Double Integral form of the nth Harmonic Number

Question

Consider the double integral $$\int_0^1 \int_0^1 n(1-xy)^{n-1}dx dy.$$ Entering this double integral into Wolfram Alpha gives that it is equal to the $n^{th}$ harmonic number, where the $n^{th}$ harmonic number denotes $\sum_{k=1}^n 1/k$. Now when I first saw this result, I didn't initially believe it and I tested it out by trying discrete, integer values of n until I was finally convinced. I wanted to ask and see if anyone could help me understand the intuition behind why this integral results in the Harmonic series, perhaps similar to this question. Essentially, it's just hard for me to see the connection between this integral and the series.

Answer 1

The double integral can be evaluated by repeated integration. \begin{gather*} \int_{0}^{1}\int_{0}^{1}n(1-xy)^{n-1}\,dxdy = \int_{0}^{1}\left[\dfrac{(1-xy)^{n}}{-x}\right]_{0}^{1}\, dx = \int_{0}^{1}\dfrac{1-(1-x)^{n}}{x}\,dx =[x=1-z] = \\[2ex] \int_{0}^{1}\dfrac{1-z^{n}}{1-z}\, dz = H_{n} \end{gather*} where I have used your suggested link.