I have a question that might be silly, but I really don't understand what is going on. I have to solve the following integral:
$$ \sum_{n \in \mathbb{Z}} \int_{m}^{m+1}e^{inx}dx $$
However if we try to solve it in normal fashion, its solution is divergent when $n=0$. The solution in the paper that I am reading is:
$$ \sum_{n \in \mathbb{Z}} \int_{m}^{m+1}e^{inx}dx = \sum_{n \in \mathbb{Z}}\delta_{n,0} + \frac{1-\delta_{n,0}}{in} \left(e^{in(m+1)}-e^{in(m)}\right) $$
However I think that the second term on the RHS is still divergent in $n=0$. Moreover, I do understand that the $\delta^{'}$s are introduced as an identity to cure the divergence, but I really don't understand how that is equal to the LHS.
Hopefully you can give me some insight :)
Ok, I guess I'll spell it out completely.
$$S_m=\sum_{n\in\Bbb{Z}}\int_{m}^{m+1}e^{inx}\mathrm{d}x$$ If $n=0$, the integrand is just $1$. In other words $$\int_{m}^{m+1}e^{i\cdot0\cdot x}\mathrm{d}x=\int_{m}^{m+1}1\mathrm{d}x=m+1-m=1.$$ Otherwise, $$\int_m^{m+1}e^{inx}\mathrm{d}x=\left(\frac{e^{inx}}{in}\right)\bigg|^{m+1}_m=\frac{e^{in(m+1)}-e^{inm}}{in}$$ So our sum can be written as $$S_m=1+\sum_{n\in\Bbb{Z}\backslash \{0\}}\frac{e^{in(m+1)}-e^{inm}}{in}$$ The sum given in your question is $$\sum_{n \in \mathbb{Z}}\delta_{n,0} + (1-\delta_{n,0})\frac{e^{in(m+1)}-e^{inm}}{in} $$ The zero term is $$\delta_{0,0}+(1-\delta_{0,0})\cdot(...)=1-(1-1)\cdot(...)=1$$ And the other terms are $$\delta_{n,0}+(1-\delta_{n,0})\frac{e^{in(m+1)}-e^{in(m)}}{in}=\frac{e^{in(m+1)}-e^{inm}}{in}$$ So the sum is $$S_m=1+\sum_{n\in\Bbb{Z}\backslash \{0\}}\frac{e^{in(m+1)}-e^{inm}}{in}.$$ Anything unclear?