Convergence of Integral

**EinStone** · May 8th 2011, 08:34 AM

Show that for $s\in \mathbb{C}$ and $Re(s) > 0$ the integral $\int_1^\infty \lfloor x\rfloor^{-s} - x^{-s}dx$ converges.

**girdav** · May 8th 2011, 10:18 AM

You can begin to show that the sequence $\left\{I_n\right\}$ where $I_n = \int_1^n\left(\lfloor x\rfloor^{-s}-x^{-s}\right) dx$ is convergent.

**EinStone** · May 8th 2011, 11:51 AM

Well that is exactly what I want to show, I need some hint or maybe some criterion to use.

**girdav** · May 8th 2011, 12:46 PM

If we denote s =u+iv we have $\lfloor x\rfloor^{-s}-x^{-s} =\int_x^{\lfloor x\rfloor}(-s)t^{-1-s}dt = \int_{\lfloor x\rfloor}^xst^{-1-s}dt$ hence $|\lfloor x\rfloor^{-s}-x^{-s}|\leq \int_{\lfloor x\rfloor}^x|s||t^{-1-s}|dt = \int_{\lfloor x\rfloor}^x|s||t^{-1-s}|dt =\int_{\lfloor x\rfloor}^x|s||t^{-1-u-iv}|dt=\int_{\lfloor x\rfloor}^x|s|t^{-1-u}dt =|s|\frac{-1}u(x^{-u}-\lfloor x\rfloor^{-u})$ hence we only have to show the result if $s$ is a positive real number.
It's true if $s>1$ because $\sum_{n=1}^{+\infty}n^{-s}$ and $\int_1^{+\infty}t^{-s}dt$ are convergent. If $s=1$ use the fact that the sequence $\left\{\sum_{k=1}^n\frac 1k-\ln n\right\}$ is convergent. For $s<1$ , use inequalities involving $\int_1^n t^{-s}dt$ and $\sum_{k=1}^nk^{-s}$ .

**EinStone** · May 8th 2011, 02:26 PM

Ok, I derived something similar already as well, but for $0 < s < 1$ I really don't see how to show that $\int_1^\infty \lfloor x\rfloor^{-s} - x^{-s}dx$ converges.

**girdav** · May 9th 2011, 02:07 AM

The function $t\mapsto t^{-1-u}$ is decreasing hence $|\lfloor x\rfloor^{-s}-x^{-s}|\leq \frac{|s|}u(\lfloor x\rfloor^{-u}-x^{-u}|)=|s|\int_{\lfloor x\rfloor}^xt^{-1-u}dt\leq |s|(x-\lfloor x\rfloor ){\lfloor x\rfloor}^{-1-u}$ . We have
$\begin{align*} \int_1^{n+1}|\lfloor x\rfloor^{-s}-x^{-s}|dx&\leq |s|\sum_{k=1}^n\int_k^{k+1} \left(x\lfloor x\rfloor^{-1-u}-\lfloor x\rfloor^{-u}\right)dx\\ &=|s|\sum_{k=1}^n\int_k^{k+1} \left(xk^{-1-u}-k^{-u}\right)dx\\ &=|s|\left(\sum_{k=1}^nk^{-1-u}\left(\frac{(k+1)^2}2-\frac{k^2}2\right)-k^{-u}\right)\\ &= |s|\left(\sum_{k=1}^nk^{-1-u}\left(k+\frac 12\right)-k^{-u}\right)\\ &= \frac{|s|}2\sum_{k=1}^nk^{-1-u} \end{align*}$
which is convergent (we don't have to distinguish the case $s>1$ or not, so it's better than my previous work).

**EinStone** · May 9th 2011, 02:16 AM

ok, this looks good, but what are the 's and what is exactly is u?

**girdav** · May 9th 2011, 02:44 AM

I don't know what 's are (maybe this come from latex issue). s=u+iv, I forgot it.

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