<s>[INST] Asymptotic behavior of the function $f(x) = x \int_x^\infty \frac{1 - e^{-\frac{1}{4}y^2}}{y^3}dy.$ In one of my analysis course, we considered the function
$$f(x) = x \int_x^\infty \frac{1 - e^{-\frac{1}{4}y^2}}{y^3}dy.$$
Then my teacher told us that $f$ had the following behavior
$$
f(x)\sim\begin{cases}
\frac{-1}{4}x\log x & \text{as } x \to 0^+,\\
\frac{2}{x} & \text{as } x \to \infty.
\end{cases}$$
For the case $x \to \infty$ I understand as in this case $e^{-\frac{1}{4}y^2}$ is negligeable and then
$$f \sim x \left[\frac{-1}{2y^2}\right]^\infty_x = \frac{1}{2x}.$$
However I do not have any intuition of the form of
$$\int_x^\infty \frac{1 - e^{-\frac{1}{4}y^2}}{y^3}dy,$$
which should behave as a $\log$. I thought maybe we could expand $e^{-\frac{1}{4}y^2}$ to the first order so that
$$\frac{1 - e^{-\frac{1}{4}y^2}}{y^3}dy = \frac{1}{4}\frac{1}{y} + O(y)$$
and by integrating we indeed get a $\log$. However I don't get how exactly we should get this behavior of $f$. Is there someone more skilful than me who could answer to this question ?
 [/INST] First note that $$ x \int_1^{\infty} \frac{1 - e^{-\frac{1}{4}y^2}}{y^3}dy=o(x\log{x})$$
as $x\to 0^{+}$. Indeed, the integral $\int_1^{\infty} \frac{1 - e^{-\frac{1}{4}y^2}}{y^3}dy$ is convergent and $\lim_{x\to 0^{+}}\frac{x}{x\log{x}}=0$.
Second, using l'Hopital's rule one can compute
$$\lim_{x\to 0^{+}} 
\frac{x \int_x^1 \frac{1 - e^{-\frac{1}{4}y^2}}{y^3}dy}{x\log{x}}=
-\lim_{x\to 0^{+}} 
\frac{1 - e^{-\frac{1}{4}x^2}}{x^2}=-\lim_{x\to 0^{+}} 
\frac{\frac{1}{2}x e^{-\frac{1}{4}x^2}}{2x}=-\frac{1}{4}.$$
This shows that $f(x)\sim -\frac{1}{4} x\log{x}$
as $x\to 0^{+}$.
</s>