$ H(t):= \begin{cases} \dfrac{1}{2}-t & (0<t<1) \\ 0 & (t=0) \\ \text{periodic of period 1} &(\text{otherwise}) \end{cases}$
$ \mu (x):=\sum_{n=0}^{\infty} \displaystyle\int_0^1 \dfrac{\frac{1}{2}-t}{t+x+n} dt.$
My mathematic book says
\begin{align} \mu (x) &=\sum_{n=0}^{\infty} \displaystyle\int_0^1 \dfrac{\frac{1}{2}-t}{t+x+n} dt \\ &=\sum_{n=0}^{\infty} \displaystyle\int_0^1 \dfrac{H(t)}{t+x+n} dt \\ &=\sum_{n=0}^{\infty} \displaystyle\int_n^{n+1} \dfrac{H(t)}{t+x} dt \\ &=\lim_{m \to \infty} \displaystyle\int_0^{m}\dfrac{H(t)}{t+x} dt . \end{align}
But I wonder if $\sum_{n=0}^{\infty} \displaystyle\int_0^1 \dfrac{\frac{1}{2}-t}{t+x+n} dt =\sum_{n=0}^{\infty} \displaystyle\int_0^1 \dfrac{H(t)}{t+x+n} dt $ holds, because
$H(t)$ is not equal to $\dfrac{1}{2}-t$ at the point of $t=0$ and $t=1.$
So I think I should write
\begin{align} \mu (x) &=\sum_{n=0}^{\infty} \displaystyle\int_0^1 \dfrac{\frac{1}{2}-t}{t+x+n} dt\\ &=\sum_{n=0}^{\infty} \lim_{\epsilon \to +0, \delta \to +0}\displaystyle\int_{\delta}^{1-\epsilon} \dfrac{\frac{1}{2}-t}{t+x+n} dt \\ &=\sum_{n=0}^{\infty} \lim_{\epsilon \to +0, \delta \to +0}\displaystyle\int_{\delta}^{1-\epsilon} \dfrac{H(t)}{t+x+n} dt \\ &=\sum_{n=0}^{\infty} \lim_{\epsilon \to +0, \delta \to +0}\displaystyle\int_{\delta +n}^{1-\epsilon +n} \dfrac{H(t-n)}{t+x} dt \\ &=\sum_{n=0}^{\infty} \lim_{\epsilon \to +0, \delta \to +0}\displaystyle\int_{\delta +n}^{1-\epsilon +n} \dfrac{H(t )}{t+x} dt \\ &=\lim_{m\to \infty} \sum_{n=0}^{m} \lim_{\epsilon \to +0, \delta \to +0}\displaystyle\int_{\delta +n}^{1-\epsilon +n} \dfrac{H(t )}{t+x} dt .\\ \end{align}
I caannot show that this is equal to $\lim_{m \to \infty} \displaystyle\int_0^{m}\dfrac{H(t)}{t+x} dt .$
How should I do? I would like you to give me some ideas.
