# Questions Regarding Convergence of Formulas for Riemann Zeta Function $\zeta(s)$

I've been told assuming $\{x\}=x-\lfloor x\rfloor$, the following integral (1) evaluates to $\zeta(s)$ in the interval $\Re(s)\in(0,1)$.

(1) $\quad\zeta(s)=-s\int_0^\infty \{x\}\,x^{-s-1}dx\,,\quad 0<\Re(s)<1$

Since $x-\lfloor x\rfloor=SawtoothWave(x)$ and $SawtoothWave(x)=\frac{1}{2}-\frac{1}{\pi}\sum_{k=1}^\infty \frac{\sin\,(2\,\pi\,k\,x)}{k}$, this leads to the following.

(2) $\quad\zeta(s)=-s\int_0^\infty \left(\frac{1}{2}-\frac{1}{\pi}\sum_{k=1}^K \frac{\sin\,(2\,\pi\,k\,x)}{k}\right)\,x^{-s-1}dx\,,\quad 0<\Re(s)<1\ \&\ K\to\infty$

I can't seem to get integral (2) above to converge when evaluating the integral along the critical line.

Including the saw-tooth wave offset of $\frac{1}{2}$ is one problem. Integral (3) below doesn't converge, so I assume the offset is really not supposed to be included in the evaluation and am taking the approach of evaluating integral (4) below instead of integral (2) above.

(3) $\quad -s\int_0^\infty \left(\frac{1}{2}\right)\,x^{-s-1}dx\,,\quad 0<\Re(s)<1$

(4) $\quad\zeta(s)=-s\int_0^\infty \left(-\frac{1}{\pi}\sum_{k=1}^K \frac{\sin\,(2\,\pi\,k\,x)}{k}\right)\,x^{-s-1}dx\,,\quad 0<\Re(s)<1\ \&\ K\to\infty$

Question 1: Is it correct to omit the saw-tooth wave and evaluate integral (4) instead of integral (2)?

I'm using formula (5) below to evaluate integral (4) above. Increasing the evaluation limit $K$ leads to another problem. The more I increase the evaluation limit $K$, the more formula (5) seems to diverge.

(5) $\quad\zeta(s)=2^s\pi^{s-1}\,\Gamma(1-s)\sin\left(\frac{\pi\,s}{2}\right)\sum _{k=1}^K k^{s-1}\,,\quad 0<Re(s)<1\ \&\ K\to\infty$

Formula (5) above is based on formula (6) below.

(6) $\quad-s\int_0^\infty \left(-\frac{\sin(2\,\pi\,k\,x)}{\pi\,k}\right)\,x^{-s-1}dx=2^s\pi^{s-1}\,\Gamma(1-s)\sin\left(\frac{\pi\,s}{2}\right)\,k^{s-1}\,,\quad 0<\Re(s)<1$

Also, note formula (5) above is consistent with the functional equation $\zeta(s)=2^s\pi^{s-1}\,\Gamma(1-s)\sin\left(\frac{\pi\,s}{2}\right)\,\zeta(1-s)$, where $\zeta(1-s)=\sum_{k=1}^\infty k^{s-1}$.

Question 2: Why does formula (5) seem to diverge as the evaluation limit $K$ is increased?

I encounter the same problem when evaluating the following integral.

(7) $\quad\zeta(s)=\frac{\zeta(s)}{\zeta(1-s)}\sum_{k=1}^K k^{s-1}\,,\quad 0<Re(s)<1\ \&\ K\to\infty$

Since $\zeta(1-s)=\sum_{k=1}^\infty k^{s-1}$, it seems to me formula (7) above should converge as the evaluation limit $K$ is increased, but increasing the evaluation limit $K$ seems to have the opposite effect.

Question 3: Why does formula (7) above seem to diverge as the evaluation limit $K$ is increased?

