Let's explore the behavior of the integral I ( t ) I(t) as t → ∞ t→∞. The integral is:

I ( t )

∫ 1 1 / 2 log ⁡ ∣ ζ ( a + i t ) ∣ d a . I(t)=∫ 1 1/2 ​
log∣ζ(a+it)∣da. We aim to understand the asymptotic behavior of I ( t ) I(t) as t → ∞ t→∞.

Step 1: Decomposing the Integral The function ζ ( s ) ζ(s), for s

a + i t s=a+it, can exhibit very different behaviors depending on the real part a a. The integration is taken over the interval a ∈ [ 1 / 2 , 1 ] a∈[1/2,1], so we are considering a region near and slightly off the critical strip, where the zeros of the Riemann zêta function play a crucial role in its behavior.

We need to understand the nature of log ⁡ ∣ ζ ( a + i t ) ∣ log∣ζ(a+it)∣ as t → ∞ t→∞ for a a in this range.

Step 2: Asymptotics of log ⁡ ∣ ζ ( a + i t ) ∣ log∣ζ(a+it)∣ For large t t, the behavior of ζ ( s ) ζ(s) is well-understood, especially for s

a + i t s=a+it. There are different regimes for the zêta function depending on whether a a is close to 1 or close to 1/2.

For a

1 a=1: Near a

1 a=1, we use the known fact that ζ ( s ) ζ(s) has a pole at s

1 s=1, meaning that as a → 1 a→1, ζ ( s ) → ∞ ζ(s)→∞. The logarithm of ∣ ζ ( s ) ∣ ∣ζ(s)∣ grows significantly near a

1 a=1. Specifically, we know that ζ ( 1 + i t ) ∼ 1 1 + i t , ζ(1+it)∼ 1+it 1 ​
, which leads to

log ⁡ ∣ ζ ( 1 + i t ) ∣ ∼ − log ⁡ t as t → ∞ . log∣ζ(1+it)∣∼−logtast→∞. For a

1 / 2 a=1/2: Near a

1 / 2 a=1/2, ζ ( s ) ζ(s) oscillates more strongly because we are near the critical line where the non-trivial zeros of the zêta function lie. In this regime, the behavior of log ⁡ ∣ ζ ( 1 / 2 + i t ) ∣ log∣ζ(1/2+it)∣ is more complicated due to the influence of these zeros. However, it is known that the amplitude of ∣ ζ ( 1 / 2 + i t ) ∣ ∣ζ(1/2+it)∣ fluctuates as t → ∞ t→∞, but the average size of ∣ ζ ( 1 / 2 + i t ) ∣ ∣ζ(1/2+it)∣ behaves asymptotically as log ⁡ ∣ ζ ( 1 / 2 + i t ) ∣

O ( log ⁡ t ) log∣ζ(1/2+it)∣=O(logt). Step 3: Approximation of the Integral for Large t t Given the behaviors near a

1 a=1 and a

1 / 2 a=1/2, we now approximate the integral I ( t ) I(t). In the large t t limit:

Near a

1 a=1, log ⁡ ∣ ζ ( 1 + i t ) ∣ ∼ − log ⁡ t log∣ζ(1+it)∣∼−logt, which contributes negatively to the integral. Near a

1 / 2 a=1/2, log ⁡ ∣ ζ ( 1 / 2 + i t ) ∣ log∣ζ(1/2+it)∣ fluctuates but is on the order of O ( log ⁡ t ) O(logt). Thus, for large t t, we expect the dominant contribution to come from the region near a

1 a=1, where log ⁡ ∣ ζ ( a + i t ) ∣ ∼ − log ⁡ t log∣ζ(a+it)∣∼−logt. The contribution from near a

1 / 2 a=1/2 is bounded and smaller compared to the behavior near a

1 a=1.

Step 4: Asymptotic Behavior of I ( t ) I(t) Using the approximation near a

1 a=1, we find:

I ( t ) ∼ ∫ 1 1 / 2 − log ⁡ t d a

− log ⁡ t ∫ 1 1 / 2 d a

− log ⁡ t ⋅ ( 1 − 1 2 )

− 1 2 log ⁡ t . I(t)∼∫ 1 1/2 ​
−logtda=−logt∫ 1 1/2 ​
da=−logt⋅(1− 2 1 ​
)=− 2 1 ​
logt. Thus, the integral I ( t ) I(t) behaves asymptotically as t → ∞ t→∞ like:

I ( t ) ∼ − 1 2 log ⁡ t . I(t)∼− 2 1 ​
logt. Conclusion As t → ∞ t→∞, the integral I ( t ) I(t) has the following asymptotic behavior:

I ( t ) ∼ − 1 2 log ⁡ t . I(t)∼− 2 1 ​ logt. This result reflects the fact that the dominant contribution to I ( t ) I(t) comes from the region near a

1 a=1, where the zêta function behaves like 1 1 + i t 1+it 1 ​
, leading to a logarithmic decay of I ( t ) I(t).