A note on the asymptotics for incomplete Betafunctions

We determine the asymptotic behaviour of certain incomplete Betafunctions.


Introdution and results
For integers k, ≥ 1 define By repeated partial integration we obtain the representation Vietoris [5] showed that P k, ≤ 1 2 . Alzer and Kwong [1] proved that P k, ≥ 1 4 holds for all k, . Interest in bounds for P k, stems from application to statistics, see [4].
In this note we consider the asymptotic behaviour of P k, . We prove the following.

3
and Note that the first two estimates are good if one of the parameters k, is rather small, whereas the third one gives information in the general case.
Comparing (1) and (2) with Vietoris' result that P k, ≤ 1 2 we see that Note that equality is impossible as e is transcendental. A more precise version of this inequality has been asked by Ramanujan (Question 294) and was answered by Karamata [2]. For a detailed discussion of this result, Uhlmann's inequalities [4] and Vietoris bound we refer the reader to the historical notes by Vietoris [6]. From Theorem 1 we deduce the following.

Corollary 1
The only accumulation points of the set {P k, ∶ k, ∈ ℕ} are 1 2 and the real numbers Proof Suppose that (k n ), ( n ) are integer sequences such that the pairs (k n , n ) are all distinct and that the sequence (P k n , n ) converges. If both k n , n tend to infinity, then by (3) we have that P k n , n tends to 1 2 . If one of these sequences does not tend to infinity, then we can pass to a subsequence and assume that k n or n is constant. Then our claim follows from (1) or (2). ◻ Together with Vietoris bound we obtain the following.

Corollary 2 We have
Proof We have using Stirling's formula in the form k! = � k e � k√ 2 ke ∕12k with 0 ≤ ≤ 1 , and P ,k ≤ 1 2 , we conclude

Preliminary estimates
For the proof of (3) we use another representation of P k, , which is due to Raab [3].
We first compute the occurring integrals.

Lemma 1 We have for
Proof The series expansion of e x yields for t → 0 For t → ∞ this expression tends to 0, in particular, it is bounded for all positive t. Hence for x → ∞ the integral in question becomes As a function of x, the integral ∫ ∞ 1 e −xt t 2 dt defines a function that is differentiable from the right in 0 and has bounded second derivative in (0, ∞) , hence, for x ≥ 0 this integral is Finally Combining these estimates our claim follows. ◻

Lemma 2 We have
Proof From Lemma 1 we obtain inserting the Taylor series for exp and using the fact that k, ≥ 1 our claim follows. ◻ Next we compute c (x).

Lemma 3 If x ≥ 1 , then
If x ≤ 1 , then for some constant K.
Proof If x > 1 , then we apply Lemma 1 to obtain .

and we obtain
◻ We now compute V k, .

Lemma 4 We have
Proof We have thus, .
We have and for ∕k ≤ N ≤ as well as and therefore and Using these estimates we obtain For k ≤ we obtain and the proof is complete. ◻

Proof of Theorem 1
We first prove (1). Note that this inequality is only interesting if the error term is o(1), in particular we may assume that k 2 < . Under this assumption we have The proof of (2) is quite similar. Analogously to the previous case we may assume that 2 < k.
Funding Open Access funding enabled and organized by Projekt DEAL.
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