, ) t a s n (t) dt a it follows that s 2n+ (a, ) γ(a, ) s 2n (a, ). In particular, ( γ(a, ) a a ). a + n r ( ) r r r!(a + r), This, together with (5), shows that γ(a, ) /a as a + for any fied >. We now give a second series epression for γ(a, ). We will use the following well-known beta integral: for integers n, B(a, n + ) : u a ( u) n du PROPOSITION 6. For a > and >, γ(a, ) a e n n! a(a + )... (a + n). n a(a + )... (a + n). (2) Proof. Again applying termwise integration of an eponential series, we have e γ(a, ) n n n n t a e t dt t a ( t) n dt n! n! (u) a [( u)] n du n! a+n B(a, n + ) a+n a(a + )... (a + n). (substitute t u) Again, the series (2) converges for all a ecept and negative integers. Denote, temporarily, the function so defined by G(a, ). It is straightforward to show that 8

9 G(a +, ) ag(a, ) a e (we leave the details to the reader). By this and (2), it follows that G(a, ) γ(a, ) for all such a. Some integrals We now evaluate some integrals of epressions involving the incomplete gamma functions. Since these are already defined as integrals, the integrals considered will appear as double integrals, and evaluation will be achieved by reversing them (which is valid, because the integrands are positive). The answers will be in terms of the gamma function itself. First, consider p γ(a, ) d. Convergence at infinity clearly requires p >. Since, by (5), γ(a, ) a /a as +, convergence at requires p < a. The need for these conditions also shows up clearly in the following proof. PROPOSITION 7. For < p < a, γ(a, ) d Γ(a p). (22) p+ p Proof. Reversing the double integral, we have γ(a, ) d p+ p p+ t a e t dt d ( ) t a e t d p+ t t a p e t dt Γ(a p). p dt An alternative proof is by integrating by parts, using (4) for γ (a, ). For this method, (5) is needed to identify the limit at. In particular, γ(a, ) d /(a ) for a >. a Now consider p Γ(a, ) d, possibly with a <. If a >, then convergence at requires p >. For a <, we will see later (refining (7)) that Γ(a, ) a /a as +, so convergence at requires a + p >. Then PROPOSITION 8. Suppose that either (i) a > and p > or (ii) a and p > a. p Γ(a, ) d Γ(a + p). (23) p 9

10 In particular, for a >, Γ(a, ) d Γ(a + ). (24) Proof. Under either condition, p >. Reversing the double integral, we have p Γ(a, ) d p p t a e t dt d ( t ) t a e t p d dt t a+p e t dt Γ(a + p). p PROPOSITION 9. For a > and c >, e c γ(a, ) d e c Γ(a, ) d Γ(a) c Γ(a) c(c + ), a (25) ( ). (c + ) a (26) Proof. Note that the substitution bt u gives t a e bt dt Γ(a)/b a. We have e c γ(a, ) d e c t a e t dt d ( ) t a e t e c d dt t a t e ct e c t dt t a e (c+)t dt c Γ(a) c(c + ). a We deduce (26) using (3) and e c d /c. In particular, e γ(a, ) d 2 a Γ(a). By integrating (3), one can show that (26) also holds for < a < (we leave the details to the reader). For the case a, one can show (see [Jam2]) that e c Γ(, ) d log( + c). c

11 Further results for Γ(a, ) Recall from (6) that Γ(a, ) a e for a, together with the opposite inequality for a <. We now elaborate on this comparison, giving a restricted reverse inequality when a > and showing that Γ(a, ) a e as. PROPOSITION. If a and e > 2 a, then Γ(a, ) 2 a a e. (27) For all a, we have Γ(a, ) a e as. Proof. First, let a. By (9), e Γ(a, ) ( + u) a e u du. Since + u 2 when u and + u 2u when u, e Γ(a, ) (2) a e u du + < 2 a [ a + Γ(a, )], (2u) a e u du so (e 2 a )Γ(a, ) 2 a a Clearly, (27) follows. Now take any a. By (3), Γ(a, ) a e + (a )Γ(a, ). If a 2, then by (27), Γ(a, ) 2 a a 2 e for large enough. By (6), if a < 2, this holds with 2 a replaced by. Hence in both cases Γ(a, ) a e ( + O(/)) as. We now consider the behaviour of Γ(a, ) for close to. Of course, for a >, Γ(a, ) simply tends to the finite limit Γ(a) as +. The case a (the eponential integral) is rather special: we deal with it net. PROPOSITION. There is a constant c such that Γ(, ) + log c as +. Proof. Define the companion function E () e t dt. t Since < e t t for t >, this integral is well-defined and E () for >. Now E () E () e t dt log Γ(, ) + Γ(, ), t

