Verifying that a function is analytic Consider the function $$g(x)=\begin{cases} \frac{1}{x}\ln\frac{1}{1-x}\hspace{.5cm} x\neq0,\\ 1 \hspace{1.8cm} x=0.\end{cases}$$
Let $f$ be the function defined on $(-\infty,1)$ by the integral $f(x)=\int_{0}^{x}g(u)du=\int_{0}^{x}\frac{1}{u}\ln\frac{1}{1-u}du$. Find the Taylor expansion about 0 of $f(x)$ and use this to confirm that the function $f$ is analytic. 
My Taylor expansion is the following: $$0+g(0)(x)+\frac{g'(0)}{2!}x^2+\frac{g''(0)}{3!}+\cdots$$
which equals: $$f(x)=x+\frac{1}{2}\frac{x^2}{2!}+\frac{2}{3}\frac{x^3}{3!}+\frac{3}{2}\frac{x^4}{4!}$$
Which further equals: $$f(x)=\sum_{k=1}^{\infty}\frac{x^k}{k^2}.$$
Now I need to prove that this function is analytic, but I am not quite sure how to do that. 
 A: Recall that a function is analytic in $x=c$ iff there exists an expansion 
$$\sum_{n=0}^\infty a_n(x-c)^n$$
holds in a neibourhood about $x=c$. With this definition $\ln\dfrac{1}{1-x}$ is analytic about $x=0$ and indeed for $|x|<1$
$$\ln\dfrac{1}{1-x}=\sum_{n=1}^\infty \dfrac{x^n}{n}$$
also following holds
$$\dfrac1x\ln\dfrac{1}{1-x}=\sum_{n=1}^\infty \dfrac{x^{n-1}}{n} \hspace{2cm}|x|<1$$
and by integration the convergence interval doesn't change
$$f(x)=\int_0^x\dfrac1u\ln\dfrac{1}{1-u}du=\int_0^x\sum_{n=1}^\infty \dfrac{u^{n-1}}{n}du=\sum_{n=1}^\infty \dfrac{x^n}{n^2} \hspace{2cm}|x|<1$$
this expansion shows that $f(x)$ is analytic for $|x|<1$.
A: A function is real-analytic at a point if it is equal to its (convergent) Taylor series in some neighborhood of the point.
You found the Taylor series of $g$ around $x = 0$. If you can show it converges for $x$ in some interval containing $0$ then you are finished (at least as far as analyticity at $0$ is concerned.) Hint: ratio test
Also you could have found the Taylor series by using
$$\ln(1/(1-u)) = -\ln(1-u) = \sum_{k=1}^\infty \frac{u^k}{k},$$
dividing by $u$ and integrating termwise:
$$\int_0^x \frac{1}{u} \ln \left(\frac{1}{1-u} \right) \, du = \sum_{k=1}^\infty \int_0^x \frac{u^{k-1}}{k} \, du = \sum_{k=1}^\infty \frac{x^k}{k^2}$$
A: Since you have $$f(x) = \sum_{k=1}^{\infty} \frac{x^k}{k^2}$$ does it help that $$\frac{d^nf}{dx^n}=n!\sum_{k=1}^{\infty}\frac{x^{k-n}}{k^2}$$
See Change of Summation and Differentiation
I wanted to leave a comment but don't currently have enough rep. Also see
Alternative Characterizations of Analytic Functions 
