SPECIAL VALUES RIEMANN ZETA FUNCTION

Recently, We made extensive use of special values of the Riemann Zeta function and it´s derivatives without any proof. Today, the goal of this post is to prove some of them, namely:

$\begin{aligned} &\zeta(0)=-\frac12\\ & \\ &\zeta(-1)=-\frac{1}{12}\\ & \\ & \zeta^\prime(0)=-\frac12\ln 2 \pi\\ & \\ &\zeta^\prime(-1)=\frac{1}{12}-\ln A \end{aligned}$

Where A is the Glaisher–Kinkelin constant defined by

$A=\lim _{n \rightarrow \infty} \frac{K(n+1)}{n^{\frac{n^{2}}{2}+\frac{n}{2}+\frac{1}{12}} e^{-\frac{n^{2}}{4}}}$

Where $K(n)=\prod_{k=1}^{n-1}k^k$ is the K-function.

In the way the Riemann zeta function is defined

$\zeta(s)=\sum_{k=1}^\infty \frac{1}{k^s}$

clearly for s=1 the above equation blows up!

for s=0

$\sum_{k=1}^\infty 1=1+1+1+\cdots$

for s=-1

$\sum_{k=1}^\infty k=1+2+3+\cdots$

and so on. So it does not make much sense to ask it´s value for any s that is not strictly greater than one, s>1 . But, with that aid of the analytic continuation technique it´s possible to make sense for questions like what is the value of $\zeta(-1)=\,?$ for instance. One possible way to do this is by the application of the Euler Maclaurin Formula.

Therefore, we will start by introducing the Bernoulli polynomials and numbers that are essential prerequisites to derive the Euler Maclaurin summation formula which we will prove in the sequence, to than finally prove the special values of the Riemann Zeta function and it´s derivatives.

Bernoulli Polynomial´s and Numbers

The Bernoulli Polynomials can be defined by the following properties

B_0=1 (1)

$\mathrm{B}_{k}(x)=k \int_0^x\mathrm{B}_{k-1}(t)\,dt+B_k$ (2)

$\int_0^1B_k(x)\,dx=0$ (3)

Where B_k is the kth Bernoulli number.

To find B_1(x) we proceed as following

$\begin{aligned} B_1(x)&=\int B_0(x)dx+B_1 \qquad (\text{by equation (2)} )\\ &=\int 1 \,dx+B_1 \qquad (\text{by equation (1)} )\\ &=x+B_1 \end{aligned}$

To find the value of B_1 we use equation (3)

$\begin{aligned} &\int_0^1B_1(x)\,dx=0\\ &\int_0^1 \left(x+B_1 \right)\,dx=0\\ &\frac12+B_1=0\\ &B_1=-\frac12 \end{aligned}$

We then conclude that

$B_1(x)=x-\frac12$ (4)

Similarly

$\begin{aligned} B_2(x)&=2\int P_1(x)dx+B_2 \\ &=2\int\left( x-\frac12\right)dx+B_2 \\ &=2\left(\frac{x^2}{2}-\frac{x}{2}\right)+B_2\\ &=x^2-x+B_2 \end{aligned}$

Then, to find the value of the constant

$\begin{aligned} &\int_0^1B_2(x)\,dx=0\\ &\int_0^1 \left(x^2-x+B_2 \right)\,dx=0\\ &\frac{x^3}{3}-\frac{x^2}{2}+B_2x \Big|_0^1=0\\ &B_2=\frac16 \end{aligned}$

Therefore

$B_2(x)=x^2-x+\frac16$ (5)

If We keep this process we obtain for the first few Bernoulli Polynomials

$\begin{aligned} &B_{0}(x)=1 \\ &B_{1}(x)=x-\frac{1}{2} \\ &B_{2}(x)=x^{2}-x+\frac{1}{6} \\ &B_{3}(x)=x^{3}-\frac{3}{2} x^{2}+\frac{1}{2} x \\ &B_{4}(x)=x^{4}-2 x^{3}+x^{2}-\frac{1}{30} \\ &B_{5}(x)=x^{5}-\frac{5}{2} x^{4}+\frac{5}{3} x^{3}-\frac{1}{6} x \\ &B_{6}(x)=x^{6}-3 x^{5}+\frac{5}{2} x^{4}-\frac{1}{2} x^{2}+\frac{1}{42} \end{aligned}$

By setting x=0 in the B_k(x) above, we find the Bernoulli numbers. Note that, with the exception of $B_1=-\frac12$ , all other Bernoulli numbers of odd order are equal to zero.

