Can someone check if I have this right please? (Or point out where I'm wrong!)
I have a (real) impulse-train with maximum amplitude $1$ and period $R+1$, given by
$$s(x)=\frac {1}{R+1}+\sum_{k=1}^R\frac {\cos \bigl(\frac {2k\pi x}{R+1}\bigr)}{R+1}$$
The indefinite integral is (I think)
$$\int s(x) dx=\frac {x}{R+1}+\sum_{k=1}^R\frac {\sin \bigl(\frac {2k\pi x}{R+1}\bigr)}{2k\pi}+C$$
I am looking for the definite integral over the range $0$ to $R+1$, given by
$$\int_0^{R+1} s(x) dx=\int s(R+1) dx-\int s(0) dx$$
$$=\frac {R+1}{R+1}+\sum_{k=1}^R\frac {\sin \bigl(\frac {2k\pi (R+1)}{R+1}\bigr)}{2k\pi}-\frac {0}{R+1}-\sum_{k=1}^R\frac {\sin \bigl(\frac {0}{R+1}\bigr)}{2k\pi}$$
$$=1+\sum_{k=1}^R\frac {\sin 2k\pi}{2k\pi}$$
$$=1+\sum_{k=1}^R\operatorname{sinc}2k\pi$$
$$=1$$
(Because $\operatorname{sinc}2k\pi=0$ for all integer values of $k$.)
Could someone please confirm if I have this right, or point out where the error is?
It all looks good to me. With a bit less calculation, you could have also just said that the cosines all have periods that divide $R+1$ so their integral over an interval of length $R+1$ is $0$, so what you have is just $\frac1{R+1}$ integrated over an interval of length $R+1$ (which is $1$).