Evaluating $\int\frac{\mathrm dx}{\sqrt{\lfloor 1+ \sqrt{1+x}\rfloor}}$

Question

How can I solve this integral?

$$\int\frac{\mathrm dx}{\sqrt{\lfloor 1+ \sqrt{1+x}\rfloor}}$$

Answer 1

There does not exist an indefinite integral because of the intermediate value theorem for derivatives (Darboux's theorem).

If you are given a concrete interval over which to integrate, the question is trivial, because the function is piecewise constant.

Answer 2

Notice for any integer $n \ge 2$ and $x \ge 0$,

$$\begin{align} \left\lfloor 1 + \sqrt{1+x}\right\rfloor = n &\iff n \le 1 + \sqrt{1+x} < n+1\\ &\iff x \in [ (n-1)^2 - 1, n^2 - 1 ) \end{align}$$ To perform the integral over some interval $[0,X]$ where $X > 0$, you just need to split the interval $[0,X]$ to intervals of the form $[(n-1)^2-1, n^2 - 1)$. Let $N = \left\lfloor 1 + \sqrt{1+X}\right\rfloor$, we have

$$\begin{align} \int_0^X \frac{dx}{\sqrt{\left\lfloor 1 + \sqrt{1+x}\right\rfloor}} &= \left( \sum_{n=2}^{N-1}\int_{(n-1)^2-1}^{n^2-1} + \int_{(N-1)^2-1}^X\right) \frac{dx}{\sqrt{\left\lfloor 1 + \sqrt{1+x}\right\rfloor}}\\ &= \sum_{n=2}^{N-1}\frac{2n-1}{\sqrt{n}} + \frac{X - (N-1)^2 + 1}{\sqrt{N}} \end{align} $$