In which senses can an integral exist? I asked about the value of an integral here:
Hard integral that standard CAS get totally wrong

The question got downvoted and voted to close because I didn't understand (and wasn't able to answer) the following question:


  In which sense is the integral supposed to exist?


So in what senses can integrals exist? What are the options here?
In addition to other useful answers... Before giving the standard example of an integral whose convergence is problemmatical, but which does have some sense... we should ask ourselves what it is we are expecting "integrals" to do, what properties the process should have. For example, "integration" should not just produce random numerical outcomes. And which functions, on which intervals, should be expected to be acceptable inputs? At the very least, if $f,g$ are acceptable, then linear combinations $af+bg$ should be, and, letting $I$ denote the integration procedure, $I(af+bg)=aI(f)+bI(g)$. ("Linearity".)

Riemann integration works best on finite intervals with nearly-continuous functions, while Lebesgue integration accommodates very-discontinuous functions, etc. In both cases, the integral of $f$ on a set or interval is a limit of finite sums, and the set-up for the game consists of proving these limits will exist (and be finite numbers!) under various assumptions on $f$.

A simpler standard example similar to one in your other post is $\int_0^\infty \sin(x^2)\;dx$. This has the disturbing feature that the function doesn't go to $0$ at infinity, so if we think of Cauchy's criterion for convergence of a series (rather than integral), we might conclude that this would diverge, meaning that $\lim_N \int_0^N \sin(x^2)\;dx$ might be $\pm\infty$? Or not exist? However, the oscillation produces enough self-cancellation so that this doesn't happen. In fact, changing variables, replacing $x$ by $\sqrt{x}$, gives integral
$\int_0^\infty {\sin(x)\over \sqrt{x}}\;dx$. Now, at least it looks like it decays at infinity, and there is still cancellation due to the oscillation. In fact, the limit can be evaluated by various tricks: I think it is $\pi/2$ or something similar.

In fancier circumstances, it often happens that "an integral" is not meant to be taken literally, but only to indicate the structure of some operation on functions. The basic case is with Fourier or Laplace transforms on the real line. Fourier transforms expressed as integrals $\hat{f}(\xi)=\int_{\mathbb R} e^{-i\xi x}\,f(x)\;dx$ make best sense for $\int_{\mathbb R} |f(x)|\;dx&lt;\infty$, but, in fact, via the Plancherel theorem for Fourier transforms, we know that $\int_{\mathbb R} \hat{f}(\xi)\;\hat{g}(\xi)\;dx=\int_{\mathbb R}f(x)\,g(x)\;dx$ (maybe up to a constant multiple), so there is a unique extension of Fourier transform to square-integrable functions $f$, that is, such that $\int_{\mathbb R}|f(x)|^2\;dx&lt;\infty$. This extension has the same properties as the Fourier transform that is literally an integral, but is not quite given by that integral.

Similarly, Fourier transforms can be extended to ("tempered") "generalized functions" (="distributions", not in the probabilistic sense), in a way that is completely sensible structurally, but in which the integrals are wildly not-convergent. For example, $\int_{\mathbb R} x^n\cdot e^{-i\xi x}\;dx$ doesn't converge at all, but by other means we can conclude that it is (a constant multiple of) the $n$th derivative of the Dirac delta distribution. 

And, in case there were any doubt, computer algebra systems have their limitations, especially in dealing with "divergent" (not numerically docile!) integrals.
Well, there is a simple definition: an Integral $\int_{a}^{\infty} f(x)\;dx$ exists $\Leftrightarrow$ the series $\sum_{n=a}^{\infty} f(n)$ converges. The hard part might be proving this. Use a convergence test to proof, if it converges or not.

For Integrals with both bounds of integration infinite you should look up the Residue Theorem.