Periodic Bernoulli Functions

We define the periodic Bernoulli polynomials P_k(x) for it´s importance in the Euler Maclaurin Formula. The periodic Bernoulli functions are defined as

$P_k(x)=B_k(x- \lfloor x \rfloor)$ (6)

The first Bernoulli function is defined as

$P_1(x)=\begin{cases}x- \lfloor x \rfloor-\frac12 & \text{if}\,\, x\neq \text{integer}\\ 0 & \text{if}\,\, x= \text{integer}\end{cases}$ (7)

Because of it´s periodicity, P_k(x) , has the constant value P_k(0)=P_k(n) at all integers.

Properties of the Bernoulli Periodic Functions. $(k \geq 1):$

$\mathrm{P}_{1}(x)=$ sawtooth function (eq. (7)).
$\mathrm{P}_{k}(x)=k \int_0^x\mathrm{P}_{k-1}(t)\,dt+B_k$ for or $x \notin \mathbb{Z}$ .
$\int_{0}^{1} \mathrm{P}_{k}(x) d x=0$ .

Euler Maclaurin Formula

We now derive the Euler Maclaurin Formula. We start analyzing the following integral:

$\begin{aligned} \int_k^{k+1} \lfloor t\rfloor f^{\prime}(t)\,dt&=k\int_k^{k+1} f^{\prime}(t)\,dt &=k\left(f(k+1)-f(k) \right)\\ \end{aligned}$

Summing the above equation from to n-1 we get

$\begin{aligned} \sum_{k=m}^{n-1}\int_k^{k+1} \lfloor t\rfloor f^{\prime}(t)\,dt&=\int_m^{n} \lfloor t\rfloor f^{\prime}(t)\,dt\\ &=\sum_{k=m}^{n-1}k\left(f(k+1)-f(k) \right)\\ &=(n-1)f(n)-mf(n)-\sum_{k=m+1}^{n-1}f(k)\\ &=nf(n)-mf(n)-\sum_{k=m+1}^{n}f(k) \end{aligned}$

Therefore

$\begin{aligned} \sum_{k=m+1}^{n}f(k)&=nf(n)-mf(n)-\int_m^{n} \lfloor t\rfloor f^{\prime}(t)\,dt\\ &=tf(t)\Big|_m^b-\int_m^{n} \lfloor t\rfloor f^{\prime}(t)\,dt\\ &=\int_m^nf(t)dt+\int_m^{n} t f^{\prime}(t)\,dt-\int_m^{n} \lfloor t\rfloor f^{\prime}(t)\,dt\\ &=\int_m^nf(t)dt+\int_m^{n} \left(t-\lfloor t\rfloor\right) f^{\prime}(t)\,dt\\ &=\int_m^nf(t)dt+\int_m^{n} \left(t-\lfloor t\rfloor-\frac12\right) f^{\prime}(t)\,dt+\frac12\int_m^n f^{\prime}(t)dt\\ &=\int_m^nf(t)dt+\int_m^{n} \left(t-\lfloor t\rfloor-\frac12\right) f^{\prime}(t)\,dt+\frac12\left(f(n)-f(m) \right)\\ \end{aligned}$

Adding f(m) to both sides of the equation we obtain

$\sum_{k=m}^{n}f(k)=\frac12\left(f(n)+f(m) \right)+\int_m^nf(t)dt+\int_m^{n} \left(t-\lfloor t\rfloor-\frac12\right) f^{\prime}(t)\,dt$ (8)

Observe that the term inside the second integral in the above equation is the first Bernoulli function

$P_1(t)=t-\lfloor t\rfloor-\frac12$

Therefore, we can write the first order Euler Maclaurin summation formula as

$\sum_{k=m}^{n}f(k)=\frac12\left(f(n)+f(m) \right)+\int_m^nf(t)dt+\int_m^{n} P_1(t) f^{\prime}(t)\,dt$ (9)

