I was wondering if one would be able to calculate the outcome of the Riemann zeta function using integrals, based upon the fact that the Riemann sums are a way to calculate an integral.

The Riemann zeta function is $$\zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s}$$ Now let's say that our input $r \in \mathbb{N}$ and $r > 1$. That way we can create the parametric function $$f_r(x) = \frac{1}{x^r}$$

Now my theory was that $\zeta(s)$ could be calculated by $$\zeta(r) \approx \int_{1}^{\infty}f_r(x)dx = \lim_{n \to \infty} F_r(n) - F_r(1)$$ For that I calculated that for $r>1$ $$\int f_r(x)dx = \int(\frac{1}{x^r})dx = \int(x^{-r})dx = \frac{1}{-r+1} * x^{-r+1}$$ and for $r=1$ $\int f_r(x)dx = \int \frac{1}{x}dx = \ln(x)$

Using the information I had now I figured that for instance $\zeta(2)$ could be calculated approximately using $$\zeta(2) \approx \lim_{n \to \infty} F_2(n) - F_2(1) = \lim_{n \to \infty} \frac{1}{-2+1}*n^{-2+1}-\frac{1}{-2+1}*1^{-2+1} = \lim_{n \to \infty} \frac{1}{-n}+\frac{1}{1} = 1$$ which is nothing close to the $\frac{\pi^2}{6}$ Wikipedia told me about.

Now my question is what went wrong. Is the theory just plainly wrong or did I miscalculate something, for example in the anti-derivative?

Thanks in advance.


The series definition implies that for any $s:\text{Re}(s)>1$ we have

$$ \zeta(s)=\sum_{n\geq 1}\frac{1}{n^s} = \sum_{n\geq 1}\int_{0}^{+\infty}\frac{u^{s-1}}{\Gamma(s)}e^{-nu}\,du = \frac{1}{\Gamma(s)}\int_{0}^{+\infty}\frac{u^{s-1}}{e^u-1}\,du$$ but in approximating $\sum_{n\geq 1}\frac{1}{n^2}$ with $\int_{1}^{+\infty}\frac{dx}{x^2}$ you still have to consider the non-zero error term coming from the Euler-Maclaurin or Abel-Plana formula.

  • $\begingroup$ So the reason why it didn't work out very well was because the approximation wasn't precise enough? $\endgroup$ – SuperSjoerdie Feb 9 '18 at 18:44
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    $\begingroup$ @SuperSjoerdie: exactly. $\sum_{n\geq 1}f(n) = \int_{1}^{+\infty}f(x)\,dx$ holds in very few occasions. $\endgroup$ – Jack D'Aurizio Feb 9 '18 at 18:47
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    $\begingroup$ It is especially rare when $f$ is strictly decreasing as is common :). $\endgroup$ – Erick Wong Feb 10 '18 at 2:48

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