A note on the asymptotics for incomplete Betafunctions

We determine the asymptotic behaviour of certain incomplete Betafunctions.


Introdution and results
For integers k, ℓ ≥ 1 define By repeated partial integration we obtain the representation Vietoris [5] showed that P k,ℓ ≤ 1 2 . Alzer and Kwong [1] proved that P k,ℓ ≥ 1 4 holds for all k, ℓ. Interest in bounds for P k,ℓ stems from application to statistics, see [4].
In this note we consider the asymptotic behaviour of P k,ℓ . We prove the following.
Theorem 1. We have Note that the first two estimates are good if one of the parameters k, ℓ is rather small, whereas the third one gives information in the general case.
Comparing (1) and (2) with Vietoris' result that P k,ℓ ≤ 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,ℓ : k, ℓ ∈ N} are 1 2 and the real numbers e −k k−1 ν=1 k ν ν! and 1 − e −ℓ ℓ ν=1 ℓ ν ν! . Proof. Suppose that (k n ), (ℓ n ) are integer sequences such that the pairs (k n , ℓ n ) are all distinct and that the sequence (P kn,ℓn ) converges. If both k n , ℓ n tend to infinity, then by (3) we have that P kn,ℓn 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 ℓ n is constant. Then our claim follows from (1) or (2).
Together with Vietoris bound we obtain the following.

Preliminary estimates
For the proof of (3) we use another representation of P k,ℓ , which is due to Raab [3].
We first compute the occurring integrals.
Proof. The series expansion of e x yields for t → 0 1 t For t → ∞ this expression tends to 0, in particular, it is bounded for all positive t.
Hence for x → ∞ the integral in question becomes 1 12 For x → 0 we use e −xt = 1 − xt + O(x 2 t 2 ) and obtain 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 Combining these estimates our claim follows.

From this we obtain
Lemma 2. We have U k,ℓ = 1 − 1 12 Proof. From Lemma 1 we obtain U k,ℓ = exp 1 12 inserting the Taylor series for exp and using the fact that k, ℓ ≥ 1 our claim follows.
If νx ≤ 1, then for some constant K.
Lemma 4. We have If k > ℓ we obtain We have √ ℓ 2π ν>ℓ Using these estimates we obtain For k ≤ ℓ 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 < ℓ. Under this assumption we have The proof of (2) is quite similar. Analogously to the previous case we may assume that ℓ 2 < k.