Taking advantage of the properties of the Periodic Bernoulli Function, We may integrate by parts the last term in the R.H.S. of (9)

$\begin{aligned} \int_m^n f^{\prime}(t)P_1(t)dt&=\frac12f^{\prime}(t)P_2(t)\Big|_m^n-\frac12\int_m^n f^{\prime \prime}(t)P_2(t)dt\\ &=\frac{P_2(0)\left(f^{\prime}(n)-f^{\prime}(m) \right)}{2}-\frac12\int_m^n f^{\prime \prime}(t)P_2(t)dt \qquad (P_2(n)=P_2(m)=P_2(0))\\ &=\frac{\left(f^{\prime}(n)-f^{\prime}(m) \right)}{12}-\frac12\int_m^n f^{\prime \prime}(t)P_2(t)dt \qquad \left(P_2(0)=B_2(0)=\frac16 \right) \end{aligned}$

And the Euler Maclaurin formula becomes

$\sum_{k=m}^{n}f(k)=\frac12\left(f(n)+f(m) \right)+\frac{\left(f^{\prime}(n)-f^{\prime}(m) \right)}{12}+\int_m^nf(t)dt-\frac12\int_m^n f^{\prime \prime}(t)P_2(t)dt$ (10)

Again, we can integrate by parts the last integral in (10)

$\begin{aligned} \int_m^n f^{\prime\prime}(t)P_2(t)dt&=\frac13f^{\prime \prime}(t)P_3(t)\Big|_m^n-\frac13\int_m^n f^{(3)}(t)P_3(t)dt\\ &=\frac{P_3(0)\left(f^{\prime\prime}(n)-f^{\prime \prime}(m) \right)}{2}-\frac13\int_m^n f^{(3)}(t)P_3(t)dt \qquad (P_3(n)=P_3(m)=P_3(0))\\ &=-\frac13\int_m^n f^{(3)}(t)P_3(t)dt \qquad \left(B_3(0)=0 \right) \end{aligned}$

Plugging in (10) we get

$\sum_{k=m}^{n}f(k)=\frac12\left(f(n)+f(m) \right)+\frac{\left(f^{\prime}(n)-f^{\prime}(m) \right)}{12}+\int_m^nf(t)dt+\frac16\int_m^n f^{(3)}(t)P_3(t)dt$ (11)

If we keep intergrating by parts the last integral and take into account that Bernoulli numbers of odd indices vanish we obtain the general Euler Maclaurin formula

$\boxed{\sum_{k=m}^{n} f(k)=\frac{f(n)+f(m)}{2}+\sum_{j=1}^{\left\lfloor\frac{p}{2}\right\rfloor} \frac{B_{2 j}}{(2 j) !}\left(f^{(2 j-1)}(n)-f^{(2 j-1)}(m)\right)+\int_{m}^{n} f(t) \,d t+\frac{(-1)^{p+1}}{p!}\int_{m}^{n} f^{(p)}(t)P_p(t)\,dt}$
(12)

Now lets choose $f(x)=\frac{1}{x^s}$ , then we have

$\begin{aligned} &f^\prime(x)=-\frac{s}{x^{s+1}}\\ &f^{\prime \prime}(x)=\frac{s(s+1)}{x^{s+2}}\\ &f^{(3)}(x)=-\frac{s(s+1)(s+2)}{x^{s+3}}\\ \end{aligned}$

and

$\begin{aligned} &f(\infty)=0\\ &f^\prime(\infty)=0\\ &f^{\prime \prime}(\infty)=0\\ &f^{(3)}(\infty)=0\\ & \\ &f(1)=1\\ &f^\prime(1)=-s\\ &f^{\prime \prime}(1)=s(s+1)\\ &f^{(3)}(1)=-s(s+1)(s+2)\\ \end{aligned}$

Finally, for $\Re(s)>1$

$\int_1^\infty\frac{dx}{x^s}=\frac{1}{s-1}$

We can now apply the Euler Maclaurin formula we just derived to obtain an analytic continuation for the Riemann Zeta function. Therefore:

$\zeta(s)=\frac{1}{s-1}+\frac12-s\int_1^\infty\frac{P_1(x)}{x^{s+1}}\,dx$ (13)