• See any proof of the functional equation and any book on $\zeta(s)$. And (1) is true because $\zeta(s) = s \int_1^\infty \lfloor x \rfloor x^{-s-1}dx$ for $Re(s) > 1$ (Abel summation formula) – reuns Mar 7 '17 at 4:13
• Of course $\lim_{K \to \infty} \sum_{k=1}^K k^{s-1}$ diverges for $Re(s) > 0$. See any course on real analysis/Riemann series – reuns Mar 7 '17 at 4:25

## 2 Answers

See this proof of the functional equation

Even if we have the Fourier series $$x -\lfloor x \rfloor = \{x\} =\frac12 - \sum_{k=1}^\infty \frac{\sin 2 \pi k x}{\pi k}$$and for $Re(s) \in (0,1)$ $$\zeta(s) = -s \int_0^\infty \{x\} x^{-s-1}dx$$ It it not true that for $Re(s) \in (0,1)$ $$\zeta(s) = \lim_{K \to \infty}-s \int_0^\infty \left(\frac12 - \sum_{k=1}^K \frac{\sin 2 \pi k x}{\pi k}\right) x^{-s-1}dx \quad \text{nor } \lim_{K \to \infty}s \int_0^\infty \sum_{k=1}^K \frac{\sin 2 \pi k x}{\pi k} x^{-s-1}dx$$

(can you even prove the RHS converges ?)

This is what I'm explaining to you since the beginning :

in general it is not true that $$\lim_{K \to \infty} \int_a^b f_K(x)dx=\int_a^b \lim_{K \to \infty} f_K(x) dx$$ and proving they are equal for some particular $f_K$ is one of the main subject of real analysis.

• I understand your point in general that $\lim_{K \to \infty} \int_a^b f_K(x)dx=\int_a^b \lim_{K \to \infty} f_K(x) dx$ is not always true, but I disagree with your specific point that relationship (2) never converges to $\zeta(s)$ in the interval $Re(s)\in(0,\,1)$. Your specific point may be true for the initial formula I was using to evaluate the integral, but it is not true for the formula illustrated in the answer that I posted. – Steven Clark Mar 8 '17 at 15:15

I originally thought that perhaps the reason I was having problems getting relationship (1) to converge was it was incorrect, and relationship (A) below was the correct relationship. However as Peter Humphries pointed out, relationship (1) is equivalent to relationship (A) in the interval $\Re(s)\in(0,1)$. In any event, relationship (A) below can be used to derive a formula which converges under the more general condition $\Re(s)>0$ with $s\ne 1$.

(A) $\quad\zeta(s)=\frac{s}{s-1}+s\int_1^{\infty }(\lfloor x\rfloor-x)\,x^{-s-1}\,dx\,,\quad \Re(s)>0\ \&\ s\ne 1$

The following plot illustrates a formula which I derived from relationship (A) above converges nicely for $\Re(s)=\frac{1}{2}$. The formula is illustrated in blue and the function provided by the Wolfram Language for $\zeta(s)$ is illustrated in orange as a reference, but since the formula evaluates so closely to $\zeta(s)$ the reference function in orange hides the evaluation of the formula in blue. So the green curve was added to the plot to illustrate the difference between the evaluations of the formula and the reference function. The plot below illustrates the real part of $\zeta(s)$ evaluated along the critical line. I'll note I derived the formula illustrated above using the Fourier series for the SawtoothWave(x) function, and the plot above illustrates the evaluation of the first 10 harmonics.

The following plot illustrates how the evaluation of the real part of the first harmonic of the derived formula (blue) tracks the real part of $\zeta(s)$ (orange). Note there's a one-to-one correspondence between the positive peaks of the real parts of the first harmonic and $\zeta(s)$. • Note that $-s \int_{0}^{\infty} \{x\} x^{-s} \, \frac{dx}{x} = -s \int_{0}^{1} x^{-s} \, dx - s\int_{1}^{\infty} \{x\} x^{-s} \, \frac{dx}{x}$, and as $0 < \Re(s) < 1$, the first integral is $\left. -s \frac{x^{1 - s}}{1 - s} \right|_{0}^{1} = \frac{s}{s - 1}$. So your first statement is false; relationship (1) is correct in the strip $0 < \Re(s) < 1$. – Peter Humphries Mar 7 '17 at 21:57
• On the other hand, relationship (A) is actually valid for all $\Re(s) > 0$ with $s \neq 1$, whereas (1) diverges for $\Re(s) > 1$. – Peter Humphries Mar 7 '17 at 22:00
• Peter: Thanks for the clarification which illustrates the equivalence of relationship (1) and relationship (A) in the interval $0<\Re(s)<1$. – Steven Clark Mar 7 '17 at 23:50