Prove/Disprove (here $ f $ is Riemann Integrable on $ [0,\infty) $ ):
If a function $ f $ is continuously differentiable and $ \int_0^{\infty} f(x)dx$ converges then $ f $ is bounded.
I'm not sure but I think the theorem is correct ,however, I have a difficulty trying to prove it. I thought maybe to use Lagrange's Intermediate value theorem ( but I don't know where that'd lead me ) or even using the fundamental theorem of calculus by looking at $ F(x) = \int_0^{x} f(t)dt $ and knowing that $ \lim_{ x \to \infty} F(x) = \int_0^{x} f(t)dt $ exists then I'd somehow have to proceed, maybe by assuming that $ f $ is not bounded and then find a contradiction.
Best Answer
The claim is not true. You can define a function whose graph consists of ever higher triangular spikes with ever narrower bases. For example, define $f$ in such a way that for each positive integer $m\geq 2$, the graph of $f$ looks like an isosceles triangle of height $m$ on the interval $[m-m^{-3},m+m^{-3}]$. Otherwise, $f$ is defined to be $0$. Then, $$\int_{m-\frac{1}{m^3}}^{m+\frac{1}{m^3}}f(x)\,\mathrm dx =\frac{1}{2}\times\underbrace{\frac{2}{m^3}}_{\text{base}}\times \underbrace{m\vphantom{\frac{2}{m^3}}}_{\text{height}}=\frac{1}{m^2}.$$ Clearly, $f$ is unbounded, but $$\int_{0}^{\infty}f(x)\,\mathrm dx=\sum_{m=2}^{\infty}\frac{1}{m^2}=\frac{\pi^2}{6}-1<\infty.$$ If you insist that $f$ be continuously differentiable, you can smooth out the spikes a little bit at the top and at the bottom without qualitatively changing the conclusion.