Approximation of a non-trivial gaussian integral

Question

I want to show that the following integral is $O(T)$ for $T\rightarrow0$

$$ \int_0^{+\infty}dz\frac{1}{\sqrt{2\pi}}e^{-\frac{z^2}{2}}e^{-\frac{2z}{T}\,\sqrt{q(T)}}$$

where $q(T)=1-\sqrt{\frac{2}{\pi}}T+O(T^2)$. I tried with the expansion of $e^{-\frac{z^2}{2}}$ inside the integral and then reconstructing the integral representation of the Euler Gamma function, but I'm not convinced it is the right thing to do. Can anyone help me?

Answer 1

Let the integral be $f(T)$. Then, substituting $z=\frac{Tx}{2\sqrt{q(T)}}$, you get (by DCT) $$\frac{f(T)}{T}=\frac1{2\sqrt{2\pi q(T)}}\int_0^\infty\exp\left(-x-\frac{T^2 x^2}{8q(T)}\right)dx\color{blue}{\underset{T\to 0}{\longrightarrow}}\frac1{2\sqrt{2\pi}}\color{lightgray}{\int_0^\infty e^{-x}\,dx}.$$

Answer 2

$$q(T)=1-\sqrt{\frac{2}{\pi }} T+O\left(T^2\right)$$ $$\sqrt{q(T)}=1-\frac{T}{\sqrt{2 \pi }}+O\left(T^2\right)$$ Completing the square, for any $T$ positive and smaller than $2\pi$ $$I=\int_0^{+\infty}\frac{1}{\sqrt{2\pi}}e^{-\frac{z^2}{2}}e^{-\frac{2z}{T}\,\sqrt{q(T)}}\,dz$$ $$I=\frac 12 \exp\left( \frac{\left(T\sqrt{\frac{2}{\pi }} -2\right)^2}{2 T^2} \right)\left(1+\text{erf}\left(\frac{1}{\sqrt{\pi }}-\frac{\sqrt{2}}{T}\right)\right)$$ leading to the limit already provided by @metamorphy.