$ 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.

  • $\begingroup$ The nth terms in the two sums are equal, i.e. the differences $=0$. A countably infinite sum of zeroes $=0$.. $\endgroup$ – herb steinberg Jan 17 at 4:34
  • $\begingroup$ The value of an integral does not change if you change the values at a finite number of points. $\endgroup$ – Kavi Rama Murthy Jan 17 at 4:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.