Integral from infinity to infinity

Solution 1:

This is not necessarily true. Take the following example; $$\int_a^{2a}\frac1x\mathrm{d}x=[\ln{|x|}]_a^{2a}=\ln{(2)}$$ If we take $a\to\infty$ then the integral becomes $$\int_\infty^\infty\frac1x\mathrm{d}x=\ln{(2)}$$ as the integral is constant for all $a\in\mathbb{R}$. What I guess your professor meant was that $$\lim_{a\to\infty}\int_a^a f(x)\mathrm{d}x=0$$ which is trivially true as the LHS is constantly zero.

Solution 2:

An improper integral with an endpoint of $\infty$ means a limit of proper integrals where the endpoint approaches $\infty$. Thus a reasonable definition of $\int_{\infty}^\infty f(x)\; dx$ would be $$ \int_{\infty}^\infty f(x)\; dx = \lim_{a, b \to \infty} \int_a^b f(x)\; dx $$ This is $0$ if and only if $\int_a^\infty f(x)\; dx$ converges for some $a$.

EDIT: If the double limit is $0$, there is $N$ such that $\left|\int_a^b f(x)\; dx\right| < 1$ for all $N < a < b$. For any $\epsilon > 0$ there is $M > N$ such that for $b, c > M$, $$ \left|\int_b^c f(x)\; dx \right| = \left| \int_a^c f(x)\; dx - \int_a^b f(x)\; dx \right|< \epsilon$$
and this implies that $\lim_{b \to \infty} \int_a^b f(x)\; dx$ exists, i.e. $\int_a^\infty f(x)\; dx$ converges.

Conversely, if $\int_a^\infty f(x)\; dx = L$ converges, then for any $\epsilon > 0$ there is $N$ such that $\left|\int_a^b f(x)\; dx - L\right| < \epsilon/2$ whenever $b > N$. Then if $b > N$ and $c > N$, $$ \left| \int_b^c f(x)\; dx\right| = \left|\int_a^c f(x)\; dx - \int_a^b f(x)\; dx \right| < \epsilon $$

Solution 3:

As has been pointed out by other answers, this is not always true because the symbol $\infty$ can hide many things, even if we work with the extended reals. The actual meaning of the $\infty$ is a limiting process as a certain variable becomes arbitrarily large. The upper and lower limits in the $$\int_a^b f(x)\mathrm d x$$ however may approach $\infty$ at different rates, and this is the key point. Your professor's argument works only when $a$ and $b$ approach infinity equally fast, or in other words, when they are of equal order at infinity. In particular it is always valid if $a=b.$

So what about your professor's "proof"? Well, the ambiguity should now be obvious -- he uses the same symbol $\infty$ for things that may behave differently. I think it's safe to assume he's thinking only of the case when the variables are of equal order at infinity. Otherwise his proof breaks down since $\infty-\infty$ can then be anything.

PS. However, you say an integral from $\infty$ to $\infty$ has no meaning to you. Well, I see you're thinking of the usual ordering of the reals here. But note that we're not just dealing with the reals here, but the extended reals. As explained above, the best way to think of it is to think of the limits of the integral as approaching infinity (at not necessarily equal rates). Then it's easy to make sense of it. Another way may be to think of the one-point compactification of the real axis.