What do improper integrals represent?

Question

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?

Answer 1

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.

Answer 2

If $f : [0,\infty) \to \Bbb R$ then: $$\int_0^\infty f(x) \ \mathrm dx := \lim_{b \to \infty} \int_0^b f(x) \ \mathrm dx$$

Answer 3

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 $\int_0^{(2n+1)\pi} \sin x \ dx = 2$