I tried to show that $$\int_{0}^{1}\left(\sum_{k=1}^{\infty}\frac{x^{k}}{k}\ln{x}\right)\,dx=\sum_{k=1}^{\infty}\left(\int_{0}^{1}\frac{x^{k}}{k}\ln{x}\,dx\right).$$
How to show that the series of functions $F_{n}(x):=\sum_{k=1}^{n}\frac{x^{k}}{k}\ln{x}$ converges uniformly on $[0,1]$?
Because of the factor $\ln{x}$, it is hard to me.
Give some advice. Thank you!
On
The map $x \mapsto -x \log x$, together with $0 \log 0 := 0$, is non-negative and continuous on $[0, 1]$. From this, we find that $-F_n$ is continuous, non-negative and pointwise increasing in $n$. Moreover, its pointwise limit is
$$ \lim_{n\to\infty} (-F_n(x)) = \log(1-x)\log x, $$
which is continuous on $[0, 1]$. So by Dini's theorem, this convergence is uniform.
If your ultimate goal is simply interchanging the order of integral and summation, uniform convergence is a bit redundant, in view of the monotone convergence theorem.
The maximum value of $|(x^k\ln x)/k|$ on $[0,1]$ occurs at $x=(1/e)^{1/k},$ giving a value of $\dfrac{1}{ek^2}.$ Since $\sum 1/k^2 < \infty,$ the given series converges uniformly by the Weierstrass M test.