I know that the integral $\int_{-\infty}^{\infty} \frac{\sin(x)}{x}dx = \pi$.
Just for fun, I thought if it would be possible evaluate the same integral but only consider the area below the curve and above the x-axis? Approximately would be fine, if an exact expression is not available.
To make this precise one could define the integral as $$\int_{-\infty}^{\infty} \frac{\sin(x)}{x} \mu_0(\frac{\sin(x)}{x})dx $$
where $\mu_0(x)$ is the unit step function such that
$x<0 \implies \mu_0(x) = 0$
and $x \geq 0 \implies \mu_0(x) = 1$
The integral will diverge. Without doing the math in detail, each positive section between 2n*pi and (2n+1)*pi will integrate to k/n, for k=2 in the limit as n increases. So the integral on the positive side will be approximately the sum (n=1 to infinity) of 2/n . Which is divergent. (same on the -ve side).