Equivalent of $ I_{n}=\int_0^1 \frac{x^n \ln x}{x-1}\mathrm dx, n\rightarrow \infty$ I would like to show that
$$ I_{n}=\int_0^1 \frac{x^n \ln x}{x-1}\mathrm dx \sim_{n\rightarrow \infty} \frac{1}{n}$$
Using the change of variable $u=x^n$:
$$ I_{n}=\frac{1}{n^2} \int_0^1 \frac{u^{1/n} \ln u}{u^{1/n}-1} \mathrm du=\frac{1}{n^2}\left(\int_0^1 \ln x \mathrm dx+\int_0^1 \frac{\ln x}{x^{1/n}-1} \mathrm dx \right)=\frac{-1}{n^2}+\frac{1}{n^2}\int_0^1 \frac{\ln x}{x^{1/n}-1} \mathrm dx=o(1/n)+\frac{1}{n^2}\int_0^1 \frac{\ln x}{x^{1/n}-1} \mathrm dx$$
So I have to show that 
$$ \int_0^1 \frac{\ln x}{x^{1/n}-1} \mathrm dx \sim_{n \rightarrow \infty} n$$
Could you help me?
 A: $$\frac{x^n}{x-1} = \frac{x^n-1}{x-1} + \frac{1}{x-1}= \left(1+ x+ \cdots + x^{n-1}\right) +\frac{1}{x-1}.$$
So $$I_n = \int^1_0 (1+x+\cdots + x^{n-1}) \log x + \int^1_0 \frac{\log x}{x-1} dx.$$
Integration by parts shows $ \int^1_0 x^k \log x = -1/(k+1)^2$ and expanding $\log$ by Taylor series will show $\displaystyle \int^1_0 \frac{\log x}{x-1} dx = \frac{\pi^2}{6}$ so $$I_n = \sum_{k=n+1}^{\infty} \frac{1}{k^2}.$$
Thus $$nI_n = \frac{1}{n} \sum_{k=n+1}^{\infty} \frac{1}{(k/n)^2} \to \int_1^{\infty} \frac{1}{x^2} dx=1$$
so $I_n \sim 1/n.$

You can skip showing $\displaystyle \int^1_0 \frac{\log x}{x-1} dx = \frac{\pi^2}{6}$ if you expand $x^n/(x-1)$ as a geometric series from the start. 
Instead of using Riemann sums we could also have noted that $1/x^2$ is monotonically decreasing and use the well known theorem that if $f$ is monotone then $\displaystyle \int^n_1 f(x) dx \sim \sum_{k=1}^n f(n).$
A: I propose different approach.
First, consider integral:
$$J_n = \int_0^1 \frac{x^n -1}{x-1} \, dx$$
than we have $I_n = \tfrac{dJ_n}{dn}$. Observe that:
$$J_n = \int_0^1 \frac{x^n - 1 }{x-1} \, dx = H_n \sim \ln n + \ldots$$
So it's now easy to note that:
$$I_n \sim \frac{1}{n}$$
for large $n$.
A: Actually, a closed form solution may be given: 
$$I_{n}=\int_0^1 \frac{x^n \ln x}{x-1}dx=-\int_0^1[x^n(1+x^2+x^3+...)\ln x]dx=$$  
$$=-\sum_{k=0}^\infty \int_0^1 (x^{k+n}\ln x)dx=\sum_{k=0}^\infty\frac{1}{(k+n+1)^2}=$$  
$$=\frac{1}{(n+1)^2}+\frac{1}{(n+2)^2}+...=\zeta(2,n+1) $$ where $$\zeta(s,a)=\sum_{k=0}^\infty\frac{1}{(k+a)^s}$$ is so called Hurwitz Zeta Function
