# What do improper integrals represent? [closed]

So I was wondering what is the "meaning" behind the improper integral, say $\int_0^{\infty}f(x) dx$. Usually, an integral of a function over a certain region is simply the area under the function, however, in this case, it doesn't seem fully the case. After all, if it was the sum of the areas then $\int_0^{\infty} \sin(x)$ would converge, as the areas form a telescoping series. Hence my confusion, what does the definition of this kind of improper integral really mean?

## closed as off-topic by Mark Viola, Namaste, Xander Henderson, Leucippus, max_zornAug 1 '18 at 5:52

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• The integrals still represent areas. By definition $\int_0^\infty \sin x dx = \lim_{n \to \infty} \int_0^n \sin x dx$, which does not converge just as $\lim_{n \to \infty} \cos n$ or $\lim_{n \to \infty} (-1)^n$ do not converge. – JavaMan Jul 31 '18 at 19:27
• Also, did you try looking anything up online before asking here? What did you find? – JavaMan Jul 31 '18 at 19:28
• @JavaMan I know what the definition means. I know why it does not converge. But if you consider the telescoping series of the areas under and above $f(x)=\sin(x)$ you get a convergent series. – Sorfosh Jul 31 '18 at 19:32
• If you understand the definition of the improper integral, then your misunderstanding is coming from the definition of convergence. Improper integrals still represent area. Divergence can mean that the quantity tends to infinity, or it can mean that the limit never settles around a single number. The second type of divergence is what happens here. – JavaMan Jul 31 '18 at 19:35
• @Sorfosh The "areas under and above $\sin(x)$" do NOT represent a telescoping series. And even if they did, telescoping does not imply convergence. – Mark Viola Jul 31 '18 at 19:36

I still assert that $\int_0^\infty f(x)\,dx$ represents the signed area between $f$ and the $x$ axis. You complain that this cannot be right, because the signed area between $\sin x$ and the positive $x$ axis is zero, being a telescoping sum, while $\int_0^\infty \sin x\,dx$ does not converge. However, this is not a telescoping sum any more than $$1-1+1-1+1-\dots$$ is. The above sum does not converge, as the partial sums are $1,0,1,0,\dots$. In the same way that this series does not have a sum, the region under $\sin x$ and the positive $x$ axis does not have an area, which agrees with the statement that $\int_0^\infty \sin x\,dx$ diverges.
If $f : [0,\infty) \to \Bbb R$ then: $$\int_0^\infty f(x) \ \mathrm dx := \lim_{b \to \infty} \int_0^b f(x) \ \mathrm dx$$
• @Sorfosh : What's the difficulty? If the area up to $b$ is $A(b)$, then of course $A(\infty)$ would mean the limiting area as $b$ increases without bound (if that exists). It's completely analogous to understanding $\sum_{n=0}^{\infty}a_n$ as a limit of partial sums, since there's no such thing as an infinite sum. Do you have problems understanding infinite series? – MPW Jul 31 '18 at 19:44
If the integral converges then for all $n> N \implies \int_0^{n} f(x) \ dx$ is within $\epsilon$ of some constant.
But $\int_0^{2n\pi} \sin x \ dx = 0$ while $\int_0^{(2n+1)\pi} \sin x \ dx = 2$