What's the antiderivative of |sinc(x)|? I need to calculate the following expression
$$f(x)=\frac{1}{x}\int_0^x|\operatorname{sinc}(t)|dt,$$
but don't know how to find the antiderivative of $|\operatorname{sinc}|$. I am not sure whether there exit an closed form for $f(x)$
 A: I get this, in terms of $\operatorname{Si}(x) := \int_0^x\frac{\sin t}{t}\;dt$ :
$$
\int_0^x \left|\frac{\sin t}{t}\right|\;dt = \operatorname{sgn}\left(\frac{\sin x}{x}\right)\;\operatorname{Si(x)} -
2\sum_{k=0}^{\lfloor x/\pi \rfloor}(-1)^k \operatorname{Si}(k\pi)
$$
Here is the graph:


I did this by first asking Maple for the antiderivative. That answer was
$$
\operatorname{sgn}\left(\frac{\sin x}{x}\right)\;\operatorname{Si(x)}
$$
In fact, that has derivative $\big|\frac{\sin x}{x}\big|$ except for
integer multiples of $\pi$, where it has jumps.  So I corrected by subtracting the jumps between $0$ and $x$.
A: If there was an antiderivative it would be so on (0,π/2) , But then sincx=sinx/x which is a well known function that has no antiderivative in closed form.
A: I'll assume that $\operatorname{sinc}$ is the traditional, unnormalized function $x \mapsto \frac{\sin x}{x}$ (and $0 \mapsto 1)$, not the normalized function described by $x \mapsto \frac{\sin \pi x}{\pi x}$. Translating the below answer for the latter just requires adjusting constants appropriately.
One can't say much. By definition, $$\int_0^x \operatorname{sinc} t \,dt = \operatorname{Si}(x) := \int_0^x \frac{\sin t\,dt}{t} .$$
The function $\operatorname{Si}$ is often called the sine integral.
The function $\operatorname{sinc}$ is positive at $0$ and has simple roots $k \pi$, $k \in \Bbb Z \setminus \{0\}$, so for $x \geq 0$,
$$\int_0^x | \operatorname{sinc} t | \,dt = (-1)^{\left\lfloor \frac{x}{\pi} \right\rfloor} \operatorname{Si}(x) + C_{\left\lfloor \frac{x}{\pi} \right\rfloor},$$ where the constants $C_k$ are determined by the condition that the quantity is continuous.
For example:

*

*For $x \in [0, \pi]$, $\int_0^x | \operatorname{sinc} t | \,dt = \int_0^x \operatorname{sinc} t \,dt = \operatorname{Si}(x)$ (so $C_0 = 0$).

*For $x \in [\pi, 2\pi]$, $\int_0^x | \operatorname{sinc} t | \,dt = \int_0^\pi \operatorname{sinc} t \,dt - \int_\pi^x = \operatorname{sinc} t \,dt = -\operatorname{Si}(x) + 2 \operatorname{Si}(\pi),$ so $C_1 = 2 \operatorname{Si}(\pi)$. The quantity $G' := \operatorname{Si}(\pi) \approx 1.85913\ldots$ is called the Wilbraham-Gibbs constant.

By induction we can see that
$C_k = 2 [\operatorname{Si}(\pi) - \operatorname{Si}(2 \pi) + \operatorname{Si}(3 \pi) - \cdots \pm \operatorname{Si}(k \pi)]$.
Since $\operatorname{sinc}$ is even, $\int_0^x \operatorname{sinc} t \,dt$ is odd in $x$. In summary, for $x \geq 0$ we have
$$\int_0^x |\operatorname{sinc} t| \,dt = (-1)^k \operatorname{Si}(x) - 2 \sum_{i = 1}^k (-1)^i \operatorname{Si}(\pi i) , \qquad k := \left\lfloor \frac{x}{\pi} \right\rfloor .$$
For large $x$ we have $\int_0^x |\operatorname{sinc} t| \,dt \sim \frac{2}{\pi} \log x + o(\log x) .$
