Actually, for a small $d\in (0,1)$, I am interested in the asymptotic for $$\int_0^1 z e^{-\frac{(d+z)^2}{a}} \text d z$$ as $a\to 0$. To simplify this a bit I wanted to look at the upper bound $$\int_0^1 e^{-\frac{(d+z)^2}{a}} \text d z,$$ which decreases faster than $e^{-\frac{d^2} a}$ and slower than $e^{-\frac{(d+\varepsilon)^2} a}$ for any $\varepsilon \in (0,1)$.
Is there a "closed form" of the asymptotic behaviour of those integrals?
I did not try to calculate the integral value, because the actual integral involves the additional factor $1/(d+z)^2$ under integration.
$$ \int_0^1 z e^{-(d+z)^2/a} dz\sim e^{-d^2/a}\left( \frac{a^2}{4d^2} - \frac{3a^3}{8d^4} + \frac{15a^4}{16d^6} + \sum_{s=3}^\infty c_s \, (s+1)! \,a^{s+2}\right).$$
Apply Laplace's method. (See II.Theorem 1 in Asymptotic Approximations Of Integrals by R. Wong; the proof is based on Watson's lemma.)