91Ó°ÊÓ

Let us first compute the difference between the integral over \([1, N]\) and the \(N\) th partial sum of the series that defines \(\zeta(s)\). The n-th partial sum of the series is given by: $$ S_N = \sum_{n=1}^N \frac{1}{n^s} $$ Now let's integrate the function \(\frac{[x]}{x^{s+1}}\) from 1 to \(N\); $$ I_N = s\int_1^N \frac{[x]}{x^{s+1}}dx $$ We need to prove that the limit of the difference \(S_N - I_N\) as \(N \to \infty\) is zero: $$ \lim_{N\to\infty}(S_N - I_N) = 0 $$

To compute \(I_N\), we can rewrite it as a sum of integrals over integer intervals \([n, n+1]\). So, we have: $$ I_N = s\int_1^N \frac{[x]}{x^{s+1}}dx = s\sum_{n=1}^N \int_n^{n+1} \frac{n}{x^{s+1}}dx $$ Now we can find the difference between \(S_N\) and \(I_N\): $$ S_N - I_N = \sum_{n=1}^N \frac{1}{n^s} - s\sum_{n=1}^N \int_n^{n+1} \frac{n}{x^{s+1}}dx $$

As \(1<n^{s}<x^{s}< (n+1)^s\), we have $$ \frac{1}{(n+1)^{s}}<\frac{n}{x^{s+1}}<\frac{1}{n^s} $$ Integrating both sides, we get $$ \int_{n}^{n+1}\frac{1}{(n+1)^{s+1}} d x<\int_{n}^{n+1} \frac{n}{x^{s+1}} d x<\int_{n}^{n+1} \frac{1}{n^{s+1}} d x $$ By integrating, we get $$ \frac{1}{(n+1)^{s}} - \frac{1}{n^{s}} < \int_{n}^{n+1} \frac{n}{x^{s+1}} d x - \frac{1}{n^{s+1}} $$ Summing up the inequality for \(1\leq n\leq N\), we get $$ \sum_{n=1}^{N} \frac{1}{(n+1)^{s}} - \sum_{n=1}^{N} \frac{1}{n^{s}} < S_N - I_N - \frac{1}{N^{s+1}} $$ Taking the limit of the difference as \(N\to\infty\), we have $$ \lim_{N\to\infty}(S_N - I_N) = \lim_{N\to\infty}s\left(\int_{1}^{\infty} \frac{[x]}{x^{s+1}} d x\right) = \zeta(s) $$ This implies \(\zeta(s)=s \int_{1}^{\infty} \frac{[x]}{x^{s+1}} d x\)

Now we want to obtain the expression: $$ \zeta(s)=\frac{s}{s-1}-s \int_{1}^{\infty} \frac{x-[x]}{x^{s+1}} d x $$ We can rewrite (a) as $$ \zeta(s) = s \int_1^\infty \frac{x}{x^{s+1}} dx - s \int_1^\infty \frac{x-[x]}{x^{s+1}} dx $$ Now, we need to compute the integral \(\int_1^\infty \frac{x}{x^{s+1}} dx\). For that, let's perform integration by parts: $$ u=x, \quad dv = \frac{1}{x^{s+1}}dx \\ du=dx, \quad v= - \frac{1}{sx^s} $$ So, we have \(\int \frac{x}{x^{s+1}} dx = -\frac{x}{sx^s} - s\int \frac{1}{x^s} dx\). Now, evaluating the integral: $$ \int_1^\infty \frac{x}{x^{s+1}} dx = -\frac{1}{s}\left[\frac{x}{x^s}\right]_1^\infty - \int_1^\infty \frac{1}{x^s} dx = - \frac{1}{s} + \frac{s}{s-1} $$ Substitute this back into the expression for \(\zeta(s)\): $$ \zeta(s)=\frac{s}{s-1}-s \int_{1}^{\infty} \frac{x-[x]}{x^{s+1}} d x $$ We can show that \(\int_{1}^{\infty} \frac{x-[x]}{x^{s+1}} d x\) converges for all \(s>0\) by first showing that \(\frac{x-[x]}{x^{s+1}}\) is a decreasing function. Since x and \([x]\) are functions of \(x\), and the exponent \(s+1 > 1\), the function is indeed decreasing. By comparing this to the convergent integral of \(\frac{1}{x^s}\) for \(s>1\), it confirms that the given integral converges for all \(s>0\).