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.

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.

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.