12 so Γ(, ) E () log + c, where c Γ(, ) E (). The statement follows. Note. One can show that c γ: e.g. see [Jam]. Recall from (7) that Γ(a, ) a e /a for a <. asymptotic estimate for close to. We now develop this into an PROPOSITION 2. For a <, Γ(a, ) a /a as +. Proof. By (3), a a Γ(a, ) a Γ(a +, ) e. The stated result follows if we can show that a Γ(a +, ) as +. If < a <, this follows from the fact that Γ(a +, ) tends to Γ(a + ). If a, it follows from Proposition, since log as +. If a <, then (7) gives a+ /(a + ): again, the required limit follows. Γ(a +, ) We now formulate what (4) says when applied to Γ(a, ) where a is not a positive integer. The outcome is a system of approimations to Γ(a, ) that are effective for large. Before stating the result for general a, we describe the case a. Eample. Consider Γ(, ). We have f() /, so f (k) () ( ) k / k+. Let S k () + 2 (k )! + 2 ( )k. 3 k If k is even, then f (k) (), hence also J f (k)(), is positive, so by (4), Γ(, ) e S k (). The opposite holds if k is odd. Now consider general a. To simplify the statement, we write P k (a) (a )(a 2)... (a k) for k (also P (a) ). (Here I am departing from the more-or-less standard notation, which would be ( ) k ( a) k.) Note that if f() a, then f (r) () P r (a) a r. Also, once k is larger than a, the numbers P k (a) are alternately positive and negative. and let PROPOSITION 3. Suppose that a is not a positive integer. Define P k (a) as above, Then S k (a, ) a + P (a) a P k (a) a k (28) ( a + P (a) + + P ) k (a). (29) k Γ(a, ) e S k (a, ) + R k (a, ), (3) 2

13 where R k (a, ) P k (a)γ(a k, ). So Γ(a, ) S k (a, ) if P k (a) >, and Γ(a, ) S k (a, ) if P k (a) <. In particular, if P k (a) > and P k+ (a) <, then e S k (a, ) Γ(a, ) e S k+ (a, ). (3) Equivalently, Γ(a, ) e [S k (a, ) + r k (a, )], (32) where r k (a, ) P k (a) a k. Proof. Apply (4) with f() a. We just need to note that J f (k)() P k (a)γ(a k, ). The stated inequalities follow from the fact that Γ(a k, ) >. It is easy to be more specific about the parity of P k (a). If a is positive (and not an integer), let k be the integer such that a < k < a. If a, put k. In both cases, P k (a) >. Now let k > k. The negative factors in P k (a) are a j for k + j k, so P k (a) is positive when k k is even, negative when k k is odd. So if k k even, then (3) applies. Eample. Consider Γ(, ). The statement S 2 4(, ) 2 e Γ(, ) S 2 3(, ) equates to 2 ( Γ(, ) 2 e / ) 4 r 3(), 2 where r 3 () 5/(8 3 ). For 4, this gives bounds.85 and.845. It is important to realise that (28) is not the beginning of a convergent series n P n(a) a n. Indeed, P n (a) grows like n!, so for any, P n (a) n tends to infinity as n. This is an asymptotic series, not a convergent one. A judicious choice of k may make the error term r k (a, ) small, but ultimately it grows large again. Now recall Euler s identity: for < a <, Γ( a)γ(a) π/(sin πa). Our last result gives an integral epression for Γ( a)γ(a, ) which implies Euler s identity in the case. PROPOSITION 4. For < a < and, Γ( a)γ(a, ) e Proof. The substitution tu v gives u a e tu du t a Γ( a). u a ( + u) e u du. (33) 3

14 Substituting this integral epression for t a Γ( a), we obtain The case gives Γ( a)γ(a, ) Γ( a) Γ( a)γ(a) t a e t dt e t u a e tu du dt u a e t(+u) dt du a e (+u) u + u du. u a ( + u) du π sin πa. For >, the substitution u t in (33) gives the alternative epression Γ( a)γ(a, ) a e e t dt. (34) t a ( + t) References [Erd] A. Erdélyi, W. Magnus, F. Oberhettinger and F. G. Tricomi, Higher Transcendental Functions, vol. II, McGraw-Hill (953). [Jam] G. J. O. Jameson, The incomplete gamma functions, Math. Gazette (26), [Jam2] [Olv] G. J. O. Jameson, The real eponential integrals, at jameson F. W. J. Olver, D. W. Lozier, R. F. Boisvert and C. W. Clark (eds.), NIST Handbook of Mathematical Functions, Cambridge Univ. Press (2); also at: dlmf.nist.gov updated 7 September 27 4