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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.

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

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  • $\begingroup$ Months later I revisited my notes and I realized the interchange of limits may be justified using Lebesgue dominated convergence theorem. It is still subtle because the integration limit also changes, but apparently rigorous proof can be done along this line. $\endgroup$ – noir Mar 17 '17 at 1:03

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