If a function $ f $ is continuously differentiable and $ \int_0^{\infty} f(x)dx$ converges then $ f $ is bounded. 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.
 A: 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.
A: Another counterexample is given by the integral
$$\int_0^\infty x\sin(e^x)dx$$
It does indeed converge (see the proof below) and $x\sin(e^x)$ is continuously differentiable, but $x\sin(e^x)$ is unbounded over $[0,\infty)$.
Proof of convergence: Notice that
\begin{align}
\int_0^t x\sin(e^x)dx &= \int_0^t \frac{\ln(e^x)\sin(e^x)}{e^x}e^xdx\\
&= \int_1^{e^t} \frac{\ln(x)\sin(x)}{x}dx
\end{align}
so proving the convergence of $\int_0^\infty x\sin(e^x)dx$ boils down to proving the convergence of
$$\int_1^\infty \frac{\ln(x)\sin(x)}{x}dx$$
Notice that $\frac{\ln(x)}{x}$ is strictly positive and eventually decreases monotonically to $0$. Since $\sin(x)$ oscillates between $1$ and $-1$ periodically, this suggests that proving the convergence of $\int_0^\infty \frac{\ln(x)\sin(x)}{x}dx$ is analogous to proving the convergence of the alternating series
$$\sum_{n=0}^\infty (-1)^n\frac{\ln(n)}{n}$$
We can put these ideas into action by splitting the integral into distinct intervals that capture regions where the sine is non-negative and non-positive. Assuming $t$ is sufficiently large, we write
\begin{align}
\int_1^t \frac{\ln(x)\sin(x)}{x}dx &= \int_1^\pi \frac{\ln(x)\sin(x)}{x}dx+\int_\pi^{3\pi} \frac{\ln(x)\sin(x)}{x}dx+\int_{3\pi}^{5\pi} \frac{\ln(x)\sin(x)}{x}dx+\cdots\\
&+\int_{\left(2\left\lfloor t\right\rfloor-1\right)\pi}^{\left(2\left\lfloor t\right\rfloor+1\right)\pi} \frac{\ln(x)\sin(x)}{x}dx+\int_{\left(2\left\lfloor t\right\rfloor+1\right)\pi}^t \frac{\ln(x)\sin(x)}{x}dx\\
&= \int_1^\pi\frac{\ln(x)\sin(x)}{x}dx+\int_{\left(2\left\lfloor t\right\rfloor+1\right)\pi}^t\frac{\ln(x)\sin(x)}{x}dx+\sum_{n=1}^{\left\lfloor t\right\rfloor}\int_{(2n-1)\pi}^{(2n+1)\pi}\frac{\ln(x)\sin(x)}{x}dx\\
&= \int_1^\pi\frac{\ln(x)\sin(x)}{x}dx+\int_{\left(2\left\lfloor t\right\rfloor+1\right)\pi}^t\frac{\ln(x)\sin(x)}{x}dx\\
&+ \sum_{n=1}^{\left\lfloor t\right\rfloor}\left(\int_{(2n-1)\pi}^{2n\pi}\frac{\ln(x)\sin(x)}{x}dx+\int_{2n\pi}^{(2n+1)\pi}\frac{\ln(x)\sin(x)}{x}dx\right)\\
&= \int_1^\pi\frac{\ln(x)\sin(x)}{x}dx+\int_{\left(2\left\lfloor t\right\rfloor+1\right)\pi}^t\frac{\ln(x)\sin(x)}{x}dx\\
&+ \sum_{n=1}^{\left\lfloor t\right\rfloor}\left(\int_{2n\pi}^{(2n+1)\pi}\frac{\ln(x)}{x}\left|\sin(x)\right|dx-\int_{(2n-1)\pi}^{2n\pi}\frac{\ln(x)}{x}\left|\sin(x)\right|dx\right)\\
&= \int_1^\pi\frac{\ln(x)\sin(x)}{x}dx+\int_{\left(2\left\lfloor t\right\rfloor+1\right)\pi}^t\frac{\ln(x)\sin(x)}{x}dx+\sum_{n=1}^{2\left\lfloor t\right\rfloor}(-1)^n\int_{n\pi}^{(n+1)\pi}\frac{\ln(x)}{x}\left|\sin(x)\right|dx\\
\end{align}
All that remains is to show that
$$\lim_{t\to\infty}\int_{\left(2\left\lfloor t\right\rfloor+1\right)\pi}^t\frac{\ln(x)\sin(x)}{x}dx=0\text{ and }\lim_{t\to\infty}\sum_{n=1}^{2\left\lfloor t\right\rfloor} (-1)^n\int_{n\pi}^{(n+1)\pi}\frac{\ln(x)}{x}\left|\sin(x)\right|dx\text{ exists}$$
The first limit is provably true (I'll omit the proof; the post is long enough). To prove the second, it is sufficient to prove that the alternating series
$$\sum_{n=0}^\infty (-1)^n\int_{n\pi}^{(n+1)\pi}\frac{\ln(x)}{x}\left|\sin(x)\right|dx$$
converges. We do this by showing that the terms $a_n:=\int_{n\pi}^{(n+1)\pi}\frac{\ln(x)}{x}\left|\sin(x)\right|dx$ eventually decrease to $0$, which implies the convergence of the series from the alternating series test.

