Let $\phi$ denote the standard Gaussian pdf, i.e., $\phi(x) = \frac{1}{\sqrt{2\pi}}e^{-x^2/2}$ for $x\in \mathbb{R}$; and $\alpha,\delta\in(0,1]$. Define $t\in\mathbb{R}$ as $$ \alpha = \int_t^\infty \phi\,. \tag{1} $$ I want to (upper) bound the quantity $\int_{t-\delta}^\delta\phi$ as a function of $\delta,\alpha$; assuming $\alpha < 1/2$ is such that $t\geq \delta$. What I can get is $$ \int_{t-\delta}^\delta\phi \leq \delta \phi(t-\delta) = \delta e^{-\delta^2/2}e^{t\delta}\phi(t) \leq \delta \cdot e^{-\delta^2/2}e^{\delta\sqrt{2\ln(1/\alpha)}}\cdot\alpha \sqrt{2\log(1/\alpha)} \tag{2} $$ using the fact that, from (1), $t \leq \sqrt{2\ln(1/\alpha)}$, and the fact that (I think I can prove that) $\phi(t) \leq \alpha \sqrt{2\log(1/\alpha)}$.
My issue is with the middle term $e^{-\delta^2/2}e^{\delta\sqrt{2\ln(1/\alpha)}}$, and more precisely the factor $e^{\delta\sqrt{2\ln(1/\alpha)}}$, which I would really like not to be there to have the $\delta$ and $\alpha$ parts "decoupled."
Is it necessary? Can one obtain a better bound than (2) (without assuming anything on the relation between $\delta$ and $t$ besides $t\geq \delta$?)