In my probability class we were asked to show that $\int_0^\infty \left\vert\frac{\sin(x)}{x}\right\vert = \infty$. The hint states to find a function below that is easier to integrate and show it diverges. This is a pretty fun problem to solve. I was curious as to what would be the absolutely "easiest" (or simplest) function to prove this problem.
The function used in class was $$T: [2\pi, \infty ) \rightarrow [0,\infty)$$ where we define $$T(2\pi k) = 0$$ $$ T((2k+1)\pi) = 1$$ and interpolate linearly for $x\in[2k\pi,(2k+1)\pi]$ and $x\in[(2k+1)\pi,(2k+2)\pi]$.
for all $k \in \mathbb{N}$. Then he proceeded to show that $\int_{2\pi}^\infty \frac{T(x)}{x} = \infty$
Note: Constructed solutions are the most interesting ones.
I find pretty handy the standard observation $$\left\vert\frac{\sin x}{x}\right\vert\ge \frac{\vert\sin x\vert}{\pi\left\lceil \frac{x}{\pi}\right\rceil}=F(x)$$
The denominator is $(k+1)\pi$ for $x\in(k\pi,(k+1)\pi]$, therefore
$$\int_0^\infty F(x)\,dx=\frac1\pi\left(\int_0^\pi\vert\sin x\vert\,dx\right)\sum_{k=1}^\infty \frac1n=+\infty$$