Can I integrate from $+\infty$ to $+\infty$ and get a nonzero result? In my book there is such example but the solution shows that they meant $\int_{-\infty}^{\infty}$ not $\int_{\infty}^{\infty}$. Is there any use to such integral? I can imagine having $$\lim_{n\to\infty}\int_{n\pi}^{(n+1)\pi}f(\sin(x))\,{\rm d}x$$
 A: Forgetting about measure theory, the basic properties of the integral $I(f)$ of a real function are


*

*It is a real number.

*It is linear $I(a f + b g) = a I(f) + b I(g)$

*It is positive: if $f\ge 0$ then $I(f)\ge 0$


In functional analysis these properties characterize a state, a positive linear form (actually states are more restricted because there is a norm condition on them). Obviously your $\int_\infty^\infty$ must not change if $f$ is modified on a bounded interval $[a, b]$, so it is a state at infinity. Such objects exist in some theories such as $C^*-$algebras. They are not called integrals but it's probably where to look if you are interested in such peculiarities.
A: There are certain useful ways to do integrals over the complex numbers that involve integrals from $+\infty$ to 0 and back to $+\infty$.  For example, one can find the value of the real integral
$$
\int_0^\infty \frac{\sqrt{x}}{x^2 + 6x + 8} \, dx 
$$
by integrating along a "keyhole contour" that includes an integral along the real axis from $+\infty$ to $0$ and another integral from $0$ to $+ \infty$.  See here for details.  These integrals don't cancel each other out because of the properties of the square root function in the complex plane;  technically, we're taking the integrals between $\pm i \epsilon$ and $+ \infty \pm i \epsilon$, and the function has different values as we approach the real axis from above and below in the complex plane.
A: As a different perspective from Gribouillis:
This $\int_{\infty}^\infty$ itself can never be non-zero. If $\lim_{(s,t)\rightarrow(\infty,\infty)}\int_s^t f(x)dx = I = \int_\infty^\infty f(x)dx$, then $\lim_{(s,t)\rightarrow(\infty,\infty)}\int_t^s f(x)dx = -I = \int_\infty^\infty f(x)dx$. Thus, $\int_\infty^\infty f(x)dx = -\int_\infty^\infty f(x)dx \Rightarrow \int_\infty^\infty f(x)dx = 0$.
A: The only nontrivial interpretation of $\int_{\infty}^\infty f(x)\,\textrm{d}x$ I can think of is as an improper integral of sorts: $\lim_{a,b\to\infty}\int_{a}^b f(x)\,\textrm{d}x$. But if this limit exists, it must be zero, because $\lim_{a\to \infty} \int_a^a f(x)\,\textrm{d}x=0$. On the other hand, $\int_\infty^\infty 1\,\textrm{d}x$ doesn't exist.
As suggested by Gribouillis, it might be interpreted as a functional dependent on the germ at infinity, but I think this is most likely just a typo in your book.
