<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] $$f(x) = x \int_x^\infty \frac{1 - e^{-\frac{1}{4}y^2}}{y^3}\,dy$$
Let $y=2 \sqrt{t}$ to make
$$f(x)=\frac 18 x \int^\infty_{\frac {x^2}4} \frac{1-e^{-t}}{t^2}\,dt$$ Using the incomplete gamma function
$$\int t^{-n} e^{-t}\,dt=-\Gamma (1-n,t)$$
wich makes
$$f(x)=\frac{1}{2 x}-\frac{x}{8}\, \Gamma \left(-1,\frac{x^2}{4}\right)$$ Using the expansion for small positive $x$
$$f(x)=-\frac{1}{8} x \,(2 \log (x)+\gamma -1-2\log(2))+O\left(x^3\right)$$
Now, if $x$ is large, you would find
$$f(x)=\frac 1{2x}-\frac{2}{x^3}  e^{-\frac{x^2}{4}}+\cdots$$
Just trying

*

*for $x=\frac 1{10}$, the approximation will give $0.0801781$ while the exact value is $0.0801937$.

*for $x=10$, the approximation will give $0.0499999999999722$ while the exact value is                   $0.0499999999999742$
</s>