Is it possible to interchange summation and integral on $\int^{1}_0 \sum^{\infty}_{i=1} \frac{(-1)^{i+1}x^{i-1}}{i} dx$? Is it possible to interchange summation and integral on
$$\int^{1}_0 \sum^{\infty}_{i=1} \frac{(-1)^{i+1}x^{i-1}}{i} dx ?$$
The Fubini's theorem says that for general $f_n$, if $\int \sum |f_n| < \infty$ or $\sum \int |f_n| < \infty$, then $\int \sum f_n = \sum \int f_n$. I tried with this theorem, but it doesn't work. 
Could someone tell me the method to be able to interchange the summation and integral in this case? Or just a hint is sufficient.
 A: Abel's Theorem states that if $\sum_n^\infty  a_n$ is a convergent series of constants, then the power series $\sum_n^\infty  a_nx^n$ converges uniformly for $0\le x\le 1$.  
In addition, we have that if a sequence of functions $f_n(x)$ converges uniformly to $f(x)$ on $x\in[a,b]$, then 
$$\lim_{n\to \infty}\int_a^b f_n(x)\,dx=\int_a^b f(x)\,dx$$
Therefore, we see that $\sum_{i=1}^{\infty}\frac{(-1)^{i+1}x^{i-1}}{i}$ converges uniformly on $[0,1]$ and 
$$\bbox[5px,border:2px solid #C0A000]{\int_0^1 \sum_{i=1}^{\infty}\frac{(-1)^{i+1}x^{i-1}}{i}\,dx=\sum_{i=1}^{\infty}\frac{(-1)^{i+1}}{i}\int_0^1 x^{i-1}\,dx=\sum_{i=1}^{\infty}\frac{(-1)^{i+1}}{i^2}}$$

It is straightforward to show that 
$$\sum_{i=1}^{\infty}\frac{(-1)^{i+1}x^{i-1}}{i}=\frac{\log(1+x)}{x}$$
Putting all of this together, we note that 
$$\begin{align}
\int_0^1\frac{\log (1+x)}{x}\,dx&=\sum_{i=1}^{\infty}\frac{(-1)^{i+1}}{i^2}\\\\
&=-\text{Li}_2(-1)\\\\
&=\frac{\pi^2}{12}
\end{align}$$
where $\text{Li}_2(x)$ is the dilogarithm function.