$\zeta(s)=\frac{1}{s-1}+\frac12+\frac{s}{12}-\frac{s(s+1)}{2}\int_1^\infty\frac{P_2(x)}{x^{s+2}}\,dx$ (14)

and

$\zeta(s)=\frac{1}{s-1}+\frac12+\frac{s}{12}+\frac{s(s+1)(s+2)}{2}\int_1^\infty\frac{P_3(x)}{x^{s+3}}\,dx$ (15)

Of course we can keep this process further and further, but for our goal in this post, (15) in enough. (15) extends the domain of the Riemann Zeta function to $-2<\Re(s)$ with a simple pole @ . Now it makes sense to ask what is $\zeta(0)=\,?$ and $\zeta(-1)=\,?$ . To answer this question, let $s \to 0$ in (15) to immediately get

$\zeta(0)=-\frac12$ (16)

Letting $s \to -1$ in (15) we obtain

$\zeta(-1)=-\frac{1}{12}$ (17)

Differentiating (13) w.r. to s

$\zeta^\prime(s)=-\frac{1}{(s-1)^2}- \int_1^{\infty}\frac{P_1(x)}{x^{s+1}}\,dx+s\int_1^{\infty}\frac{P_1(x) \ln x}{x^{s+1}}\,dx$ (18)

Letting $s \to 0$ in (18)

$\begin{aligned}\zeta^\prime(0)&=-1-\int_1^{\infty}\frac{P_1(x)}{x}\,dx\\ &=-1-\int_1^{\infty}\frac{x-\lfloor x \rfloor -\frac12}{x}\,dx\\ &=-1-\int_{1}^{\infty}\frac{\left\{ x\right\}-\frac{1}{2}}{x}dx \end{aligned}$ (19)

Let´s now compute the integral

$\begin{aligned}I&=\int_{1}^{\infty}\frac{\left\{ x\right\}-\frac{1}{2}}{x}\,dx\\ &=\sum_{k=1}^{\infty}\int_{k}^{k+1}\frac{x-k-\frac{1}{2}}{x}\,dx\\ &=\sum_{k=1}^{\infty}\int_{k}^{k+1}\frac{x-k-\frac{1}{2}}{x}\,dx\\ &=\sum_{k=1}^{\infty}\int_{0}^{1}\frac{t-\frac{1}{2}}{t+k}dt \qquad (x-k \to t)\\ &=\sum_{k=1}^{\infty}\int_{0}^{1}\frac{t-\frac{1}{2}}{t+k}dt\\ &=\sum_{k=1}^{\infty}\left(\int_{0}^{1}\frac{t}{t+k}dt-\frac{1}{2}\int_{0}^{1}\frac{1}{t+k}dt\right)\\ &=\sum_{k=1}^{\infty}\left(\int_{0}^{1}\frac{t+k}{t+k}dt-\left(k+\frac{1}{2}\right)\int_{0}^{1}\frac{1}{t+k}dt\right)\\ &=\sum_{k=1}^{\infty}\left(1-\left(k+\frac{1}{2}\right)\ln\left( t+k\right)\Big|_{0}^{1}\right)\\ &=\sum_{k=1}^{\infty}\left(1-\left(k+\frac{1}{2}\right)\ln\left( \frac{k+1}{k}\right)\right)\\ &=\sum_{k=1}^{\infty}\left(\ln e-\ln\left(\left( \frac{k+1}{k}\right)^{\left(k+\frac{1}{2}\right)\right)}\right)\\ &=\sum_{k=1}^{\infty}\left(\ln\left( e \cdot\left(\frac{k}{k+1} \right)^k \cdot\sqrt{\frac{k}{k+1}}\right)\right)\\ &=\ln\prod_{k=1}^{\infty}\left( e \cdot\left(\frac{k}{k+1} \right)^k \cdot\sqrt{\frac{k}{k+1}}\right) \end{aligned}$

