What sort of bounds (explicit of preference) can one give for
$$\int\_T^{2 T} \frac{dt}{|\zeta(1+i t)|^2} \;\;\;\;\;?$$
Some obvious points:


* One can give a pointwise bound $\frac{1}{|\zeta(1+ it)|} \leq C \log t$ (with $C\leq 42.9$ for $t\geq 2$) and deduce a bound of the form $\leq K T (\log T)^2$ on the integral above. If possible, we would like to do better (asymptotically and also as far as the constants involved are concerned).
* To know that the integral is always finite, one needs to know that $\zeta(1+it)\ne 0$ for all $t$. Thus, it is not enought to just apply a mean-value theorem.



Bound on $L^2$ norm of $1/\zeta(1+i t)$?
Problems like this are classical (as noted in Terry's answer), and have been considered more recently with attention to uniformity in the moments. To give a quick indication, one can show that
$$ 
\zeta(1+it) \approx \prod\_{p\le y} \Big( 1-\frac{1}{p^{1+it}}\Big)^{-1} 
$$
for all but a set of values $T\le t\le 2T$ of small measure, provided $y$ is bigger than some power of $\log T$. Such ideas go back to Littlewood (on RH, which can then be removed via zero-density results). See Lemmas 2.1 and 2.2 in [Extreme values of $|\zeta(1+it)|$](https://arxiv.org/pdf/math/0501232.pdf) for example. From this one can work out uniform moments of $\zeta(1+it)$ (positive, negative, complex ...) and understand its distribution. Look in the cited paper, and also work of Lamzouri.


The result is
$$ 
\frac 1T \int\_T^{2T} |\zeta(1+it)|^{-2} dt \sim \sum\_{n=1}^{\infty} \frac{\mu(n)^2}{n^2} = \frac{\zeta(2)}{\zeta(4)}, 
$$
as one would expect.


An alternative (which may be useful if one wishes to get explicit results, and avoid zero-density estimates) is to use a (very) long approximation to $1/\zeta(1+it)$. For example, first show that
$$ 
\frac{1}{\zeta(1+it)} \approx \sum\_{n\le \exp((\log T)^3)} \frac{\mu(n)}{n^{1+it}}. 
$$
This would need an explicit zero free region -- I would guess something off-the-shelf from the work of Kadiri et al should be good, and you can make the sum even longer. Usually such a long approximation would be useless, but on the $1$-line you can do whatever you want. Now one can show (easiest to smooth and unsmooth) that
$$ 
\int\_T^{2T} \frac{dt}{|\zeta(1+it)|^2} = 
\sum\_{n\le \exp((\log T)^3) } (T+O(n)) \frac{\mu(n)^2}{n^2} 
= \frac{\zeta(2)}{\zeta(4)} T + O((\log T)^3). 
$$
It shouldn't be too bad to make this explicit, but not something I'd care to do!


