Trying to get the average distance between two random points inside a circle I came accross this function, which I have no idea how to go around this.
The question is to simplify $$\lim_{n\to \infty } \sum _i^n \sqrt{\left(\frac{h-x}{n}\right)^2+\left(\frac{i}{n^2}\right)^2\left(1-x^2\right)}$$
knowing that $h,x\in\mathbb{R}$ and $i,n\in\mathbb{Z}$.
In a unit circle, a random point is identified by $\rho\cdot(\cos\theta,\sin\theta)$ where $\theta$ has a uniform distribution over $[0,2\pi]$ and the PDF of $\rho$ is supported on $(0,1)$ and given by $2x$. It follows that the average distance between two random points in a unit circle is given by
$$ \mathbb{E}[d]=\frac{1}{\pi}\int_{0}^{\pi}\int_{0}^{1}\int_{0}^{1}4\rho_1\rho_2\sqrt{\rho_1^2+\rho_2^2-2\rho_1\rho_2\cos\theta}\,d\rho_1\,d\rho_2\,d\theta $$ that is an elliptic integral, whose numerical value is $\approx 0.905415$. It follows that the average distance between two random points in a unit circle is just a bit less than $1$. That can be proved by convexity, too: $$ \mathbb{E}[d]\leq \int_{0}^{1}\int_{0}^{1}4ab\sqrt{a^2+b^2}\,da\,db = \frac{8}{15}(\sqrt{8}-1)<1.$$
Late addendum: it can be shown through interesting techniques that $\mathbb{E}[d]=\color{blue}{\frac{128}{45\pi}}$.
Have a look at pages $135$ and $136$ of these notes.