# How to take the integral of a series?

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 $\int\frac{1}{1-x}dx=-\ln(1-x)+C$

• Term by term integration? Mar 2, 2018 at 18:35
• well, you have your answer, that is $$\int\sum_{k=0}^\infty x^k=-\ln |1-x|+C$$ for $|x|<1$. Mar 2, 2018 at 18:39
• What about exploit the linearity of summation and integration and exchanging the order of the integral and summation signs? But do it fast, before a mathematician warns you about some trouble with a guy called Lebesgue or something like that ;) Mar 2, 2018 at 18:41
• @AndrewLi We don't have uniform convergence in the case of $\sum_{i = 0}^\infty x^n$, so we can't be sure that the integral of the sum is the same as the sum of the integral of each term Mar 2, 2018 at 18:43
• @AndrewLi the series indeed converges on that interval, but it does not do so uniformly. See math.stackexchange.com/questions/2105693/… Mar 2, 2018 at 18:50

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.

• There are some hypothesis to check to make this shift.
– user371663
Mar 2, 2018 at 18:47
• Actually, one can shift only on $[a,b]$ with $-1<a<b<1$
– user371663
Mar 2, 2018 at 18:53
• @Lucas Please see my update! Mar 2, 2018 at 19:02
• I read your answer; but you could just use normal convergence of the series on such a $[a,b]$.
– user371663
Mar 2, 2018 at 19:32