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Imagine it is known the value of the next integral:

$$ \int_{1}^{n}{ \left \lfloor{x}\right \rfloor a^{x}dx}$$

For any value of $a$. Is there any way to use this value to calculate, approximate or bound:

$$ B= \int_{1}^{n}{ \left \lfloor{x}\right \rfloor x^{s}dx}$$

It is clear that B could be rewritten as:

$$ B= \int_{1}^{n}{ \left \lfloor{x}\right \rfloor (e^{s})^{\log{x}}dx}$$

And making $a=e^s$ gets a bit closer, but I don't have any clue on how to deal with the integer part, how to move with the proper variable transformation or even if this is possible. Any bounding could work as well.

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John Wayland Bales gave a nice solution which, using generalized harmonic numbers, can write $$\int_1^n\lfloor x\rfloor x^s\,dx=\frac{n^{s+2}-H_n^{-(s+1)}}{s+1}$$ If you just want to bound the integral, why not to simply use $x-1 \leq \lfloor x\rfloor \leq x$ and get $$\frac{((n-1) s+n-2) n^{s+1}+1}{(s+1) (s+2)}\leq\int_1^n\lfloor x\rfloor x^s\,dx\leq \frac{n^{s+2}-1}{s+2}$$ $$\left( \begin{array}{cccc} s & n &\text{lower} & \text{exact} & \text{upper} \\ \sqrt{2} & 5 & 51.2608 & 60.4096 & 71.0158 \\ \sqrt{2} & 6 & 101.687 & 116.173 & 132.595 \\ \sqrt{2} & 7 & 179.612 & 200.903 & 224.641 \\ \sqrt{2} & 8 & 292.247 & 321.910 & 354.562 \\ \sqrt{2} & 9 & 447.273 & 486.967 & 530.221 \\ \sqrt{2} & 10 & 652.805 & 704.272 & 759.898 \\ & & & & \\ \sqrt{3} & 5 & 79.1761 & 92.5852 & 108.536 \\ \sqrt{3} & 6 & 166.04 & 188.542 & 214.591 \\ \sqrt{3} & 7 & 307.508 & 342.251 & 381.677 \\ \sqrt{3} & 8 & 521.43 & 571.949 & 628.413 \\ \sqrt{3} & 9 & 827.739 & 897.942 & 975.472 \\ \sqrt{3} & 10 & 1248.38 & 1342.54 & 1445.52 \\ & & & & \\ \sqrt{5} & 5 & 159.328 & 184.453 & 215.499 \\ \sqrt{5} & 6 & 365.206 & 411.503 & 466.787 \\ \sqrt{5} & 7 & 729.562 & 806.926 & 897.048 \\ \sqrt{5} & 8 & 1321.33 & 1441.79 & 1579.51 \\ \sqrt{5} & 9 & 2223.45 & 2401.25 & 2601.56 \\ \sqrt{5} & 10 & 3533.32 & 3784.99 & 4065.18 \end{array} \right)$$

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If $n>1$ is an integer then

\begin{eqnarray} \int_1^n\lfloor x\rfloor x^s\,dx&=&\sum_{k=1}^{n-1}\int_k^{k+1}kx^s\,dx\\ &=&\frac{1}{s+1}\sum_{k=1}^{n-1}\left[kx^{s+1}\right]_k^{k+1}\\ &=&\frac{1}{s+1}\sum_{k=1}^{n-1}k\left[(k+1)^{s+1}-k^{s+1}\right]\\ &=&\frac{1}{s+1}\left[1\left(2^{s+1}-1\right)+2\left(3^{s+1}-2^{s+1}\right)+\cdots+(n-1)\left(n^{s+1}-(n-1)^{s+1}\right)\right]\\ &=&\frac{1}{s+1}\left[(n-1)n^{s+1}-\left(1+2^{s+1}+3^{s+1}+\cdots+(n-1)^{s+1}\right)\right]\\ &=&\frac{1}{s+1}\left[(n-1)n^{s+1}-\sum_{k=1}^{n-1}k^{s+1}\right]\\ &=&\frac{1}{s+1}\left[n^{s+2}-\sum_{k=1}^{n}k^{s+1}\right] \end{eqnarray}

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  • $\begingroup$ Thanks for answer. It is a nice expansion of Abel’s summation lemma. I however was trying to get a way to avercome summing anything of the shape $x^s$, since there is no explicit formula for non integer $s$ $\endgroup$ – abr Apr 7 '18 at 0:31

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