Question

The n-th harmonic number is defined by

Hn=∑n, i=1 (1/i) =1+1/2+1/3+…+1/n.

Show that

Hn=∫(1>0) 1−x^n/1−x dx.

Answer 1

$x=1-u$

$\int^1_0\frac{1-x^n}{1-x}dx=-\int^0_1\frac{1-(1-u)^n}{u}du=\int^1_0\frac{1-(1-u)^n}{u}du=$

$=\int^1_0[\displaystyle{\sum_{k=1}^n(-1)^{k-1}\begin{pmatrix} n \\ k \end{pmatrix}u^{k-1}}]du=\displaystyle{\sum_{k=1}^n(-1)^{k-1}}\begin{pmatrix} n \\ k \end{pmatrix}\int^1_0u^{k-1}du=$

$=\displaystyle{\sum_{k=1}^n(-1)^{k-1}}\frac{1}{k}\begin{pmatrix} n \\ k \end{pmatrix}$