On the convergence of certain integrals

Let M(r) := max|z|=r |f(z)|, where f(z) is an entire function. Also let α > 0 and β > 1. We discuss the behavior of the integrand M(r)e−α(log r) β as r → ∞ if ∫∞ 1 M(r)e−α(log r) β dr is convergent. 1. Convergence of integrals vis-à-vis convergence of series. There is one fundamental property of a convergent infinite series in regard to which the analogy between infinite series and infinite integrals breaks down. If ∑∞ n=1 θ(n) is convergent, then θ(n)→ 0 as n→∞; but it is not always true, even when θ(r) is always positive, that if ∫∞ a θ(r) dr, a > 0, is convergent, then θ(r)→ 0 as r →∞. As a counterexample, we can consider the function given by

It is however true that if ∞ a θ(r) dr converges and θ(r) is non-negative, then where k r is the k-th iterate of log r. If this was not true, then there would exist positive numbers c and R 0 such that for all R > R 0 , we would have and then e R R θ(r) dr could not be made arbitrarily small by taking R sufficiently large ([2, p. 376]), contradicting the convergence of the integral ∞ a θ(r) dr. On the other hand, it is well known that if θ(r) is positive and non-increasing, then ∞ a θ(r) dr can converge only if r θ(r) → 0 as r → ∞. The same conclusion can be drawn if θ(r) is the product of a monotonic function ϕ(r) and a non-negative function L(r) which is continuous and L(cr) ∼ L(r) as r → ∞ (i.e.
for every fixed positive k. Lemma 6]) that r 2Q M (r) e −πr → 0 as r → ∞. In order to prove it he does not require anything more than the fact that M (r) is a non-decreasing function of r. However, M (r) is not just a non-decreasing function of r but also log M (r) is a downward convex function of log r. Thus r 2Q M (r) = o (e πr ) was not expected to be all that the convergence of ∞ 1 r 2Q M (r) e −πr dr would imply. Recently, Qazi [5] has proved the following stronger result, which is "essentially" best possible. If f is a transcendental entire function, then (see a remark following Theo- however, M (r) e −α(log r) β may tend to zero as r → ∞ for some α > 0 and some β > 1. This can happen if f is an entire function of order 0, that is, if In connection with Theorem 2.1, one may then ask the following question: What can we say about the behavior of M (r) as r → ∞ if f is an entire function such that ∞ 1 M (r) e −α(log r) β dr converges for some α > 0 and some β > 1?
We give an answer to this question. The proof of Theorem 2.1 as given by Qazi [5] is based on the use of the well-known Stirling's formula for Euler's Gamma function. This was somehow natural because of the integrand in ∞ 0 r α M (r) e −βr dr having e −βr as a factor. Since the integrand does not anymore have such a factor, the use of Stirling's formula is more or less out of the question. So, we have to use some other ideas. In addition to Stirling's formula, Qazi's proof of Theorem 2.1 uses Hadamard's threecircles theorem. That remains available to us and we have tried to use it as efficiently as we could.
Hadamard's three-circles theorem [8, p. 172] can be stated as follows: Theorem 3.1. Let f (z) be an analytic function, regular for r 1 ≤ |z| ≤ r 3 . Furthermore, let r 1 < r 2 < r 3 , and let M 1 , M 2 , M 3 be the maxima of |f (z)| on the three circles |z| = r 1 , r 2 , r 3 , respectively. Then Since we may write (3.1) in the form Hadamard's three-circles theorem may be interpreted by saying that log M (r) is a convex function of log r. If f is a transcendental entire function, then inequality (3.2) leads to the existence of a positive number r 0 such that r → log M (r) log r is a strictly increasing and unbounded function of r, for r ≥ r 0 . Now we can state our theorem:

Proof of Theorem 3.2.
We present the proof in several steps.
Step I. First we prove that Take any ε > 0 and note that is an increasing function of r for all large r. Hence, if R is large enough, then which implies (4.1).
Step II. Next, we prove that for all large r, If this was not true, then for all t ∈ (r , r + r/(log r) γ ), which in the case where 1 < β ≤ 2 means "for all t ∈ (r , 2r)", we would have This would imply that It is easily checked that the last expression is equal to log 2 if γ is zero and is 1 + o (1) if γ is positive. Thus the integral ∞ 1 M (t) e −α (log t) β dt would not be convergent, contradicting our hypothesis. Hence (4.2) holds. This means that for all large r, M (λr) < (log λr) γ λr e α (log λr) β for some λ ∈ 1 , 1 + 1 (log r) γ .
Step III. Since log M (r) is a convex function of log r, we have This is our main tool. We use (4.1) and (4.3) in (4.4) to conclude that (M (r)) 2 ≤ c 1 (r) (log(r/λ)) β−1 r/λ e α (log(r/λ)) β · (log(λr)) γ λr e α (log(λr)) β , where c 1 (r) = o(1) as r → ∞. Now, note that where c 2 (r) = o(1) as r → ∞, and where we have used (4.6) in the last line. Note that β − 2γ − 2 is negative if 1 < β < 2 and zero if β ≥ 2. Hence (log r) β−2γ−2 = O(1) as r → ∞. This allows us to conclude that where c 3 (r) = o(1) as r → ∞, which is equivalent to (4.5). Inequality (4.5) is considerably stronger than (4.1) and provides a better estimate for M (r/λ) in (4.4). Using (4.5) and (4.3) in (4.4) the way (4.1) and (4.3) were used above in (4.4), we obtain which may in turn be used to conclude that Clearly, (4.8) is stronger than (4.7). Since this process can go on indefinitely, we see that for any positive integer k, we have from which the desired result follows. 2 Remark. The property of the function M (r) by which log M (r) is a convex function of log r is shared by some other functions associated with an entire function f, which makes the proof of Theorem 3.2 applicable to these associated functions. In fact, let f (z) := ∞ ν=0 a ν z ν be an entire function; for any r > 0, we define the function M p (r) = M p (f ; r) by  [3] for log M p (r). The reader might find [6] to be of some interest in this connection.