Solution
= Solution
Fubini's theorem gives
$$
\frac1u\int_{-u}^u(1-\varphi(t))dt
=\mathbb E\left[\frac1u\int_{-u}^u(1-e^{itX})dt\right].
$$
The odd imaginary part integrates to zero, while
$$
\int_{-u}^ue^{itX}dt=\frac{2\sin(uX)}{X}.
$$
Therefore the expression equals
$$
2\mathbb E\left(1-\frac{\sin(uX)}{uX}\right),
$$
with the quotient interpreted continuously at $X=0$.