# 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
I agree with your thoughts on to exist. We speak as if a limit is an object which may or may not "exist." But really a limit is a statement about a function. – Matthew Leingang May 14 '14 at 13:31

Does infinite exist? Maybe your question is asking to simply to integrate i.e. the answer is (x^3)/3. Often questions are phrased in the way you wrote, but essentially they are asking to integrate with respect to x.

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But if it is a definite integral where interval is the entire real line then it evaluates to infinity...so does that mean it exists or not? – dukenukem Sep 24 '12 at 10:36
If the definite integral evaluates to infinite i.e. integrate (from x= 0 to x=1) y=1/x with respect to x. Then the answer is infinite. So yes, then the answer is infinite. – student101 Sep 24 '12 at 10:41
Want to explain more about what you mean by 'exist?' In the end your question seems to be asking 'does infinite exist'. – student101 Sep 24 '12 at 10:43

You can unambiguously say either the integral $\displaystyle\int_{-\infty}^\infty x^2\,dx = \infty$ or the integral diverges.

It's been said that the best definitions are the ones which make the theorems easiest to state and prove. The trouble with counting an infinite integral as “existing” is that they have to be ruled out from various theorems about combinations of integrals. For instance, the statement that $\int_a^b (f(x)-g(x))\,dx = \int_a^b f(x)\,dx - \int_a^b g(x)\,dx$ results in something indeterminate if both of the integrals on the right are infinite. So you have to preface that equation with the condition that both integrals exist and are finite. Simpler to just include only finite limits as “existing.”

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