mathimatikoi.org

Let $n \in \mathbb{N} \cup \{0\}$. Evaluate:

$$\mathcal{S}=\sum_{n=0}^{\infty} \int_0^1 \int_0^1 \left|x - y \right|^n \, {\rm d}(x, y)$$

Well,

\begin{align*}
\sum_{n=1}^{\infty} \int_0^1 \int_{0}^{1}\left | x-y \right |^n \, {\rm d}(x, y) &= \sum_{n=1}^{\infty} \frac{2}{(n+1)(n+2)}\\
&= 2\sum_{n=1}^{\infty}\left [ \frac{1}{n+1} - \frac{1}{n+2} \right ] \\
&=2 \cdot \frac{1}{2} \\
&= 1
\end{align*}

since $\displaystyle \iint \limits_{[0,1]^2} \left | x-y \right |^n \, {\rm d}(x, y) = \frac{2}{(n+1)(n+2)}$. Proof ?

If \(\displaystyle{y\in\left[0,1\right]}\), then

\(\displaystyle{\begin{aligned}\int_{0}^{1}|x-y|^n\,\mathrm{d}x&=\int_{0}^{y}(y-x)^n\,\mathrm{d}x+\int_{y}^{1}(x-y)^n\,\mathrm{d}x\\&=\left[-\dfrac{(y-x)^{n+1}}{n+1}\right]_{0}^{y}+\left[\dfrac{(x-y)^{n+1}}{n+1}\right]_{y}^{1}\\&=\dfrac{y^{n+1}}{n+1}+\dfrac{(1-y)^{n+1}}{n+1} \end{aligned}}\)

and then

\(\displaystyle{\begin{aligned}\int_{[0,1]^2}|x-y|^n\,d(x,y)&=\int_{0}^{1}\left(\dfrac{y^{n+1}}{n+1}+\dfrac{(1-y)^{n+1}}{n+1}\right)\,\mathrm{d}y\\&=\left[\dfrac{y^{n+2}}{(n+1)\,(n+2)}-\dfrac{(1-y)^{n+2}}{(n+1)\,(n+2)}\right]_{0}^{1}\\&=\dfrac{1}{(n+1)\,(n+2)}+\dfrac{1}{(n+1)\,(n+2)}\\&=\dfrac{2}{(n+1)\,(n+2)}\end{aligned}}\)

Am I right ?