Prove or disprove the limit of a definite integral

I am trying to reproduce the work of a published paper where I need to evaluate a limit of a definite integral $$\lim_{\xi\to\infty}\xi\int_0^1\exp\left[\frac{\xi^2}{4}\frac{2y^3-3y^2}{{(1-y)}^2}\right]\mathrm{d}y\,.$$

The author of the paper argues that because $$\lim_{\xi\to\infty}\xi\int_0^1\exp\left[\frac{\xi^2}{4}\frac{2y^3-3y^2}{{(1-y)}^2}\right]\mathrm{d}y=\lim_{\xi\to\infty}\xi\int_0^1\exp\left[-\frac{\xi^2}{4}\sum_{n=3}^\infty ny^{n-1}\right]\mathrm{d}y\,,$$ the contribution from higher orders of $y$ is negligible when $\xi\to\infty$, so $$\lim_{\xi\to\infty}\xi\int_0^1\exp\left[\frac{\xi^2}{4}\frac{2y^3-3y^2}{{(1-y)}^2}\right]\mathrm{d}y=\lim_{\xi\to\infty}\xi\int_0^1\exp\left[-\frac{\xi^2}{4}3y^2\right]\mathrm{d}y=\sqrt{\frac{\pi}{3}}\lim_{\xi\to\infty}\mathrm{erf}\left(\frac{\sqrt{3}\xi}{2}\right)=\sqrt{\frac{\pi}{3}}\,.$$

I am not sure if it is OK just to throw away all the higher order terms of $y$, but numerical evaluation shows the limit is correct. Is there some better way to obtain this limit? It looks that if we can find a function $f(y)$ such that $$f(y)\leq\frac{2y^3-3y^2}{{(1-y)}^2}\leq-3y^2$$ and $$\lim_{\xi\to\infty}\xi\int_0^1\exp\left[\frac{\xi^2}{4}f(y)\right]\mathrm{d}y=\sqrt{\frac{\pi}{3}}\,,$$ then from the squeeze rule the limit is correct. I tried to find an appropriate $f(y)$ but did not make much progress.

• So, you have $f(y)<-3y^2$ – Claude Leibovici Jun 3 '16 at 5:27
• See what happens for $y^3$. – marty cohen Jun 3 '16 at 5:31

Here's an incomplete solution (also known as an idea, I guess). The change of variables $y=\frac1{\xi u}$ gives $$\xi\int_0^1\exp\bigg(\frac{\xi^2}{4}\frac{2y^3-3y^2}{{(1-y)}^2}\bigg) \,dy = \int_{1/\xi}^\infty \exp \bigg( {-}\frac{\xi (3 \xi u-2)}{4 u (\xi u-1)^2} \bigg) \frac{du}{u^2}.$$ If the interchange of limits can be justified somehow, then we could get \begin{align*} \lim_{\xi\to\infty} \xi\int_0^1\exp\bigg(\frac{\xi^2}{4}\frac{2y^3-3y^2}{{(1-y)}^2}\bigg) \,dy &= \lim_{\xi\to\infty} \int_{1/\xi}^\infty \exp \bigg( {-}\frac{\xi (3 \xi u-2)}{4 u (\xi u-1)^2} \bigg) \frac{du}{u^2} \\ &\underset{?}= \int_0^\infty \exp \bigg( {-}\frac3{4u^2} \bigg) \frac{du}{u^2} = \sqrt{\frac\pi3}. \end{align*}