# If the improper integral exists, must the limit exist if the function is continuous and differentiable?

Let $$f:\mathbb{R}\rightarrow \mathbb{R}$$ be a continuous differentiable function, and suppose $$\int_0^\infty f(r)dr<\infty$$ and that $$f(0)<\infty$$. Is it true that $$\lim_{r\rightarrow\infty}f(r)=0$$?

My physics class seems to assume this. I see why, if the limit exists, it must be zero, but I don't see why the limit has to exist. Perhaps one line of logic would be that, since we assume the derivative exists, we can assume $$\int_0^\infty f(r)\frac{df(r)}{dr}dr$$ exists, and by integration by parts this is equal to $$|_0^\infty f(r)^2-\int_0^\infty f(r)\frac{df(r)}{dr}dr$$, so moving everything else to one side the limit is finite, and thus zero. But by using integration by parts aren't I already assuming the limit exists?

For context, this is an introductory quantum mechanics course. They state the only properties a wavefunction $$\psi$$ must have is that $$\int_{-\infty}^\infty |\psi(x)|^2dx<\infty$$ and $$\psi$$ is continuous and have continuous derivative. To prove the momentum operator is hermitian, we need $$|_{-\infty}^\infty|\psi|^2=0$$.

No. Consider $$f(x) = \sum_{n\geq 1}n\exp\left[-n^6(x-n)^2\right].$$
$$f(r)=\sin (r^2)$$