Approximation to $\int_a^bu(t/\epsilon)\,dt$ for periodic function $u$, small $\epsilon > 0$ I'm reading a book and I'm not sure about the marked equality. 
$u_\epsilon$ is defined by $u_\epsilon (t) = u\left(\dfrac{t}{\epsilon}\right)$
where $u$ is a 1-periodic function defined on $\Bbb R$. 
Detailed explanation would be really appreciated.

Thanks a lot! 
 A: To declutter the notation, I'll drop the primes from the symbols $a'$ and $b'$, and use simply $a$ and $b$ instead.
Let $N$ denote the non-negative integer $\left\lfloor(b - a)/\epsilon\right\rfloor$.
Then $b - a = N\epsilon + \eta$, where $0 \leqslant \eta < \epsilon$.
Using a more correct notation for the integral on the left hand side (also acceptable is $\int_a^bu_\epsilon(t)\,dt$, of course):
$$
\int_a^bu_\epsilon =
\int_a^{a + N\epsilon} u_\epsilon + \int_{a + N\epsilon}^bu_\epsilon.
$$
Because $u_\epsilon$ has period $\epsilon$, we can rewrite the first term on the right hand side, as follows:
$$
\int_a^{a + N\epsilon} u_\epsilon
= \sum_{n=1}^N \int_{a + (n - 1)\epsilon}^{a + n\epsilon} u_\epsilon
= N \int_a^{a + \epsilon} u_\epsilon,
$$
and by similarly modifying the second term, we get:
$$
\int_a^bu_\epsilon =
N \int_a^{a + \epsilon} u_\epsilon + \int_a^{a + \eta}u_\epsilon.
$$
Assume that the function $u_\epsilon$ is bounded - equivalently, $u$ is bounded - say $|u_\epsilon(t)| \leqslant K$ for all $t \in \mathbb{R}$, where $K$ is independent of $\epsilon$. Then the absolute value of the second term is at most $\eta K \leqslant \epsilon K$, which is $O(\epsilon)$.
That is:
$$
\int_a^bu_\epsilon = N \int_a^{a + \epsilon} u_\epsilon + O(\epsilon).
$$
Let $M$ denote the integer $\left\lfloor a/\epsilon\right\rfloor$ (which of course may be negative).
Then $a = M\epsilon + \delta$, where $0 \leqslant \delta < \epsilon$.
By the periodicity of $u_\epsilon$:
\begin{align*}
\int_a^{a + \epsilon} u_\epsilon
& = \int_{M\epsilon + \delta}^{M\epsilon + \delta + \epsilon} u_\epsilon \\
& = \int_\delta^{\delta + \epsilon} u_\epsilon \\
& = \int_\delta^\epsilon u_\epsilon + \int_\epsilon^{\delta + \epsilon} u_\epsilon \\
& =\int_\delta^\epsilon u_\epsilon + \int_0^\delta u_\epsilon \\
& = \int_0^\epsilon u_\epsilon,
\end{align*}
giving finally:
$$
\int_a^bu_\epsilon = N \int_0^\epsilon u_\epsilon + O(\epsilon),
$$
which is what was stated.
