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Evaluate$\int_{1}^{\infty}$ $\frac{1-(x-[x])}{x^{2-\sigma}}$dx where [x] denotes greatest integer function and $0<\sigma<1$.

My try:- 1-(x-[x])$\leq 1 \Rightarrow$ $\int_{1}^{\infty}$ $\frac{1-(x-[x])}{x^{2-\sigma}}$dx $\leq$ $\int_{1}^{\infty}$ $\frac {1}{x^{2-\sigma}}$dx= $\frac{1}{1-\sigma}$

$\int_{1}^{\infty}$ $\frac{1-(x-[x])}{x^{2-\sigma}}$dx= $\int_{1}^{2}$ $\frac{1-(x-1)}{x^{2-\sigma}}$dx+$\int_{2}^{3}$ $\frac{1-(x-2)}{x^{2-\sigma}}$dx+... But on integration I am not getting a finite value.

StubbornAtom
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2 Answers2

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Trying to continue along your way.

You wrote $$\int_1^\infty(1-x+\lfloor x\rfloor )\, x^{\sigma -2}\,dx=\sum_{n=1}^\infty \int_n^{n+1}(n+1-x)\,x^{\sigma -2}\,dx=\sum_{n=1}^\infty I_n$$ $$I_n=\int_n^{n+1}(n+1-x)\,x^{\sigma -2}\,dx=\frac{ (n+\sigma )n^{\sigma }-n (n+1)^{\sigma }}{n (1-\sigma) \sigma }=\frac{n^{\sigma -1} (n+\sigma )-(n+1)^{\sigma } } {(1-\sigma)\, \sigma }$$ and here the problem starts to be difficult if you are not familiar with the zeta function.

Hoping that you are, the result should be $$\frac{1+\sigma\, \zeta (1-\sigma ) } {(1-\sigma) \,\sigma }$$

Claude Leibovici
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  • Thank you for your answer. I am not able to get how did you get the last step. I know the Riemann zeta function but still i cannot understand. Please write some steps to ger the zeta part –  Aug 21 '20 at 08:44
  • Please write the answer to get the zeta part –  Aug 21 '20 at 09:06
  • Zeta function is defined for Re(s)>1 –  Aug 21 '20 at 09:20
  • The series is telescoping so we will be left with first and last term so $ (n+1)^\sigma$ will also be present. –  Aug 21 '20 at 13:29
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This is not a complete answer, but hopefully will provide some insight:

Let $f(x) = \frac{1-(x-\lfloor x\rfloor)}{x^{2-\sigma}}$, then as $0\leqslant x-\lfloor x\rfloor<1$ and $x^{2-\sigma}>0$, $f$ is nonnegative on $[1,\infty)$. Hence by Tonelli's theorem, $$ \int_1^\infty f(x)\ \mathsf dx = \int_1^\infty \sum_{n=1}^\infty f_n(x)\ \mathsf dx = \sum_{n=1}^\infty \int_n^{n+1} f_n(x)\ \mathsf dx, $$ where $f_n(x) = f(x)\cdot\mathsf 1_{[n,n+1)}$. By induction (the tricky part), we can show that $$ \int_n^{n+1} f_n(x)\ \mathsf dx = \frac{n^{\sigma -1} (n+\sigma )-(n+1)^{\sigma }}{\sigma(1-\sigma)}. $$ Hence $$ \int_1^\infty f(x)\ \mathsf dx = \frac{1+\sigma\, \zeta (1-\sigma ) } {(1-\sigma) \,\sigma }, $$ where $$ \zeta(s) := \sum_{n=1}^\infty \frac1{n^s} $$ is the Riemann zeta function.

Math1000
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  • 1000 How do you get to zeta function. Please add some more steps –  Aug 21 '20 at 09:01
  • Zeta function is defined for Re(s)>1 –  Aug 21 '20 at 09:21
  • @ShekharSuman More precisely, the zeta function is the analytic continuation of that series (which indeed only converges for $\mathsf{Re}(s)>1$. It is well-defined on $\mathbb C\setminus{1}$. – Math1000 Aug 21 '20 at 11:04
  • So here Re (s)<1 so the series should be divergent? –  Aug 21 '20 at 11:53
  • @ShekharSuman See here: https://math.stackexchange.com/questions/437883/what-is-the-analytic-continuation-of-the-riemann-zeta-function – Math1000 Aug 21 '20 at 11:59
  • But how do you get to the zeta function from the second last step. Please explain –  Aug 21 '20 at 12:38
  • Please write the last part when we take summation of I_n –  Aug 21 '20 at 13:02
  • The series is telescoping so we will be left with first and last term so $ (n+1)^\sigma$ will also be present. –  Aug 21 '20 at 13:29