<s>[INST] Calculation of $ \int_{0}^{\sqrt{n}}\lfloor t^2 \rfloor dt\;$, where $n\in \mathbb{N}$ and $\lfloor x \rfloor = $ floor function of $x$. Calculation of $\displaystyle \int_{0}^{\sqrt{n}}\lfloor t^2 \rfloor dt\;$, where $n\in \mathbb{N}$ and $\lfloor x \rfloor = $ floor function of $x$.
$\bf{My\; Try}::$ Let $t^2 = u$ and $\displaystyle dt = \frac{du}{2\sqrt{u}}$ and changing Limits, we get $\displaystyle \frac{1}{2}\int_{0}^{n}\frac{\lfloor u \rfloor }{\sqrt{u}}du$
Now Let 
$\displaystyle I = \int_{0}^{n}\lfloor u \rfloor \cdot u^{-\frac{1}{2}}du = \int_{0}^{1}0\cdot u^{-\frac{1}{2}}du+\int_{1}^{2}1\cdot u^{-\frac{1}{2}}du+\int_{2}^{3}2\cdot u^{-\frac{1}{2}}du+.........+\int_{n-2}^{n-1}(n-2)\cdot u^{-\frac{1}{2}}du+\int_{n-1}^{n}(n-1)\cdot u^{-\frac{1}{2}}du$
$\displaystyle =-2\left\{1\cdot \left(\sqrt{2}-1\right)+2\cdot \left(\sqrt{3}-\sqrt{2}\right)+3\cdot \left(\sqrt{4}-\sqrt{3}\right)+.......+(n-2)\cdot \left(\sqrt{n-1}-\sqrt{n-2}\right)+(n-1)\cdot \left(\sqrt{n}-\sqrt{n-1}\right)\right\}$
Now how can i calculate given sum in closed form
please Help me
Thanks
 [/INST] You can obtain $$\frac{1}{2}\int_0^n\frac{\lfloor u\rfloor}{u^{1/2}}\text{d}u =\frac{1}{2}\int_0^n\frac{u-\{u\}}{u^{1/2}}\text{d}u = \frac{1}{2}\int_0^n u^{1/2}\text{d}u - \frac{1}{2}\int_0^n\frac{\{u\}}{u^{1/2}}\text{d}u,$$ which gives
$$\frac{1}{2}\int_0^n\frac{\lfloor u\rfloor}{u^{1/2}}\text{d}u =\frac{1}{2}\left[\frac{2}{3}u^{3/2}\right]_0^n-\frac{1}{2}\int_0^n\frac{\{u\}}{u^{1/2}}\text{d}u=\frac{1}{3}n^{3/2}-\frac{1}{2}\int_0^n\frac{\{u\}}{u^{1/2}}.$$
Not sure if you can do much with the last integral, but for a "similar" integral notice that
$$\lim_{n\longrightarrow\infty}\int_0^n\frac{\{u\}}{u^{s+1}}\text{d}u = -\frac{\zeta(s)}{s},$$ where $\zeta(s)$ is the Riemann zeta function and $0<s<1$.
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