An integral representation of nth Harmonic number is
$$H_n = \displaystyle \int_0^1 \frac{1 - x^n}{1 - x}\,dx$$
Wikipedia states that for every x > 0, integer or not, we have: $$H_{n} = n \displaystyle\sum_{k=1}^\infty \frac{1}{k(n+k)}$$
How can I get to this result from integral representation?
$\endgroup$3 Answers
$\begingroup$We have $$H_n = \sum_{k=1}^n \dfrac1k = \sum_{k=1}^n \int_0^1 x^{k-1} dx = \int_0^1 \left(\sum_{k=1}^n x^{k-1} \right) dx = \int_0^1 \left(\dfrac{1-x^n}{1-x} \right)dx$$
You may also want to look at these answers [1] and [2] and , where I derived a similar/same result. The second part of your question is answered in [1].
$\endgroup$ 7 $\begingroup$For the second, $$n \sum\limits_{k=1}^{+ \infty} \frac{1}{k(n+k)}= \sum\limits_{k=1}^{+ \infty} \left( \frac{1}{k}-\frac{1}{n+k} \right)$$
$\endgroup$ $\begingroup$"How can I get to this result from integral representation?"
Here's how, taking your request literally. The trick is integrating by parts
$$ \begin{align} &\int_0^1 \frac{1-x^n}{1-x} \, dx\tag{1}\\ &=-\int_0^1 \left(1-x^n\right) \frac{d}{dx} \log (1-x) \, dx\tag{2a}\\ &=-\left(1-x^n\right) \log (1-x)\Big|^{1}_{0}-\int_0^1 n\; x^{n-1} \log (1-x) \, dx\tag{2b}\\ &=\int_0^1 n\; x^{n-1} \sum _{k=1}^{\infty } \frac{x^k}{k} \, dx\tag{2c}\\ &= \sum _{k=1}^{\infty } \frac{n}{k}\int_0^1 x^{n+k-1} \, dx\tag{2d}\\ &=\sum _{k=1}^{\infty } \frac{n}{k (n+k)}\tag{2e}\\ \end{align} $$
Explanation
(2a): rewrite the integrand to prepare integration by parts
(2b): do integration by parts. Notice that the integrated part vanishes at both ends of the interval
(2c): expand $\log (1-x)$ into a series
(2d): Exchange summation and integration
(2e): do the integral
Derivation finished
