For ease of exposition, I will stick to the simplest case: consider the Eisenstein series for $SL_2(\bf R)$ $$E(z,s)=\sum_{\gamma\in P_{\bf Z}\backslash SL_2(\bf Z)}\text{Im}(\gamma z)^s=\sum_{(c,d)\in{\bf Z}\backslash (0,0)}\frac{y^s}{|cz+d|^{2s}}$$ and its truncation used to compute the inner product of Eisenstein series (called the Maaß-Selberg relation) over $SL_2(\bf Z)\backslash H$, $$\Lambda^TE(z,s)=\begin{cases} E(z,s) & y<T\\E(z,s)-y^s-\varphi(s)y^{1-s}&y>T\end{cases}$$ (I am being sloppy with the difference between the 'naive' truncation and Arthur's truncation, but I think in this case it does not make a difference; please correct me otherwise.)

My question concerns the parameter $T$. The first condition is that $T>0$, large enough that the truncated fundamental domain is such that the curve $y=T$ cuts the fundamental domain only on the two sides that pass through the cusp, and such that the truncated function has rapid decay at cusps to be automorphic.

What is the smallest value that $T$ can take so that the inner product formula remains valid?

Why I am asking: from what I understand, Selberg in Harmonic Analysis (Collected works, p.633) says that $T=1$ will suffice, while Garrett in his note Simplest Example of Truncation and Maaß-Selberg Relations claims $T=\sqrt3/2$ (p.2). Unfortunately, a computation I am making seems to fail for such small values, but holds for some other given $T\gg0$. I would like to know if there is a computational error, or I should expect the Maass-Selberg relation to fail or alter at some limit.

EDIT: I am starting a bounty, as after making certain computations with Hejhal's calculation and my own formulas I am sure something is fishy. From Hejhal in The Selberg Trace Formula and the Riemann Zeta Function one sees that
$$\frac{1}{4\pi}\int^{\infty}_{-\infty}h(r)\int_{\Gamma\backslash\bf H}|\Lambda^T E(z,\frac{1}{2}+ir)|^2dz\ dr$$
is equal to the sum of
$$\frac{1}{4}m(\frac{1}{2})h(0)-g(0)\log\pi+\frac{1}{2\pi}\int^\infty_{-\infty}h(r)\frac{\Gamma'}{\Gamma}(1+ir)dr-2\sum_{n=1}^\infty\frac{\Lambda(n)}{n}g(2\log n)$$
where $\Lambda(n)$ is the von Mangoldt function, and
$$+\frac{1}{4\pi i}\int^\infty_{-\infty}m(\frac{1}{2}-ir)h(r)e^{2ir \log T}\frac{dr}{r}+g(0)\log T$$
where $m(s)=\xi(2-2s)/\xi(2s),$ the quotient of completed Riemann zeta functions. Then taking as a test function $h(r)=e^{-r^2}$ (which is allowable), and $T$ very close to 1, one finds that the second line is negligible, the digamma function is negative, and the resulting expression is **negative**, whereas the original integral is clearly positive. More generally, if one chooses $g(0)$ or $h(0)$ to be zero, the expression is even more likely negative.

Certainly for $T$ large enough, the expression is positive for positive test functions. I have been unable to find accurate information on what $T$ is allowable for the inner product formula to hold.

precisely(note that I am familiar with the Maass-Selberg relations). From the classical fundamental domain for $\mathrm{SL_2}(\mathbf{Z})\backslash\mathbf{H}$ it is clear that $T\geq\sqrt{3}/2$ is necessary if you want the natural projection from $\{z:\ |\Re(z)|<1/2,\ \Im(z)>T\}$ to $\mathrm{SL_2}(\mathbf{Z})\backslash\mathbf{H}$ to be injective. $\endgroup$necessary, as I should have said clearly in my note, whatever I did happen to say. As Peter Humphries quite correctly observes in his answer, a naive form of truncation can go that low, ... but it's dangerous. So, no, one ought not try to truncate below the bottom of a Siegel set that behaves well. No doubt. No contest. What is theunderlyingissue? $\endgroup$