I would like to obtain the asymptotic behavior as $x \rightarrow -\infty$ of the following function $f(x)$:
$$f(x)= c e^{\frac{a x^2}{2}} \int_{-\infty}^x e^{-\frac{a \eta^2}{2}} d \eta$$
where $a~\text{ and}~ c$ are constants. How can the asymptotic be obtained? I tried Taylor expansions for the exponential and then integrating that but I don't quite understand it.
You can apply integration by parts: $$e^{a x^2/2} \int_{-\infty}^x -\frac 1 {a \eta} d(e^{-a \eta^2/2}) = -\frac 1 {a \eta} e^{a x^2/2 - a \eta^2/2} \bigg \rvert_{\eta = -\infty}^x - e^{a x^2/2} \int_{-\infty}^x \frac 1 {a \eta^2} e^{-a \eta^2} d\eta$$ and show that the second term on the rhs is asymptotically smaller, or you can apply Laplace's method: $$e^{a x^2/2} \int_{-\infty}^x e^{-a \eta^2/2} d\eta = \int_{-\infty}^0 e^{-a x u - a u^2/2} du \sim e^{-a u^2/2} \bigg \rvert_{u = 0} \int_{-\infty}^0 e^{-a x u} du.$$