Why is the power series representation for $\ln(1-x^2)=-\sum_{n=1}^{\infty} \frac{x^{2 n}}{n}$ and not $\sum _{n=0}^\infty \frac {x^{2n+1}}{2n+1}$?

Question

I am working on finding a power series representation of $\ln(1-x^2)$ and am getting a wrong answer according to this answer key (problem 97.) I would like to understand why my answer is wrong and their answer is right. First I differentiate $\ln(1-x^2)$ to be $\frac {1}{1-x^2}$ which I get as $\sum _{n=0}^\infty x^{2n}$ using the fact that $\frac 1 {1-x^2} =\sum _{n=0} ^\infty x^n$. Then I integrate $\sum _{n=0}^\infty x^{2n}$ as $\sum _{n=0}^\infty \frac {x^{2n+1}}{2n+1}$ by integrating the generic term in the body of the sum. However, the answer is $-\int_{t=0}^{x^2} \frac{1}{1-t} d t=-\sum_{n=0}^{\infty} \int_0^{x^2} t^n d x-\sum_{n=0}^{\infty} \frac{x^{2(n+1)}}{n+1}=-\sum_{n=1}^{\infty} \frac{x^{2 n}}{n}$, which is different than my answer of $\sum _{n=0}^\infty \frac {x^{2n+1}}{2n+1}$. I have 4 questions here:

1. I have chosen to integrate $\frac 1 {1-x^2}$; how is $\int_{t=0}^{x^2} \frac 1 {1-t}dx$ related to this and why would we perform this computation rather than what I integrated?
2. How does the second expression in the equations hold? I do not understand it.
3. How does definite integration of power series work here? The textbook only discusses indefinite integration
4. Why is my answer wrong?

UPDATE: I have understood why my derivative is wrong as the comment helpfully pointed out. However, my confusion with the original answer still stands.

Answer 1

Let $F(x) = \ln(1-x)$. Then, $\ln(1-x^2) = F(x^2)$. We know that $F(x) = -\int_0^x\frac{1}{1-t}\,dt + F(0)$, and since $F(0) = 0$, we have $F(x^2) = -\int_0^{x^2}\frac{1}{1-t}\,dt$. The idea here is that if $F(x) = \sum a_nx^n$, then $F(x^2) =\sum a_nx^{2n}$, so basically, you are calculating series expansion of $\ln(1-x)$ (which comes from the integral $-\int\frac{dt}{1-t}$) and plugging in $x^2$ in the formula.

You can do it your way as well, we have $\frac d{dx}\ln(1-x^2) = \frac{-2x}{1-x^2} $, so $$\ln(1-x^2) = -2\int_0^x\frac{t}{1-t^2}\,dt = -2\sum_{n=0}^\infty \int_0^x t^{2n+1}\,dt = -\sum_{n=0}^\infty \frac{x^{2n+2}}{n+1} = -\sum_{n=1}^\infty \frac{x^{2n}}{n}.$$

The difference between definite and indefinite integration here is minor. Let's say that we want power series expansion of $F(x)$, and that we know power series expansion of $f(x) = \frac d{dx} F(x)$. Above we used the formula $F(x) = \int_0^x f(t)\,dt + F(0)$, which explains why definite integration works. If you used indefinite integration, what you would use instead is $\int f(x)\,dx = F(x)+C$. So, if you knew that $f(x) = \sum_{n=0}^\infty a_nx^n$, we get $$F(x) + C = \sum_{n=0}^\infty a_n\int x^n\,dx = \sum_{n=0}^\infty a_n \frac{x^{n+1}}{n+1}.$$ You still want to calculate $C$ and you can do it by plugging in $x = 0$ to get $F(0)+C = 0$, so $F(x) = \sum_{n=0}^\infty a_n \frac{x^{n+1}}{n+1} + F(0)$. Notice the similarity with the formula we had before with definite integration. Essentially, the difference is that definite integration automatically takes care of "$+C$" we get from indefinite integration.

Answer 2

A. Using the expansion,

\begin{aligned} \frac{1}{1-t}&=\sum_{n=0}^{\infty} t^n, \forall|t|<1, \\ \int_0^{x^2} \frac{1}{1-t} d t&=\sum_{n=0}^{\infty} \int_0^{x^2} t^n d t \\ -\ln \left(1-x^2\right)&=\sum_{n=0}^{\infty} \frac{x^{2(n+1)}}{n+1}=\sum_{n=0}^{\infty} \frac{x^{2 n}}{n} \end{aligned}

B. $$ \begin{aligned} \int \frac{1}{1-x^2} d x & =\frac{1}{2} \int\left(\frac{1}{1-x}+\frac{1}{1+x}\right) d x \\ & =\frac{1}{2} \ln \left|\frac{1+x}{1-x}\right|+C \end{aligned} $$ C. $$ \begin{aligned} \int_0^{x^2} \frac{1}{1-t} d t & =-[\ln |1-t|]_0^{x^2} \\ & =-\ln \left|1-x^2\right| \end{aligned} $$ D. Noting that for any $|x|<1$, we have $$ \begin{aligned} \frac{d}{d x}\left(\sum_{n=0}^{\infty} \frac{x^{2 n+1}}{2 n+1}\right)&=\sum_{n=0}^{\infty} \frac{d}{d x}\left(\frac{x^{2 n+1}}{2 n+1}\right) \\ & =\sum_{n=0}^{\infty} x^{2 n} \\ & =\frac{1}{1-x^2} \end{aligned} $$ Therefore $$ \sum_{n=0}^{\infty} \frac{x^{2 n+1}}{2 n+1} =\int \frac{1}{1-x^2} d x=\frac{1}{2} \ln \left|\frac{1+x}{1-x}\right|+C $$ Putting $x=0$ yields $C=0$ and hence $$ \sum_{n=0}^{\infty} \frac{x^{2 n+1}}{2 n+1}=\frac{1}{2} \ln \left|\frac{1+x}{1-x}\right| $$ Wish it helps!