2
$\begingroup$

I was recently integrating the floor of $x^2$ and had almost finished it, however this finite Sum was left unevaluated.

$$\frac{1}{3}\sum_{i=1}^{\lfloor x^2 \rfloor} {\Biggl(\Bigl(\sqrt{(i-1)}\Bigr)(2i-2)-\Bigl(\sqrt{(i-1)}\Bigl)(2i-3)\Biggl)}=?$$

It would be really nice if there was a closed form for this.

The way I came up with this sum:

$$\int_{0}^x {\lfloor t^2 \rfloor} = \int_{0}^x {t^2}-\{t^2\}$$ $$=\frac{x^3}{3}-\int_{0}^x\{t^2\}$$

and if you are Looking at the function of $\{x\}$ then you can see that this is the function $x^2$ only that it starts again at $0$ at every value of a square root of an integer.

so that $$\int_{0}^x\{t^2\}=\Biggl(\sum_{i=0}^{\lfloor x^2 \rfloor} \int_{\sqrt{i-1}}^{\sqrt{i}} k^2-i+1 dk\Biggr)+\int_{_{\sqrt{\lfloor x^2 \rfloor}}}^x v^2-\lfloor x^2 \rfloor dv$$

Now that means that: $$\int_{0}^x\{x^2\}=\frac{1}{3}\sum_{i=1}^{\lfloor x^2 \rfloor} {\Biggl(\Bigl(\sqrt{(i-1)}\Bigr)(2i-2)-\Bigl(\sqrt{(i-1)}\Bigl)(2i-3)\Biggl)}+\Bigl(-x{\lfloor x^2 \rfloor} +\frac{2{\lfloor x^2 \rfloor}^{\frac{3}{2}}}{3}+\frac{x^3}{3}\Bigr)$$

However, I was unable to find a closed form for the Sum, I would appreciate any help of you.

Moreover, I know that it can be expressed in Terms of the Zeta function and the Harmonic series, but what I am searching for i a form in Terms of some easy mathematical function.

$\endgroup$
0

1 Answer 1

1
$\begingroup$

If you instead consider the original integral as a sum you get for $x\ge0$ $$\int_0^x\lfloor t^2\rfloor \mathrm{d}t=\lfloor x^2\rfloor\left(x-\sqrt{\lfloor x^2\rfloor}\right)+\sum_{k=1}^{\lfloor x^2\rfloor -1} k\left(\sqrt{k+1}-\sqrt{k}\right)$$ which telescopes to $$\begin{align} \int_0^x\lfloor t^2\rfloor \mathrm{d}t &=x\lfloor x^2\rfloor-\sum_{k=1}^{\lfloor x^2\rfloor} \sqrt{k}\\ &=x\lfloor x^2\rfloor-H_{-1/2}\left(\lfloor x^2\rfloor\right)\\ \end{align}$$ where $H_m(n)$ denote the generalized harmonic numbers. We can also generalise this for $n\in\mathbb{N}$ to $$\begin{align} \int_0^x\lfloor t^n\rfloor \mathrm{d}t &=x\lfloor x^n\rfloor-\sum_{k=1}^{\lfloor x^n\rfloor} \sqrt[n]{k}\\ &=x\lfloor x^n\rfloor-H_{-1/n}\left(\lfloor x^n\rfloor\right)\\ \end{align}$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .