# This integral evaluates to infinity, does this mean it exists or doesn't exist?

Say I have$f(x) = x^2$ for all $x \in \mathbb{R}$

Does the integral of f(x) over the entire real line exist? It's infinity so does that mean it doesn't exist?

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As noticed in other comments, your question is highly unstable: the verb to exist is used with an unclear meaning. Let us clarify the situation for the improper Riemann integration theory. I'll add some regularity assumptions that may be relaxed, just for definiteness.

Definition. A continuous function $f \colon [a,+\infty) \to \mathbb{R}$ is integrable if the limit $\lim_{b \to +\infty} \int_a^b f(x)\, dx$ exists as a real number.

There is no verb to exist, but you can imagine that "is integrable" means "the improper integral exists". If you like this viewpoint, then $\int_0^{+\infty} x^2 \, dx$ does not exist.

Another viewpoint is that of allowing improper integrals to exist, either as finite numbers or as $\pm\infty$. In this context, $\int_0^{+\infty} x^2 \, dx$ exists and is $+\infty$, while $\int_0^{+\infty} \cos x \, dx$ does not exist. As you see, it is a matter of taste.

When you switch to more complete integration theories (Lebesgue, gauge, etc.), you will see that most mathematicians do not discard infinite integrals. I personally say that $\int_0^{+\infty} x^2 \, dx$ is infinite; but I know many people who, in the framework of elementary integration theory, say that the same integral does not exist.

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+1, for highly unstable (OK, and also for the rest of the answer...). –  Did Sep 24 '12 at 12:18