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Recently I was learning to evaluate the improper integral $$ I=\int_{-\infty}^\infty\frac{du}{u^2+2} $$ My instructor said that we could write $$ I=\lim_{t\to\infty}\int_{-t}^t \frac{du}{u^2+2}=\lim_{t\to\infty}\left(2\int_0^t \frac{du}{u^2+2}\right) $$ to be precise.

But I thought: “Hey, who said that the function was “approaching its upper and lower bound at the same rate (t)”? If you really wanted to be rigorous, shouldn’t you write $$ I=\lim_{t\to\infty}\lim_{s\to{-\infty}} \int_s^t \frac{du}{u^2+2} $$ and proceed with the calculation?” (Because the function approaches its upper and lower bounds independently?)

I know that however I write out this integral, the answer won’t be affected. I’m just curious about which argument is the more rigorous one.

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  • $\begingroup$ None. You would first prove that it converges, which is neither, and then pick your instructor's way for being simpler. $\endgroup$ Commented Oct 25 at 13:36
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    $\begingroup$ You’re quite right to be concerned in general. See en.wikipedia.org/wiki/Cauchy_principal_value $\endgroup$ Commented Oct 25 at 13:36
  • $\begingroup$ The most general definition is an iterated limit. The principle value can exist more generally, but when the iterated limit value exists it equals the principle value. $\endgroup$ Commented Oct 25 at 13:37
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    $\begingroup$ @BrevanEllefsen No, it is not. Iterated limits might depend on the order. $\endgroup$ Commented Oct 25 at 13:39
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    $\begingroup$ @BrevanEllefsen So, you are arguing that it is the "most general definition" by showing that it must be proven that it is well defined, and moreover depends on $\mathbb{R}$ having two ends? Well, what you are saying is part of the reason it is not. $\endgroup$ Commented Oct 25 at 21:29

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What you are saying is correct. To be more precise, the definition of $\int_{-\infty}^{\infty}f(t)dt$ is exactly $$\lim_{n,m\to\infty}\int_{-m}^n f(t)dt$$ which can sometimes be written as $$\lim_{n\to\infty}\int_0^nf(t)dt+\lim_{m\to\infty}\int_{-m}^0f(t)dt$$

where instead of $0$ you can take any real number $a$. If this converges, then you can take $m=n$ and get the definition you wrote. This, however, is dependent on the fact that the integral does converge. If it doesn't, then $\lim_{n\to\infty}\int_{-n}^n f(t)dt$ might converge, while $\int_{-\infty}^{\infty}f(t)dt$ does not (take for example $f(t)=\sin(2\pi t)$). In that case, we define $\lim_{n\to\infty}\int_{-n}^n f(t)dt$ to be the principal value of the divergent integral.

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    $\begingroup$ I think this should be $\lim_{m,n\rightarrow\infty}\int_{-m}^nf(t)dt$. What you've written would give $$\int_{-\infty}^\infty t^3dt=-\infty$$ (since for each specific $n$ we have $\lim_{m\rightarrow\infty}\int^n_{-m}f(t)dt=-\infty$) which I don't think you want. $\endgroup$ Commented Oct 26 at 1:03
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    $\begingroup$ Your second definition (with the arbitrary but fixed "break point") is the more restrictive one and the one taught in good standard Calculus classes. The first one can depend on the arbitrary choice which limit to take first, and leads to undesirable since arbitrary definitions as in @NoahSchweber 's comment. -1 until this gets corrected. $\endgroup$ Commented Oct 26 at 4:25
  • $\begingroup$ Another way to see that $\lim_{m,n\rightarrow\infty}\int_{-m}^nf(t)dt$ (from Noah Schweber) is different from what you wrote is first taking one limit, then another limit, whereas limits are not commutative in general. The notation $\lim_{m,n\rightarrow\infty}$ implies that the limit exists only if we get the same limit no matter how $m$ and $n$ increase (together or one at a time) as long as they each go to $\infty.$ In those cases your limits turn out to be commutative after all, but $\lim_{m,n\rightarrow\infty}$ actually implies much stronger conditions even than that. $\endgroup$ Commented Oct 26 at 6:27
  • $\begingroup$ @TorstenSchoeneberg Fixed. $\endgroup$ Commented Oct 26 at 9:34

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