How to show that the integral $\int_0^1 \ln(1-x^{1/n})dx=-\sum_{k=1}^n\frac{1}{k}$ The integral is 
$$\int_0^1 \ln(1-x^{1/n})dx=-\sum_{k=1}^n\frac{1}{k}$$
I tried integral by part, but cannot get the $\frac{1}{k}$ term. How can we do this integral? Any hint would be appreciated.
 A: Rewrite the integrand as a power series, so the integral is$$\begin{align}-\sum_{k\ge1}\frac1k\int_0^1x^{k/n}dx&=-\sum_k\frac{n}{k(n+k)}\\&=-\sum_k\left(\frac{1}{k}-\frac{1}{n+k}\right),\end{align}$$which simplifies to $-\sum_{k=1}^n\frac1k$ by telescoping series. To make this argument rigorous, one should compute the sum from $k=1$ to $k=K$ for $K\ge n$ by induction.
A: Use the Taylor series of the logarithm
$$
\ln\left(1-x^\frac{1}{n}\right)=-\sum_{k=1}^\infty\frac{x^\frac{k}{n}}{k}.
$$
Then, integrate term by term. This yields
$$
-\sum_{k=1}^\infty\frac{1}{k\left(1+\frac{n}{k}\right)}=-\psi_0(n+1)-\gamma
$$
being $\psi_0$ the digamma funtion and $\gamma$ the Euler-Mascheroni constant. By the definition of the digamma function this yields your result.
A: Start with $x^\frac{1}{n}=u$  we have 
$$I=\int_0^1\ln(1-x^{\frac1n})\ dx=n\int_0^1 u^{n-1}\ln(1-u)\ du$$
Now we integrate by parts but to avoid divergence, we write $\int n u^{n-1}\ du=u^n-1$.
So
$$I=\underbrace{(u^n-1)\ln(1-u)|_0^1}_{0}-\int_0^1\frac{1-u^n}{1-u}\ du=-\sum_{k=1}^n\frac1k=-H_n$$

Proof for the last result
$$\sum_{k=1}^n\frac1k=\sum_{k=1}^n\int_0^1 u^{n-1}\ du=\int_0^1\sum_{k=1}^n u^{n-1}\ du=\int_0^1\frac{1-u^n}{1-u}\ du$$
