Asymptotic Behavior of the Integral 
I
(
t
)
=
∫
1
1
/
2
log
⁡
∣
ζ
(
a
+
i
t
)
∣
 
d
a
I(t)=∫ 
1
1/2
​	
 log∣ζ(a+it)∣da as 
t
→
∞
t→∞
To analyze the asymptotic behavior of the integral

I
(
t
)
=
∫
1
1
/
2
log
⁡
∣
ζ
(
a
+
i
t
)
∣
 
d
a
,
I(t)=∫ 
1
1/2
​	
 log∣ζ(a+it)∣da,
as 
t
→
∞
t→∞, we begin by examining the known behavior of 
log
⁡
∣
ζ
(
a
+
i
t
)
∣
log∣ζ(a+it)∣ in different regions of 
a
a as 
t
t grows large.

1. Behavior of 
log
⁡
∣
ζ
(
a
+
i
t
)
∣
log∣ζ(a+it)∣

For large 
t
t, the behavior of 
log
⁡
∣
ζ
(
a
+
i
t
)
∣
log∣ζ(a+it)∣ varies depending on the real part 
a
a.

a. For 
a
=
1
a=1:
Near 
a
=
1
a=1, the zeta function exhibits a pole at 
s
=
1
s=1, which implies that
ζ
(
1
+
i
t
)
→
∞
ζ(1+it)→∞ as 
a
→
1
a→1. The asymptotic behavior of 
ζ
(
1
+
i
t
)
ζ(1+it) is well-approximated by:

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→∞.
Thus, as 
t
t increases, 
log
⁡
∣
ζ
(
a
+
i
t
)
∣
log∣ζ(a+it)∣ near 
a
=
1
a=1 grows negatively, contributing a large negative term to the integral.

b. For 
a
=
1
/
2
a=1/2:
Near 
a
=
1
/
2
a=1/2, the function 
ζ
(
1
/
2
+
i
t
)
ζ(1/2+it) exhibits more oscillatory behavior due to the influence of the non-trivial zeros of the Riemann zeta function. While
ζ
(
1
/
2
+
i
t
)
ζ(1/2+it) fluctuates, the average size of 
∣
ζ
(
1
/
2
+
i
t
)
∣
∣ζ(1/2+it)∣ can be described asymptotically as:

log
⁡
∣
ζ
(
1
/
2
+
i
t
)
∣
=
O
(
log
⁡
t
)
.
log∣ζ(1/2+it)∣=O(logt).
Therefore, while the function exhibits strong oscillations near the critical line
a
=
1
/
2
a=1/2, its logarithmic growth is much smaller than near 
a
=
1
a=1.

2. Approximating the Integral

Given the behaviors at 
a
=
1
a=1 and 
a
=
1
/
2
a=1/2, we can now approximate the integral 
I
(
t
)
I(t). For large 
t
t, the dominant contribution comes 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 of smaller order and does not dominate the asymptotics.

Thus, we approximate the integral as:

I
(
t
)
∼
∫
1
1
/
2
−
log
⁡
t
 
d
a
.
I(t)∼∫ 
1
1/2
​	
 −logtda.
This simplifies to:

I
(
t
)
∼
−
log
⁡
t
⋅
∫
1
1
/
2
d
a
=
−
log
⁡
t
⋅
(
1
−
1
2
)
=
−
1
2
log
⁡
t
.
I(t)∼−logt⋅∫ 
1
1/2
​	
 da=−logt⋅(1− 
2
1
​	
 )=− 
2
1
​	
 logt.
3. Conclusion

As 
t
→
∞
t→∞, the integral 
I
(
t
)
I(t) behaves asymptotically as:

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) arises from the region near 
a
=
1
a=1, where 
ζ
(
a
+
i
t
)
∼
1
1
+
i
t
ζ(a+it)∼ 
1+it
1
​	
 , leading to a logarithmic decay.