I've been wondering if an improper integral (e.g. upper bound is $\infty$) over a function that is unbounded on the integration interval can converge.
Specifically, I know the integral $\int_1^\infty \frac{1}{x}~dx$ does not converge, whereas
$\int_1^\infty\frac{1}{x^2}~dx$ converges. I also know that $\int_0^\infty\frac{1}{x^2}~dx$ does not converge, so i guess in its simplest form I'm wondering if there exists an $n$ (real?) such that $$\int_0^\infty\frac{1}{x^n}~dx$$ converges.
If that form is too simple but there exists something similar I'd be interested to hear how that works.
Thanks in advance!
Best Answer
It's not hard to modify your example to get a positive function $f(x)$ which is continuous and unbounded on $(0, \infty)$ such that $\int_0^\infty f(x) dx$ converges. One such example would be $$f(x) = \frac{e^{-x}}{\sqrt{x}};$$ we can check that \begin{align*} \int_0^\infty \frac{e^{-x}}{\sqrt{x}} dx &= \int_0^1 \frac{e^{-x}}{\sqrt{x}} dx + \int_1^\infty \frac{e^{-x}}{\sqrt{x}} dx \\ & \leq \int_0^1 \frac{dx}{\sqrt{x}} + \int_1^\infty e^{-x}dx \\ &= 2 + \frac{1}{e} \\ &< \infty. \\ \end{align*}
More generally, any function of the form $f(x) = \frac{a^{-x}}{x^p}$ with $a > 1$ and $0 < p < 1$ will furnish a suitable example of a positive unbounded function on $(0, \infty)$ with convergent improper integral over that open ray.