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.