Now lets now concentrate in the product inside the logarithm

$\begin{aligned} P&=\prod_{k=1}^{\infty} e \cdot\left(\frac{k}{k+1} \right)^k \cdot\sqrt{\frac{k}{k+1}}\\ &=\large\lim_{m\to\infty}\prod_{k=1}^m e\left(\frac{k}{k+1}\right)^{k}\sqrt{\frac{k}{k+1}}\\ &=\large\lim_{m\to\infty}e^m\left[\left(\frac{1}{2}\right)^{1}\cdot\left(\frac{2}{3}\right)^{2}\cdot\left(\frac{3}{4}\right)^{3}\cdots\left(\frac{m}{m+1}\right)^{m}\right]\sqrt{\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdots\frac{m}{m+1}}\\ &=\large\lim_{m\to\infty}e^m\left[\frac{1\cdot2\cdot3\cdots m}{(m+1)^m} \right]\sqrt{\frac{1}{m+1}}\\ &=\large\lim_{m\to\infty}e^m\left[\frac{ m!}{(m+1)^m} \cdot \frac{m+1}{m+1}\right]\sqrt{\frac{1}{m+1}}\\ &=\large\lim_{m\to\infty}e^m\left[\frac{ (m+1)!}{(m+1)^{m}\cdot(m+1)^{1/2}\cdot(m+1)} \right]\\ &=\large\lim_{m\to\infty}\frac{e^m (m+1)!}{(m+1)^{m+3/2}}\\ &=\large\lim_{m\to\infty}\frac{e^m }{(m+1)^{m+3/2}} \cdot\left[\sqrt{2\pi(m+1)}\left(\frac{m+1}e\right)^{m+1}\right]\\ &=\frac{\sqrt{2 \pi}}{e} \qquad \blacksquare \end{aligned}$

In the penultimate line we applied Stirling´s approximation to (m+1)!

Plugging back in our Integral we conclude that

$\int_{1}^{\infty}\frac{\left\{ x\right\}-\frac{1}{2}}{x}\,dx=\frac{\ln 2 \pi}{2}+1$ (20)

Plugging (20) in (19) we conclude that

$\begin{aligned}\zeta^\prime(0) &=-1-\left(\frac{\ln 2 \pi}{2}+1 \right)\\ &=-\frac12 \ln 2 \pi \qquad \blacksquare \end{aligned}$

For our last computation, consider equation (11)

$\begin{aligned} &\sum_{k=m}^{n}f(k)=\frac12\left(f(n)+f(m) \right)+\frac{\left(f^{\prime}(n)-f^{\prime}(m) \right)}{12}+\int_m^nf(t)dt+\frac16\int_m^n f^{(3)}(t)P_3(t)dt\\ &\sum_{k=m}^{n-1}f(k)+f(n)=\frac12\left(f(n)+f(m) \right)+\frac{\left(f^{\prime}(n)-f^{\prime}(m) \right)}{12}+\int_m^nf(t)dt+\frac16\int_m^n f^{(3)}(t)P_3(t)dt\\ &\sum_{k=m}^{n-1}f(k)=-\frac{\left(f(n)-f(m) \right)}{2}+\frac{\left(f^{\prime}(n)-f^{\prime}(m) \right)}{12}+\int_m^nf(t)dt+\frac16\int_m^n f^{(3)}(t)P_3(t)dt\\ \end{aligned}$

Again, if We choose $f(x)=\frac{1}{x^s}$ and now let $n \to \infty$ in the above equation, and taking into account that

$\begin{aligned} &f(m)=\frac{1}{m^s}\\ &f^\prime(m)=-\frac{s}{m^{s+1}}\\ &f^{(3)}(x)=-\frac{s(s+1)(s+2)}{x^{s+3}}\\ \end{aligned}$

