Consider the limit:
$$\lim_{x \to +\infty} \frac{\int_{-x}^x \frac{1}{y^2}dy}{x}$$
Since $\int_{-x}^x \frac{1}{y^2}dy=\int_0^x \frac{1}{y^2}dy-\int_0^{-x} \frac{1}{y^2}dy$, using Hopital's rule:
$$\lim_{x \to +\infty} \frac{\int_{-x}^x \frac{1}{y^2}dy}{x}=\lim_{x \to +\infty} \frac{\frac{2}{x^2}}{1}=0$$
However, since $1/y^2$ is unbounded around $y=0$, using the definition of improper integral:
$$\int_{-x}^x \frac{1}{y^2}dy=\int_{-x}^0\frac{1}{y^2}dy+\int_0^x \frac{1}{y^2}dy=\lim_{a\to 0^+}\left(\int_{-x}^{-a}\frac{1}{y^2}dy+\int_a^{x}\frac{1}{y^2}dy\right)=2\lim_{a\to 0^+} \left[\frac{1}{a}-\frac{1}{x}\right]$$
And, for each $x \in\mathbb{R}\setminus\{0\}$, the limit as $a\to 0^+$ is $+\infty$. Some questions:
(i) What's happening with the definition? After evaluating the integral with the definition I have to take another limit as $x\to+\infty$, but the integral already diverges to $+\infty$. Can I conclude that $\lim_{x \to +\infty}\int_{-x}^x \frac{1}{y^2}dy=+\infty$ because (very unprecisely) "I am taking the limit as $x\to+\infty$ of something that already tends to $+\infty$"? More precisely: since the integral is $+\infty$ for each $x \in \mathbb{R}\setminus\{0\}$, I would say that in particular it remains arbitrary large if $x$ is arbitrarily large. However, the $\delta$ such that $|a|<\delta \implies \int_{-x}^x \frac{1}{y^2}dy>M$ of the limit definition I must choose depends on $x$ too, because it must be $\frac{1}{a}-\frac{1}{x}>M$. Is this dependence troublesome?
(ii) Is the approach with Hopital's rule correct?
Best Answer
No, the approach with L'Hôpital is not correct, and for a very subtle reason. Notice that the integral in the numerator is improper in the sense that it is defined as
$$\int_{-x}^x\frac{\mathrm{d}y}{y^2}:=\lim_{a\uparrow0}\int_{-x}^a\frac{\mathrm{d}y}{y^2}+\lim_{b\downarrow0}\int_{b}^x\frac{\mathrm{d}y}{y^2}$$
(if we consider the integral in the Riemann sense) since $\frac{1}{y^2}$ is undefined at the origin. In particular this gives us that
$$\int_{-x}^x\frac{\mathrm{d}y}{y^2}=\infty$$
for all $x>0$ (if we work in the extended reals). The same exact thing happens if we consider it as a Lebesgue integral, so we will stick with this. But this means that
$$\frac{\int_{-x}^x\frac{\mathrm{d}y}{y^2}}{x}=\infty$$
for all $x>0$, implying that the limit as $x\to\infty$ is also $\infty$. Now the reason why you cannot apply L'Hôpital here is for several reasons. First of all, L'Hôpital only works when you have the functions in the numerator and denominator being real-valued, which is clearly not the case here as the numerator always equals $\infty$. Secondly, even if you could apply it, the differentiation you did is not allowed either, as you are implicitly interchanging the derivative with the limit operations above, which is not allowed here. In particular what you are doing is saying
$$\frac{\mathrm{d}}{\mathrm{d}x}\int_{-x}^x\frac{\mathrm{d}y}{y^2}=\frac{\mathrm{d}}{\mathrm{d}x}\left(\lim_{a\uparrow0}\int_{-x}^a\frac{\mathrm{d}y}{y^2}\right)+\frac{\mathrm{d}}{\mathrm{d}x}\left(\lim_{b\downarrow0}\int_{b}^x\frac{\mathrm{d}y}{y^2}\right)\color{red}{=}\lim_{a\uparrow0}\frac{\mathrm{d}}{\mathrm{d}x}\int_{-x}^a\frac{\mathrm{d}y}{y^2}+\lim_{b\downarrow0}\frac{\mathrm{d}}{\mathrm{d}x}\int_{b}^x\frac{\mathrm{d}y}{y^2}=\frac{2}{x^2},$$
with the problem being that the red equality is not allowed (you cannot freely interchange limiting operations, and the derivative is a limiting operation). Even on an intuitive level the above calculation is nonsense; how could the rate of change be $\frac{2}{x^2}$ when the function is always infinite? Indeed even talking about the derivative of an infinite function is nonsensical to begin with, as the definition of the derivative would contain an undefined expression of the form $\infty-\infty$.