Exercise 6.16

Question

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Exercise 16: For $1<s<\infty$, define $$\zeta(s)=\sum_{n=1}^{\infty}\frac{1}{n^s}\quad\hbox{(Riemann's zeta function).}$$ Prove that
    \begin{align*}
        \hbox{(a)}\quad\zeta(s) &= s\int_1^{\infty}\frac{[x]}{x^{s+1}}\,dx \
        \hbox{(b)}\quad\zeta(s) &= \frac{s}{s-1}-s\int_1^{\infty}\frac{x-[x]}{x^{s+1}}\,dx,
    \end{align*}
    where $[x]$ denotes the greatest integer $\le x$.

Prove that the integral in (b) converges for all $s>0$.

analambanomenos · Accepted Answer

(a) We have \begin{align*} s\int_1^N\frac{[x]}{x^{s+1}}\,dx &= s\sum_{n=1}^{N-1}n\int_n^{n+1}x^{-s-1}\,dx \ &= \sum_{n=1}^{N-1}n\biggl(\frac{1}{n^s}-\frac{1}{(n+1)^s}\biggr) \ &= 1\biggl(\frac{1}{1^s}\biggr)+\Biggl(\sum_{n=2}^{N-1}\bigl(n-(n-1)\bigr)\frac{1}{n^s}\Biggr)-\frac{N-1}{N^s} \ &= \Biggl(\sum_{n=1}^N\frac{1}{n^s}\Biggr)-\frac{1}{N^s}-\frac{N-1}{N^s} \ &= \Biggl(\sum_{n=1}^N\frac{1}{n^s}\Biggr)-\frac{1}{N^{s-1}} \end{align*} Taking the limit as $N ightarrow\infty$, we get the desired result. (b) From the work done in part (a), we get \begin{align*} \frac{s}{s-1}-s\int_1^N\frac{x-[x]}{x^{s+1}}\,dx &= \frac{s}{s-1}-s\int_1^Nx^{-s}\,dx+s\int_1^N\frac{[x]}{x^{s+1}}\,dx \ &= \frac{s}{s-1}+\frac{s}{s-1}\biggl(\frac{1}{N^{s-1}}-1\biggr)+\Biggl(\sum_{n=1}^N\frac{1}{n^s}\Biggr)-\frac{1}{N^{s-1}} \ &= \frac{1}{s-1}\biggl(\frac{1}{N^{s-1}}\biggr)+\Biggl(\sum_{n=1}^N\frac{1}{n^s}\Biggr) \end{align*} Taking the limit as $N ightarrow\infty$, we get the desired result. Note that $$0<\int_1^b\frac{x-[x]}{x^{s+1}}\,dx<\int_1^b\frac{1}{x^{s+1}}\,dx,$$ and the integral on the right converges as $b ightarrow\infty$ for all $s>0$ by the integral test of Exercise 8. Hence the integral in part (b) also converges as $b ightarrow\infty$ for all $s>0$.

Exercise 6.16

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Exercise 6.16

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