# Convergence of Integral

• May 8th 2011, 08:34 AM
EinStone
Convergence of Integral
Show that for $s\in \mathbb{C}$ and $Re(s) > 0$ the integral $\int_1^\infty \lfloor x\rfloor^{-s} - x^{-s}dx$ converges.
• May 8th 2011, 10:18 AM
girdav
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.
• May 8th 2011, 11:51 AM
EinStone
Well that is exactly what I want to show, I need some hint or maybe some criterion to use.
• May 8th 2011, 12:46 PM
girdav
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 $\sum_{k=1}^nk^{-s}$.
• May 8th 2011, 02:26 PM
EinStone
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.
• May 9th 2011, 02:07 AM
girdav
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).
• May 9th 2011, 02:16 AM
EinStone
ok, this looks good, but what are the <br/>'s and what is exactly is u?
• May 9th 2011, 02:44 AM
girdav
I don't know what <br/>'s are (maybe this come from latex issue). s=u+iv, I forgot it.