How much can an Eisenstein series be truncated? 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.
 A: I think the answer truly does depend on whether you use the naive truncation, as in Iwaniec's book, or the Arthur truncation, as in Paul Garrett's note. In particular, the method of proof for the Maaß-Selberg relation with the naive truncation is proved in Iwaniec's book using Green's identity, and as GH from MO mentioned in his comment, the geometric picture here suggests that the proof only requires that $T \geq \sqrt{3}/2$.
On the other hand, I believe that with the Arthur truncation, one really does require that $T \geq 1$, with the proof of the Maaß-Selberg relation now instead via an unfolding argument as in Paul Garrett's notes. Here the Arthur truncation is
\[\Lambda^T E(z,s) = E(z,s) - \sum_{\substack{\gamma \in \Gamma_{\infty} \backslash \mathrm{SL}_2(\mathbb{Z}) \\ \Im(\gamma z) > T}} c_0 E(\gamma z,s),
\]
where for $z = x + iy \in \mathbb{H}$,
\[c_0 E(\gamma z,s) = \int_{0}^{1} E(\gamma z,s) \, dx,\]
so that $c_0 E(z,s) = y^s + \varphi(s) y^{1-s}$.
One can easily show that if $\gamma = \begin{pmatrix} a & b \\\ c & d \end{pmatrix} \in \mathrm{SL}_2(\mathbb{Z})$ with $\gamma \notin \Gamma_{\infty}$, so that $c \neq 0$, then
\[\Im(z) \Im(\gamma z) = \frac{y^2}{(cx + d)^2 + c^2 y^2} \leq 1,\]
while $\Im(\gamma z) = \Im(z)$ for $\gamma \in \Gamma_{\infty}$.
So for $T \geq 1$, the Arthur truncation is equal to
\[\Lambda^T E(z,s) = \begin{cases}
E(z,s) & \text{if $1/T \leq y \leq T$,} \\\
E(z,s) - y^s - \varphi(s) y^{1-s} & \text{if $y > T$.}
\end{cases}\]
If $y < 1/T$, then there may be more terms (and which terms are also present will now depend on $x$ as well), as there may be more coset representatives $\gamma \in \Gamma_{\infty} \backslash \mathrm{SL}_2(\mathbb{Z})$ other than just the identity for which $\Im(\gamma z) > T$.
If $T < 1$, then $\Lambda^T E(z,s) = E(z,s) - y^s - \varphi(s) y^{1-s}$ if $y \geq 1/T$, but there may be more terms if $y < 1/T$.
In particular, the Arthur truncation coincides with the naive truncation on the standard fundamental domain if $T \geq 2/\sqrt{3}$, but if $T < 2/\sqrt{3}$ then this is no longer the case.
Now a key step in proving the Maaß-Selberg relation is showing that
\[\int_{\mathrm{SL}_2(\mathbb{Z}) \backslash \mathbb{H}} \Lambda^T E(z,s) \left(\overline{\Lambda^T E(z,r)} - \overline{E(z,r)}\right) \, d\mu(z) = 0.\]
One shows this by unfolding to find that this is
\[-\int_{T}^{\infty} \overline{c_0 E(z,r)} \int_{0}^{1} \Lambda^T E(z,s) \, dx \, dy,\]
using the fact that $c_0 E(z,r) = y^r + \varphi(r) y^{1-r}$ does not depend on $x$.
If $T \geq 1$, then $\Lambda^T E(z,s) = E(z,s) - c_0 E(z,s)$ for $T < y < \infty$ and $0 < x < 1$, and so the inner integral vanishes. But if $T < 1$, then there may be other terms, and so this is no longer the case. So the condition $T \geq 1$ truly is necessary here.
EDIT: With regards to your comment, note that the Maaß-Selberg relation states that
\[\int_{\mathrm{SL}_2(\mathbb{Z}) \backslash \mathbb{H}} \Lambda^T E(z,s) \overline{\Lambda^T E(z,r)} \, d\mu(z) = \frac{T^{s + \overline{r} - 1}}{s + \overline{r} - 1} + \overline{\varphi(r)} \frac{T^{s - \overline{r}}}{s - \overline{r}} + \varphi(s) \frac{T^{\overline{r} - s}}{\overline{r} - s} + \varphi(s) \overline{\varphi(r)} \frac{T^{1 - s - \overline{r}}}{1 - s - \overline{r}}.
