Let $I(\lambda)=\int_{\lambda}^{\infty}e^{-t^2}dt$. We want to find the dominant term of $I(\lambda)$. For this problem, I wonder whether my approach is correct or not, and where my solution fails if it does.
The first way of solving this problem is that letting $t=\lambda + s$ yields the following integral, \begin{align} I(\lambda) &= \int_{0}^{\infty}e^{-\lambda^2 -2\lambda s -s^2}ds, \\ I(\lambda) &= e^{-\lambda^2}\int_{0}^{\infty}e^{-2\lambda s} e^{-s^2}ds. \end{align} Then for another change of variable, let $2s=u$ so that \begin{align} I(\lambda) &= \frac{e^{-\lambda^2}}{2}\int_{0}^{\infty}e^{-\lambda u} e^{-u^2/4}du, \\ I(\lambda) &= \frac{e^{-\lambda^2}}{2}\int_{0}^{\infty}e^{-\lambda u}(1-u^2/4+\dots)du,\\ I(\lambda) &= \frac{e^{-\lambda^2}}{2}\bigg[\int_{0}^{\infty}e^{-\lambda u}-\frac{1}{4}\int_{0}^{\infty}u^2e^{-\lambda u}du+ \dots\bigg],\\ I(\lambda) &= \frac{e^{-\lambda^2}}{2}\bigg(\frac{1}{\lambda} - O(1/\lambda^3) \bigg), \\ I(\lambda) &\sim \frac{e^{-\lambda^2}}{2\lambda}. \end{align}
I tried to solve this problem with a different approach. So what I observe is that the maximum value of $-t^2$ over $[\lambda,\infty)$ is taken at $\lambda$. Then I substitute the Taylor expansion of $-t^2$ about $t=\lambda$ and get \begin{equation} I(\lambda) \sim \int_{\lambda}^{\infty}e^{-\lambda^2 -2\lambda(t-\lambda)}dt. \end{equation} Let $u=t-\lambda$, then we have \begin{equation} I(\lambda) \sim \int_{0}^{\infty}e^{-\lambda^2 -2\lambda u}du, \end{equation} and solving this integral gives that \begin{equation} I(\lambda) \sim \frac{e^{-\lambda^2}}{2\lambda}. \end{equation} I do not know what is wrong with my solution, I would be happy if you share your opinions.
Why do you think your answer is wrong? The fact that you got the same answer in two ways should be strong evidence that your answer is correct.
See, for example, equation 16 here:
http://mathworld.wolfram.com/Erf.html