and

$\int_m^\infty \frac{dx}{x^s}=\frac{m^{1-s}}{s-1}$

We obtain

$\begin{aligned} &\sum_{k=m}^{\infty}\frac{1}{k^s}=\frac{m^{-s}}{2}+\frac{sm^{-s-1}}{12}+\frac{m^{1-s}}{s-1}-\frac{s(s+1)(s+2)}{6}\int_m^\infty \frac{P_3(x)}{x^{s+3}}\,dx\\ &\zeta(s)-\sum_{k=1}^{m-1}\frac{1}{k^s}=\frac{m^{-s}}{2}+\frac{sm^{-s-1}}{12}+\frac{m^{1-s}}{s-1}-\frac{s(s+1)(s+2)}{6}\int_m^\infty \frac{P_3(x)}{x^{s+3}}\,dx\\ &\zeta(s)-\sum_{k=1}^{m}\frac{1}{k^s}+m^{-s}=\frac{m^{-s}}{2}+\frac{sm^{-s-1}}{12}+\frac{m^{1-s}}{s-1}-\frac{s(s+1)(s+2)}{6}\int_m^\infty \frac{P_3(x)}{x^{s+3}}\,dx\\ &\zeta(s)-\sum_{k=1}^{m}\frac{1}{k^s}=-\frac{m^{-s}}{2}+\frac{sm^{-s-1}}{12}+\frac{m^{1-s}}{s-1}-\frac{s(s+1)(s+2)}{6}\int_m^\infty \frac{P_3(x)}{x^{s+3}}\,dx\\ &\zeta(s)=\sum_{k=1}^{m}\frac{1}{k^s}-\frac{m^{-s}}{2}+\frac{sm^{-s-1}}{12}+\frac{m^{1-s}}{s-1}-\frac{s(s+1)(s+2)}{6}\int_m^\infty \frac{P_3(x)}{x^{s+3}}\,dx\\ \end{aligned}$

If we now let $\lim_{m \rightarrow \infty}$ in the above equation, the integral vanishes and we obtain

$\zeta(s)=\lim_{m \rightarrow \infty}\left(\sum_{k=1}^{m}\frac{1}{k^s}-\frac{m^{-s}}{2}+\frac{sm^{-s-1}}{12}-\frac{m^{1-s}}{1-s}\right)$ (21)

Differentiating both sides of (21) w.r. to

$\zeta^\prime(s)=\lim_{m \rightarrow \infty}\left(-\sum_{k=1}^{m}\frac{\ln k}{k^s}+\frac{m^{-s}\ln m}{2}+\frac{m^{-s-1}}{12}-\frac{s m^{-s-1} \ln m}{12}-\frac{ -\ln( m) m^{1-s}(1-s)+m^{1-s}}{(1-s)^2}\right)$ (22)

Letting $s \to -1$ in (22)

$\begin{aligned} \zeta^\prime(-1)&=\lim_{m \rightarrow \infty}\left(-\sum_{k=1}^{m}k\ln k+\frac{m\ln m}{2}+\frac{1}{12}+\frac{ \ln m}{12}-\frac{ -2\ln( m) m^{2}+m^{2}}{4}\right)\\ &=\lim_{m \rightarrow \infty}\left(-\sum_{k=1}^{m}k\ln k+\frac{m\ln m}{2}+\frac{ \ln m}{12}+\frac{ \ln( m) m^{2}}{2}-\frac{m^2}{4}\right)+\frac{1}{12}\\ &=\lim_{m \rightarrow \infty}\left(-\sum_{k=1}^{m}k\ln k+\left(\frac{m}{2}+\frac{ 1}{12}+\frac{ m^{2}}{2}\right)\ln m-\frac{m^2}{4}\right)+\frac{1}{12}\\ &=-\lim_{m \rightarrow \infty}\left(\sum_{k=1}^{m}k\ln k-\left(\frac{m}{2}+\frac{ 1}{12}+\frac{ m^{2}}{2}\right)\ln m+\frac{m^2}{4}\right)+\frac{1}{12}\\ \end{aligned}$ (23)

Now, recall the definition of the Glaisher Kinkelin constant

$A=\lim _{m \rightarrow \infty} \frac{\prod_{k+1}^{n}k^k}{m^{\frac{m^{2}}{2}+\frac{m}{2}+\frac{1}{12}} e^{-\frac{m^{2}}{4}}}$ (24)

Taking logarithms in both sides of (24) we get

$\ln A=\lim _{m \rightarrow \infty}\left( \sum_{k=1}^{m}k\ln k-\left(\frac{m}{2}+\frac{ 1}{12}-\frac{ m^{2}}{2}\right)\ln m+\frac{m^2}{4}\right)$ (25)

Comparing (25) with the last line of (23) we conclude that

$\zeta^\prime(-1)=-\ln A+\frac{1}{12} \qquad \blacksquare$

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