$$\int_{-\infty}^{\infty}x\,dx$$ According to my teacher, this improper integral diverges because "if one or both integrals diverge, the entire integral diverges." Evaluating it as a limit, however, it seems to cancel out and give $0$.

I understand this gives an indeterminate form, and that generally, it is incorrect to "cancel out infinity," but an indeterminate form doesn't mean that it can't be evaluated to diverge, as it seems to do in this case. Some intuition behind this conclusion also lies in the fact that either side is decreasing at the same rate, so it seems obvious that the area goes to $0$.

If it truly does diverge, then to what? it seems absurd to say that it blows up to $\pm\infty$.

  • $\begingroup$ odd function, it should be $0$ $\endgroup$ – Vasya Mar 7 at 3:31
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    $\begingroup$ You may find the answers here helpful. $\endgroup$ – jmerry Mar 7 at 3:36
  • $\begingroup$ Basically you have answered your own question: "generally, it is incorrect to cancel out infinity". Except for one thing: actually, it is always incorrect to cancel out infinity (when working in the real numbers), even though sometimes it will give you the right answer by accident. $\endgroup$ – David Mar 7 at 3:38
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    $\begingroup$ @Vasya: That’s not the definition of the improper integral. An improper Riemann integral over the entire real line exists if and only if each of the integrals on $(-\infty,c]$ and on $[c,\infty)$ exist, for an arbitrary $c$, which requires two limits to exist. Here, neither of those limits exists. $\endgroup$ – Arturo Magidin Mar 7 at 3:40
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    $\begingroup$ Although the improper integral diverges, it has a Cauchy principal value of $0$. $\endgroup$ – Robert Israel Mar 7 at 3:59

The issue in assigning a value to $$ \int_{-\infty}^\infty x\,\mathrm{d}x $$ really involves what it means for a function $f$ to be integrable. Unfortunately, there is no short complete answer to this question. In the Lebesgue theory of integration, a function $f : \mathbb{R} \to \mathbb{R}$ is said to be integrable on $\mathbb{R}$ if $$ \int_{-\infty}^{\infty} \left\vert f(x)\right\vert\mathrm{d}x < \infty. $$ (Here we are ignoring any assumptions that are required for the above to make sense. In any case, these will always be satisfied for a continuous function and, in particular, your function $f(x) =x$.)

The requirement that $\int_{-\infty}^\infty |f|\mathrm{d}x < \infty$ is equivalent to asking that both $$ \int_{-\infty}^{\infty} f_+(x)\,\mathrm{d}x < \infty \quad \text{and} \quad \int_{-\infty}^{\infty} f_-(x)\,\mathrm{d}x < \infty $$ where $f_{+}(x) = \max(f(x), 0)$ and $f_-(x) = \max(-f(x),0)$. If $f$ is integrable according to the definition above, we then define \begin{equation}\label{eq:star}\tag{$\star$} \int_{-\infty}^\infty f(x)\,\mathrm{d}x = \int_{-\infty}^\infty f_+(x)\,\mathrm{d}x - \int_{-\infty}^\infty f_-(x)\,\mathrm{d}x \end{equation} which will be a finite number.

Clearly, the function $f(x) = x$ does not satisfy any of these hypothesis because \begin{align*} \int_{-\infty}^{\infty} f_+(x)\,\mathrm{d}x = \int_{0}^\infty x\,\mathrm{d}x = \infty. \end{align*}

Now, we go through all of this trouble to ensure that we never end up writing something along the lines of $\infty - \infty$ in \eqref{eq:star}, which cannot be made sense of.

However, as you have observed, something interesting happens with $f(x)=x$. For each $\alpha > 0$, you have shown that $$ \int_{-\alpha}^\alpha x\,\mathrm{d}x = \frac{\alpha^2 - \alpha^2}{2} = 0. $$ Hence, the limit $$ \lim_{\alpha \to \infty} \int_{-\alpha}^\alpha x\,\mathrm{d}x = 0 $$ exists and is well defined. This means that the number given by \begin{equation}\label{eq:dagger}\tag{$\dagger$} \int_{-\infty}^\infty x\,\mathrm{d}x \stackrel{?}{=} \lim_{\alpha \to \infty} \int_{-\alpha}^\alpha x\,\mathrm{d}x = 0 \end{equation} exists. Thus, the integral $\int_{-\infty}^\infty x\,\mathrm{d}{x}$ only exists in the improper sense (in this case, we are forced to use the Cauchy principle value as our definition of improper). In other words, $\int_{-\infty}^\infty x\,\mathrm{d}{x}$ should be interpreted as an improper integral (and even then, we need the Cauchy principle value). Although these do not make much sense in the Lebesgue sense (in which we require that $|f|$ be integrable), there are theories of integration that deal with these improper integrals (see the Gauge integral, for instance).

Short answer: Whether or not $\int_{-\infty}^\infty x\,\mathrm{d}x$ exists as an integral depends on the context. It does not exist as a Lebesgue (or Riemann) integral, but it does exist if you want to talk specifically about the value $$ \lim_{\alpha \to \infty} \int_{-\alpha}^\alpha x\,\mathrm{d}x $$

Edit: We also point out that the expression in \eqref{eq:dagger} would make a "bad" definition for an integral. To see why, consider first a (Lebesgue) integrable function $f : \mathbb{R} \to \mathbb{R}$. Then, $f$ will also be integrable on any interval $[a,b] \subset \mathbb{R}$. In fact, $f$ will be integrable on any interval of the form $(c,\infty)$. Moreover, the following additive rule would hold: $$ \int_{-\infty}^\infty f(x)\,\mathrm{d}x = \int_{-\infty}^c f(x)\,\mathrm{d}x + \int_{c}^\infty f(x)\,\mathrm{d}x. $$ Now, both of these properties are to be expected of an integral (after all, they are fundamental and very intuitive properties). However, despite existing as a limit, the "integral" $\int_{-\infty}^\infty x\,\mathrm{d}x$ fails both of these properties. Indeed, $$ \int_{c}^\infty x\,\mathrm{d}x = \infty \quad \text{and} \quad \int_{-\infty}^c x\,\mathrm{d}x = - \infty $$ for every $c \in \mathbb{R}$. Consequently, the additive rule $$ \int_{-\infty}^\infty x\,\mathrm{d}x \stackrel{?}{=} \int_{-\infty}^c x\,\mathrm{d}x + \int_{c}^\infty x\,\mathrm{d}x $$ also fails. In short, incorporating \eqref{eq:dagger} into our definition of the integral would cause us to lose many of the nice properties the integral satisfies. So, although we can partially avoid having $\infty - \infty$ in this case, we still end up breaking several familiar properties the integral should satisfy.

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    $\begingroup$ Okay, thanks, this makes sense, but I am still not completely understanding the purpose of these distinctions. Also, why do we at all "go through the trouble of avoiding $\infty - \infty$," in a case such as this where it is so easily simplifiable and really poses no problem at all and can indeed be made sense of? $\endgroup$ – Roshan Mar 7 at 7:14
  • $\begingroup$ Please see my edit. $\endgroup$ – rolandcyp Mar 7 at 15:26

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