Proof that the terms are strictly decreasing: notice that the substitution $u=x-\pi n$ yields
$$a_n=\int_{n\pi}^{(n+1)\pi}\frac{\ln(x)}{x}\left|\sin(x)\right|dx=\int_{0}^{\pi}\frac{\ln(x+\pi n)}{x+\pi n}\left|\sin(x+\pi n)\right|dx=\int_{0}^{\pi}\frac{\ln(x+\pi n)}{x+\pi n}\left|\sin(x)\right|dx$$
It follows from the monotonicity of $\frac{\ln(x)}{x}$ over $[\pi,\infty)$ that for every $n\geq 1$,
$$a_{n+1}=\int_{0}^{\pi}\frac{\ln(x+\pi n+\pi)}{x+\pi n+\pi}\left|\sin(x)\right|dx\leq \int_{0}^{\pi}\frac{\ln(x+\pi n)}{x+\pi n}\left|\sin(x)\right|dx=a_n$$


Proof that the terms converge to 0: from $|\sin(x)|\leq 1$ and the monotonicity of $\frac{\ln(x)}{x}$, it follows that
$$0\leq a_n=\int_{n\pi}^{(n+1)\pi}\frac{\ln(x)}{x}\left|\sin(x)\right|dx\leq\int_{n\pi}^{(n+1)\pi}\frac{\ln(x)}{x}dx\leq \int_{n\pi}^{(n+1)\pi}\frac{\ln(n\pi)}{n\pi}dx=\pi\cdot\frac{\ln(n\pi)}{n\pi}$$
We know that $\lim_{x\to\infty}\frac{\ln(x)}{x}=0$, so $\lim_{n\to\infty}\frac{\ln(\pi n)}{\pi n}=0$. Thus, from the squeeze theorem,
$$\lim_{n\to\infty}a_n=0$$

We have shown that
$$\sum_{n=0}^\infty (-1)^n\int_{n\pi}^{(n+1)\pi}\frac{\ln(x)}{x}\left|\sin(x)\right|dx$$
converges. Thus, from previous results,
$$\lim_{t\to\infty}\int_1^t\frac{\ln(x)\sin(x)}{x}dx=\int_1^\pi\frac{\ln(x)\sin(x)}{x}dx+\sum_{n=0}^\infty (-1)^n\int_{n\pi}^{(n+1)\pi}\frac{\ln(x)}{x}\left|\sin(x)\right|dx$$
showing that $\int_0^\infty x\sin(e^x)dx$ converges.
