Let $0<\sigma\leq 1$. Let $T$ be large. How can we give good explicit $L^2$ bounds on the tails of $\zeta(\sigma+it)$? That is, we want to bound the quantity $$\int\_{\sigma-i\infty}^{\sigma-iT} + \int\_{\sigma+iT}^{\sigma+i\infty} \frac{|\zeta(s)|^2}{|s|^2} ds.$$
Morally, we should expect something about as good as $1/T$ for $\Re(s)\geq 1/2$, and presumably $1/T^{2\sigma}$ for $0<\Re(s)\leq 1/2$ (or maybe I ate a $\log$). In particular, we should certainly expect to do better than what we can get from convexity bounds on individual values of $\zeta(s)$. (Such bounds are, incidentally, explicit (Backlund)).




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How to go about this? One inviting approach is to use the fact that the Mellin transform is an isometry. (That is, if $F$ is the Mellin transform of $f$, then $\frac{1}{2\pi} \int\_{-\infty} |F(\sigma+it)|^2 dt=\int\_0^\infty |f(x)|^2 x^ {2\sigma-1}$.) Recall that $\zeta(s)/s$ is the Mellin transform of $x\mapsto \lfloor 1/x\rfloor$. An obvious (even crass) sub-approach would be to let $$g(x) = \frac{1}{2 \pi i} \int\_{\sigma-iT}^{\sigma+iT} \frac{\zeta(s)}{s} x^{-s} ds,$$ and then note that the Mellin transform of $\lfloor 1/x\rfloor - g(x)$, evaluated at $\sigma+it$, equals $0$ when $|t|< T$ and $\zeta(s)/s$ when $|t|>T$. (I'm papering over the fact that we should check that there is no pole at $s=1$, so that we can analytically continue the Mellin transform to the left of $\Re s=1$.) It follows, by isometry, the quantity in the first displayed equation above (the integral over the tails) equals $$\int\_0^\infty |\lfloor 1/x\rfloor - g(x)|\cdot x^{2\sigma-1} dx.$$ We can write down $g(x)$ explicitly, though it is not completely trivial; if I understand correctly (see Theorem 5.2 in Montgomery-Vaughan), $$g(x) = \frac{1}{\pi} \sum\_{n\geq 1} \mathrm{si}(T \log n x),$$ where $\mathrm{si}$ is the sine integral function. Of course the problem here is that $g(x)$ is not a particularly pleasant function to work with.


My instinct is to work instead with a function of the form $$\frac{1}{2\pi i} \int\_{\sigma-iT}^{\sigma+iT} \frac{\zeta(s)}{s} \phi(s) x^{-s} ds,$$ where $\phi(s)$ is close to $0$ for $s = \sigma+i t$, $|t|\geq T$, and is close to $1$ on most of the segment $\lbrack \sigma-iT,\sigma+iT\rbrack$. I haven't hit on the right choice of weight $\phi$, though. It should give some sort of advantage over $g$ above, or else it is pointless.


Any other approaches? Is this all known (explicitly)?



$L_2$ bounds for tails of $\zeta(s)$ on a vertical line
Let $\sigma>0$ be fixed, and let $T\geq 2$ be a parameter.


By Theorem 4.11 in Titchmarsh: The theory of the Riemann zeta-function,
$$ \zeta(\sigma+it)=\sum\_{n\leq T}n^{-\sigma-it}+O(T^{-\sigma}),\qquad T<|t|\leq 2T.$$
It follows that
$$ \int\_{\sigma+iT}^{\sigma+2T}|\zeta(s)|^2\,ds\ll T^{1-2\sigma}+\int\_T^{2T}\left|\sum\_{n\leq T}n^{-\sigma-it}\right|^2\,dt.$$
We estimate the integral on the right hand side by Theorem 9.1 in Iwaniec-Kowalski: Analytic number theory. We obtain
$$ \int\_{\sigma+iT}^{\sigma+2iT}|\zeta(s)|^2\,ds\ll T\sum\_{n\leq T}n^{-2\sigma}\ll\begin{cases}T,&\sigma>1/2;\\T\log T,&\sigma=1/2;\\T^{2-2\sigma},&\sigma<1/2.\end{cases}$$
Finally, applying a dyadic decomposition in the original integral and using the last bound, we conclude
$$ \int\_{\sigma+iT}^{\sigma+i\infty}\frac{|\zeta(s)|^2}{|s|^2}\,ds\ll\begin{cases}T^{-1},&\sigma>1/2;\\T^{-1}\log T,&\sigma=1/2;\\T^{-2\sigma},&\sigma<1/2.\end{cases}$$
As you expected. (Note: implied constants depend on $\sigma$.)


