How to take the integral of a series?

Question

How do you take an integral of a series like $$\int\frac{1}{1-x}dx=\int\sum_{n=0}^\infty x^ndx$$ Can someone help me how to do this?

Note:I know that $\int\frac{1}{1-x}dx=-\ln(1-x)+C$

Answer 1

Assuming that $|x|<1$, then we have a geometric progression, which is convergent. If $|x|<1$ then $\int_{y}^{x}{\frac{1}{1-x}}dx=\int_{y}^{x}\sum_{n=0}^\infty x^ndx=\sum_{n=0}^\infty \int_{y}^{x} x^ndx.$ where $x,y\in(-1,1)$.

The proof of my statement goes as this: Consider $s(x)=u_1(x)+u_2(x)...u_n(x)+r_n(x)$ where $r_n(x)$ is the remaining. Since the series converge there is $\epsilon>0$ so that for an $n>N$ $|r_n(x)|<\epsilon$

$\int_{a}^{x}s(x)dx=\int_{a}^{x}u_1(x)dx+\int_{a}^{x}u_2(x)dx...\int_{a}^{x}u_n(x)+\int_{a}^{x}r_n(x)dx$, once we can write the sum of finite terms equal to the sum of its integrals. Since $r_n\to 0$, as $n\to\infty$ we have:

$\lim_{n\to\infty}(\int_{a}^{x}s(x)dx-(\int_{a}^{x}u_1(x)dx+\int_{a}^{x}u_2(x)dx...\int_{a}^{x}u_n(x)))=0\implies $$\int_{a}^{x}s(x)dx=\int_{a}^{x}u_1(x)dx+\int_{a}^{x}u_2(x)dx...\int_{a}^{x}u_n(x)$ as we wanted to prove.