I will take $\sigma^2 = 1$, and ignore some fine details in what follows, that is, not all $i$'s will be crossed nor will all $t$'s be dotted.
Let $X$ and $Y$ denote independent standard normal random variables. Then,
the result that you are trying to calculate looks very much like the
probability that the random point $(X, Y)$ lies inside the circle of
radius $d$ centered at $(0, \alpha)$, that is, in the disc of radius
$d$ centered at $(0, \alpha)$. For, conditioned on $X = x$,
where $\vert x \vert < d$, the line $x = d$ crosses the circle at $y = \alpha - \sqrt{d^2 - x^2}$ and at $y = \alpha + \sqrt{d^2 - x^2}$. Thus,
$$
\begin{align*}
P\{(X, Y) \in \text{disc} ~ \mid X = x\}
&= P\{\alpha - \sqrt{d^2 - x^2} < Y < \alpha + \sqrt{d^2 - x^2}\}\\
&= \Phi\left (\alpha + \sqrt{d^2 - x^2}\right)
- \Phi\left (\alpha - \sqrt{d^2 - x^2}\right)
\end{align*}
$$
and it follows that
$$
P\{(X, Y) \in \text{disc} \}
= \int_{-d}^d \left [ \Phi\left (\alpha + \sqrt{d^2 - x^2}\right)
- \Phi\left (\alpha - \sqrt{d^2 - x^2}\right) \right ]\phi(x) \mathrm dx.
$$
This looks pretty much like the integral you want to evaluate.
To the best of my knowledge, there is no closed-form expression for this integral
except when $\alpha = 0$ when it should work out to $1 - \exp(-d^2)$. For
$\alpha \neq 0$, I suggest bounding the desired probability from above by the
probability that $(X,Y)$ is in the circumscribing square of side $2d$, viz.
$$
\begin{align*}
P\{\vert X \vert < d, \alpha - d < Y < \alpha + d\}
&= P\{\vert X \vert < d\}P\{\alpha - d < Y < \alpha + d\}\\
&=\left [\Phi(d) - \Phi(-d)\right ]
\left [\Phi(\alpha + d) - \Phi(\alpha -d)\right ]
\end{align*}
$$
and bounding it from below by the probability that $(X, Y)$
is in the inscribed square of side $\sqrt{2}d$. I will leave
the details to you.