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).