Determine the primitives of $ f:(0,\infty )\rightarrow \mathbb{R},f(x)=\frac{\sqrt{[x]}+\sqrt{\left \{ x \right \}}}{\sqrt{x}} $ , where $ \left \{ x \right \} $ represents the fractional part and $ [x] $, the floor of $ x $.

I didn't find a suitable method to integrate the function.

  • $\begingroup$ Honestly, I don't even know how to start it. I didn't find a suitable method to integrate the function. $\endgroup$ – ztefelina Dec 13 '16 at 21:51

Hint 1: When dealing with the floor function, it is often useful to check how it behaves on intervals where the floor function is constant. Also notice that $\{x\} = x-[x]$. So for $x\in [n,n+1)$, $n \in \mathbb{N}$ we have $[x] = n$ and the function simplifies to $$ f(x) = \frac{\sqrt{n}+\sqrt{x-n}}{\sqrt{x}}. $$ Then you want to integrate $$\int f(x)\,dx=\int \frac{\sqrt{n}+\sqrt{x-n}}{\sqrt{x}}\,dx = \sqrt{n}\int\frac{1}{\sqrt{x}}\,dx+\int\sqrt{1-\frac{n}{x}}\,dx$$

and use the fact that $(0,\infty)$ is union of all such intervals $[n,n+1)$ (excluding $0$).

Hint 2: The second integral in the expression above is a bit tricky. Consider case $n=1$ and substitution $t^2=1-\frac{1}{x}$. Then you have $x \in [1,2)$ and $t\in \left[0,\frac{1}{\sqrt{2}}\right)$. You can write \begin{align} x&=\frac{1}{1-t^2}\\ dx&=\frac{2t}{(1-t^2)^2}dt \end{align} Finally plugging this into the integral $$ \int\sqrt{1-\frac{1}{x}}\,dx = \int \frac{2t^2}{(1-t^2)^2}\,dt $$ and you should be able to solve this using parcial fractions (be careful with the signs, especially because $t-1< 0$). You can generalize this to a generic $n$, it will just get messy.

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