Evaluate $$\int_{0}^{[x]}\left(\int_{0}^{[t]}[s]-\left[s-\frac{1}{2}\right] \mathrm{d}s\right) \mathrm{d}t$$
Where $[x]$ denotes the greatest integer function . But I don't know how to approach this kind of problem. So I want to know how to think to solve this type of problems.
If there are more than one way to approach, I would like to gather the knowledge.
Thanks in advance.
Edit: Edited after the comments.
Since $x$ only appears in a floor function in a limit of the outer integral, we can assume $x$ is an integer: $$\int_0^x\int_0^{[t]}[s]-[s-1/2]\,ds\,dt$$ $[s]-[s-1/2]$ is $0$ when the fractional part of $s$ is greater than $\frac12$ and $1$ when it is not. Since the bounds of the inner integral are also integers, it simply evaluates to half of the distance between the bounds: $$=\int_0^x\frac12[t]\,dt$$ Now assuming $x>0$ (the other case is similar), $t$ assumes the same value for each of $x$ intervals of length $1$: the values from $0$ to $x-1$ inclusive. Thus the integral becomes a finite sum. $$=\frac12(0+1+\dots+(x-1))=\frac{x(x-1)}4$$ The final answer is $\frac{[x]([x]-1)}4$.