\]
Here
\[\varphi(s) = \frac{\Lambda(2 - 2s)}{\Lambda(2s)}, \qquad \Lambda(s) = \pi^{-s/2} \Gamma\left(\frac{s}{2}\right) \zeta(s),\]
so that $|\varphi(1/2 + it)|^2 = 1$ for all $t \neq 0$, and taking logarithmic derivatives shows that
\[\frac{\varphi'}{\varphi}\left(\frac{1}{2} + it\right) = -2\Re\left(\frac{\Gamma'}{\Gamma}\left(\frac{1}{2} + it\right)\right) - 4\Re\left(\frac{\zeta'}{\zeta}(1 + 2it)\right) + 2 \log \pi.\]
Setting $s = r = 1/2 + it + \varepsilon$ for $t \neq 0$ and $\varepsilon > 0$, taking the limit as $\varepsilon \to 0$, and using the Laurent expansions of each term, we find that
\[\int_{\mathrm{SL}_2(\mathbb{Z}) \backslash \mathbb{H}} \left|\Lambda^T E\left(z,\frac{1}{2} + it\right)\right|^2 \, d\mu(z) = 2 \log T - \frac{\varphi'}{\varphi}\left(\frac{1}{2} + it\right) - \Im \left( \varphi\left(\frac{1}{2} + it\right) \frac{T^{-2it}}{t}\right).
\]
Now it's not obvious to me either that this is nonnegative, and of course if $T$ is very close to zero then you wouldn't expect it to be. But for $T$ near $1$ (because as mentioned earlier, even using the naive truncation we can't have $T < \sqrt{3}/2$) it's not obvious to me either whether this is negative, so I'm not sure if there's truly a problem here; certainly for large $t$, the digamma function should dominate. Of course, I could be missing something here.
SECOND EDIT: I still don't understand your objection; as I have already stated, one requires that $T \geq 1$ for the Arthur truncation, and this is also sufficient (i.e. with the Arthur truncation, the Maaß-Selberg relation holds for all $T \geq 1$). Indeed, I already showed that
\[\int_{\mathrm{SL}_2(\mathbb{Z}) \backslash \mathbb{H}} \Lambda^T E(z,s) \left(\overline{\Lambda^T E(z,r)} - \overline{E(z,r)}\right) \, d\mu(z) = 0\]
when $T \geq 1$, so it remains to show that
\[\int_{\mathrm{SL}_2(\mathbb{Z}) \backslash \mathbb{H}} \Lambda^T E(z,s) \overline{E(z,r)} \, d\mu(z) = \frac{T^{s + \overline{r} - 1}}{s + \overline{r} - 1} + \overline{\varphi(r)} \frac{T^{s - \overline{r}}}{s - \overline{r}} + \varphi(s) \frac{T^{\overline{r} - s}}{\overline{r} - s} + \varphi(s) \overline{\varphi(r)} \frac{T^{1 - s - \overline{r}}}{1 - s - \overline{r}}.
\]
By the definition of the Arthur truncation, the left-hand side is
\[ \int_{\mathrm{SL}_2(\mathbb{Z}) \backslash \mathbb{H}} \sum_ {\substack{\gamma \in \Gamma_{\infty} \backslash \mathrm{SL}_2(\mathbb{Z}) \\ \Im(\gamma z) \leq T}} \Im(\gamma z)^s \overline{E(z,r)} \, d\mu(z) - \varphi(s) \int_{\mathrm{SL}_2(\mathbb{Z}) \backslash \mathbb{H}} \sum_{\substack{\gamma \in \Gamma_{\infty} \backslash \mathrm{SL}_2(\mathbb{Z}) \\ \Im(\gamma z) > T}} \Im(\gamma z)^{1-s} \overline{E(z,r)} \, d\mu(z),
\]
which, by the $\mathrm{SL}_2(\mathbb{Z})$-invariance of $E(z,r)$, unfolds to
\[ \int_{0}^{T} y^s \int_{0}^{1} \overline{E(z,r)} \, \frac{dx \, dy}{y^2} - \varphi(s) \int_{T}^{\infty} y^{1-s} \int_{0}^{1} \overline{E(z,r)} \, \frac{dx \, dy}{y^2}.
\]
By the definition of the constant term of $E(z,r)$ (namely that it is $y^r + \varphi(r) y^{1-r}$), one can evaluate these integrals, with the result being the Maaß-Selberg relation.
Of course, to ensure convergence of all the integrals involved, this is initially only valid for $\Re(s), \Re(r) > 1$ with $\Re(s) - \Re(r) > 1$, but then by analytic continuation this extends to all $s,r \in \mathbb{C}$ with $s \neq \overline{r}$ and $s + \overline{r} \neq 1$ provided $\varphi(s), \varphi(r)$ are well-defined. As before, one can extend this to $s = r = 1/2 + it$ with $t \neq 0$ by setting $s = r = 1/2 + it + \varepsilon$ and taking the limit as $\varepsilon \to 0$.
All this is valid if $T \geq 1$, so one certainly should be able to take $T = 1$ in the Maaß-Selberg relation, unless I'm missing something.
With regards to the integral
\[\int_{-\infty}^{\infty} h(r) \int_{\mathrm{SL}_2(\mathbb{Z}) \backslash \mathbb{H}} \left|\Lambda^T E\left(z,\frac{1}{2} + ir\right)\right|^2 \, d\mu(z) \, dr,\]
to me it looks like there may be an issue at $r = 0$, and one must be careful when breaking up the integral and blindly using Weil's explicit formula. (In particular, I don't see why you say the second line is negligible - doesn't the integrand blow up at $r = 0$?) But I haven't looked too closely at this.
