Asymptotic behavior of the function $f(x) = x \int_x^\infty \frac{1 - e^{-\frac{1}{4}y^2}}{y^3}dy.$

Question

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 ?

Answer 1

$\newcommand{\d}{\,\mathrm{d}}$In what sense is $\sim$ being used? It matters. Anyway, we can avoid just saying "neglect" quite easily (and it is important to avoid "neglecting" things unless you know what you're doing).

We can make precise the asymptotics at $\infty$ as follows: $$f(x)=\frac{2}{x}-x\int_x^\infty y^{-3}e^{-y^2/4}\d y$$And we want to bound the error term. Well, $e^{-y^2/4}\le e^{-x^2/4}$ for all $y>x$, so we can just say that: $$\left|x\int_x^\infty y^{-3}e^{-y^2/4}\d y\right|\le x e^{-x^2/4}\cdot\int_x^\infty y^{-3}\d y=\frac{2}{x}e^{-x^2/4}$$So $|f(x)-2/x|$ is actually of class $o(e^{-x})$, i.e. it decays verrrrrry quickly as $x\to\infty$. This agrees with, and is much stronger than, the usual meaning of "$\sim$".

Now as $x\to0^+$. Define $\psi:(0,\infty)\to\Bbb R$ by $t\mapsto\frac{1-e^{-t^2/4}}{t^3}-\frac{1}{4t}$. We have: $$f(x)=kx+x\int_x^1\frac{1}{4y}\d y+x\int_x^1\psi(y)\d y=kx-\frac{x}{4}\ln x+x\int_x^1\psi(y)\d y$$With: $$k:=\int_1^\infty\frac{1-e^{-y^2/4}}{y^3}\d y$$Being a finite constant. Now, $\psi$ is certainly continuous on $(0,1]$. Can it be continuously extended at $0$? $$\lim_{t\to0^+}\psi(t)=\lim_{t\to0^+}\frac{4-4e^{-t^2/4}-t^2}{4t^3}=\lim_{t\to0^+}\frac{e^{-t^2/4}-1}{4t}=0$$Using L'Hopital's rule and the definition of derivative. So, yes, $\psi(0):=0$ continously extends $\psi$. So there is a constant $M$, that for all $0<x<1$, $\left|\int_x^1\psi(y)\d y\right|\le M$. I conclude: $$\left|f(x)+\frac{x}{4}\ln x\right|\le(k+M)x$$For $0<x<1$, hence: $$f(x)=-\frac{x}{4}\ln x+\mathcal{O}(x)$$

Showing that $f(x)\sim-\frac{x}{4}\ln x$ in the sense of asymptotic equivalence is similar. We just need to observe that: $$-\mathcal{O}(x)\cdot\frac{4}{x\ln x}\to0,\,x\to0^+$$Since $\frac{1}{\ln x}\to0,\,x\to0^+$, this is clear.

Asymptotic equivalence is about more than just neglecting terms. You need to be aware precisely how fast / slow the terms that you neglect actually vanish.

Answer 2

For a rough argument, you can argue that when $x$ is near zero, the majority of the contribution to the integral will come from the portion near the origin. Therefore

$$f(x)=x\int_x^1 g(y)\mathrm dy+\underbrace{x\int_{1}^\infty g(y)\mathrm dy}_{\text{neglect}} \\ \approx -x\int_1^xg(y)\mathrm dy$$ Where $g(y)=\frac{1-\exp(-y^2/4)}{y^3}$. Now by using the Taylor series for $\exp$ you can see $$g(y)=\frac{1-\left(1-\frac{y^2}{4}+\mathrm O(y^4)\right)}{y^3}=\frac{1}{4y}+\mathrm O(y) \\ \text{as}~y\to 0$$ Hence $$f(x)\approx -x\int_1^xg(y)\mathrm dy \approx -x\int_1^x\frac{1}{4y}\mathrm dy=\frac{x\log x}{4}$$

You can probably make this more rigorous with Watson's lemma or something.

Answer 3

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 $x\to 0^{+}$.

Answer 4

$$f(x) = x \int_x^\infty \frac{1 - e^{-\frac{1}{4}y^2}}{y^3}\,dy$$

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 $0.0499999999999742$