The following seems believable and quasi-intuitive to me, but it also doesn't quite seem trivial, and I'm not sure whether I've seen it stated before.
Let $f$ be a complex-valued integrable function on $[a,b]$. We then have $$ \left|\int_a^b f(x) dx\right| \le \int_a^b |f(x)| dx $$ and furthermore (this is the part I care more about) the inequality is an equality if and only if $$ f(x) = e^{ic}|f(x)| \qquad \mbox{almost everywhere, where $c$ is a constant.} $$ Speaking very informally, if the integral of $f(x)$ "adds up" to have the same modulus as the integral of $|f(x)|$, it seems like this could only happen if $f(x)$ always "points the same way", i.e. if $f(x)$ maintains the same argument (almost everywhere).
My questions are: 1. Does anyone know a reference for this fact? (And does it have a name)? 2. Does it have a simple self-contained proof? (Or would a proof from more fundamental principles be somewhat involved, approximating an integral with a sum and so forth?)
The fact that this is complex valued is not actually useful; this is more generally true in inner product spaces, so let's prove it there. In particular, let $$v=\int_{a}^bf(x)\,dx.$$ and let $\hat{v}$ be the unit vector in the same direction as $v$. Notice that the left side of your inequality is then $\hat{v}\cdot v=\|v\|$. Thus, we have that your inequality is the same as (bringing the dot product inside the integral by linearity) $$\int_{a}^b\hat v \cdot f(x)\,dx\leq \int_{a}^b|f(x)|\,dx$$ But this is obvious because $\hat v \cdot f(x) \leq |f(x)|$ for all $x$. Then, in order to obtain equality, you obviously need that $\hat v \cdot f(x) = |f(x)|$ for almost all $x$, as if there were a set of positive measure for which it were less, the former integral would be less too. This is essentially what your condition stated.