## anonymous 5 years ago When integrating a series, is the constant, C, from the integration evaluated as part of the summation or is it as a constant "outside" of the whole summation? That's, is there a summation from 0 to infinity of the C or is it just adding the C after the main summation?

1. anonymous

Say I have: $\sum_{n=0}^{\infty }x^{n}dx = \sum_{n=0}^{\infty }\frac{x^{n+1}}{n+1}+C$ Is the result this: $\sum_{n=0}^{\infty }x^{n}dx=\sum_{n=0}^{\infty }\left (\frac{x^{n+1}}{n+1}+C \right )=\sum_{n=0}^{\infty }\frac{x^{n+1}}{n+1}+ \sum_{n=0}^{\infty }C$ OR Is the result this: $\sum_{n=0}^{\infty }x^{n}dx=C+\sum_{n=0}^{\infty }\left (\frac{x^{n+1}}{n+1} \right )$ Thanks

2. anonymous

Got a missing integration in the previous post. I mean this: Say I have: $\int\limits \sum_{n=0}^{\infty }x^{n}dx = \sum_{n=0}^{\infty }\frac{x^{n+1}}{n+1}+C$ Is the result this: $\int\limits \sum_{n=0}^{\infty }x^{n}dx=\sum_{n=0}^{\infty }\left (\frac{x^{n+1}}{n+1}+C \right )=\sum_{n=0}^{\infty }\frac{x^{n+1}}{n+1}+ \sum_{n=0}^{\infty }C$ OR Is the result this: $\int\limits \sum_{n=0}^{\infty }x^{n}dx=C+\sum_{n=0}^{\infty }\left (\frac{x^{n+1}}{n+1} \right )$

3. nowhereman

It's actually not an 'one or the other'-answer ;-) Let us for the moment assume that those integrals exist and you can pull it to the inside. Then for every n there is a constant C(n) and for any choice of C(n) you get a solution. But you must notice that in order for the solution to exist the sum over the C(n) must be finite! So you are forced to choose the C(n) in such a way and then you can simply define: $C := \sum_{n=0}^∞{C(n)}$ to get a constant for the whole integral. Thus you see that it makes no mathematical difference if you have a constant for every single element or for the whole series. Many different choices of C(n) will yield the same sum, but in the end, that is the only information you need.