Introduction to L-Functions (Part II)

In the last post, we dived into the history and motivation and surrounding L-Functions in this post we continue with the exploration. We shall dedicate this post to make an in-depth study of the Riemann Zeta Function.
In the last post we ended up with the proof that ζ(s)\zeta(s) converges for any s∈Rs \in \mathbb{R} and s>1s > 1. This is an extremely profound statement. This opens up the possibilities of asking questions about what values these sums evaluate to and how to determine that. For answering such questions we need some prerequisites.

Bernoulli Numbers

Bernoulli was well aware of the formulas for the partial sums of the powers of the integers, the first few being:
∑k=1nk=n(n+1)2\begin{aligned} \sum_{k=1}^{n} k = \frac{n(n+1)}{2} \end{aligned}

∑k=1nk2=n(n+1)(2n+1)6\begin{aligned} \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} \end{aligned}

∑k=1nk3=n2(n+1)24\begin{aligned} \sum_{k=1}^{n} k^3 = \frac{n^2(n+1)^2}{4} \end{aligned}

Given the parallel nature between increasing exponents in the summands and the exponents in the explicit formulas, it is natural to search for some generalisation. The result gave rise to an interesting sequence of numbers, called the Bernoulli Numbers; named by Euler in honour of his teacher. We have the general formula as we denote the kthk^{th} Bernoulli Numbers as BkB_k.

∑k=1n−1km=1m+1[∑j=0m(m+1j)nm+1−jBj](2.1)\sum_{k=1}^{n-1} k^m = \frac{1}{m+1}\bigg[ \sum_{j=0}^{m}\binom{m+1}{j} n^{m+1-j}B_j \bigg] \tag{2.1}

From here, we can produce the recursive relationship between the Bernoulli numbers by plugging n=1n = 1 into equation (2.1)(2.1), resulting in

0=∑j=0m(m+1j)Bj0 = \sum_{j=0}^m \binom{m+1}{j}B_j

Using the recursive formula and B0=1B_0 = 1 we have a fairly straight forward, although not very speedy, method for computing the Bernoulli numbers. For reference, the first couple Bernoulli numbers are:

B0=1,B1=−12,B2=16,B3=0,B4=−130,B5=0B_{0}=1, \quad B_{1}=-\frac{1}{2}, \quad B_{2}=\frac{1}{6}, \quad B_{3}=0, \quad B_{4}=-\frac{1}{30}, \quad B_{5}=0

We see that the third and fifth Bernoulli number is 00, and its also can be showed that B2k+1=0B_{2k+1} = 0 for all k≥1k \ge 1 .

Fourier Series

The theory of Fourier Series is extremely important in mathematics as a fact I feel its crucial in our understanding nature as it pops up everywhere when we try to model something.
The whole idea behind Fourier series is to write a periodic function f(x)f(x) as an infinite sum of sines and cosines:
f(x)=a02+∑n=1∞(ancos⁡(nx)+bnsin⁡(nx))f(x)=\frac{a_{0}}{2}+\sum_{n=1}^{\infty}\left(a_{n} \cos (n x)+b_{n} \sin (n x)\right) where the coefficients of the sines and cosines are given by Euler’s formulas:
an=1π∫−ππf(x)cos⁡(nx)dxbn=1π∫−ππf(x)sin⁡(nx)dx(2.2)\begin{aligned} a_{n} &=\frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos (n x) d x \tag{2.2}\\ b_{n} &=\frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin (n x) d x \end{aligned}Now, the function f(x)f(x) can be represented as a Fourier series provided the following properties hold:

For our analysis of Riemann Zeta function we require a special product formula for the sinc\textbf{sinc} function. To start off our derivation we will find the Fourier series of cos⁡(μx)\cos(\mu x) on the interval (−π,π)(−\pi, \pi) where μ\mu is a real number that is not an integer. Note that all three properties are satisfied, so our quest for a Fourier series is feasible. Since cos⁡(μx)\cos(\mu x) is an even function we immediately know that our Fourier series will be a cosine series, thus bn=0b_n = 0.

Now from (2.2)(2.2) we have
an=1π∫−ππcos⁡(μx)cos⁡(nx)dx=1π∫0πcos⁡((μ+n)x)+cos⁡((μ−n)x)dx=1π[sin⁡((μ+n)π)μ+n+sin⁡((μ−n)π)μ−n]=1π[(μ−n)sin⁡((μ+n)π)+(μ+n)sin⁡((μ−n)π)μ2−n2]=1π[2μsin⁡(μx)cos⁡(nx)μ2−n2]=2μ(−1)nπ(μ2−n2)sin⁡(μπ)\begin{aligned} a_{n} &=\frac{1}{\pi} \int_{-\pi}^{\pi} \cos (\mu x) \cos (n x) d x \\ &=\frac{1}{\pi} \int_{0}^{\pi} \cos ((\mu+n) x)+\cos ((\mu-n) x) d x \\ &=\frac{1}{\pi}\left[\frac{\sin ((\mu+n) \pi)}{\mu+n}+\frac{\sin ((\mu-n) \pi)}{\mu-n}\right] \\ &=\frac{1}{\pi}\left[\frac{(\mu-n) \sin ((\mu+n) \pi)+(\mu+n) \sin ((\mu-n) \pi)}{\mu^{2}-n^{2}}\right] \\ &=\frac{1}{\pi}\left[\frac{2 \mu \sin (\mu x) \cos (n x)}{\mu^{2}-n^{2}}\right]\\ &=\frac{2 \mu(-1)^{n}}{\pi\left(\mu^{2}-n^{2}\right)} \sin (\mu \pi) \end{aligned}We have used basic trigonometric identities and Now that we have found our coefficients, we have the Fourier series representation:
cos⁡(μx)=sin⁡(μπ)μπ+2μsin⁡(μπ)π∑n=1∞(−1)ncos⁡(nx)μ2−n2\cos (\mu x)=\frac{\sin (\mu \pi)}{\mu \pi}+\frac{2 \mu \sin (\mu \pi)}{\pi} \sum_{n=1}^{\infty}(-1)^{n} \frac{\cos (n x)}{\mu^{2}-n^{2}}
Since our function satisfied the property that the value at every point (specifically the endpoints −π−\pi and π\pi) was the average of the limits from both sides, we can plug x=πx=\pi into our Fourier expression without any trouble. After doing this and dividing by sin⁡(μx)\sin(\mu x) (which we can do since μ\mu is not an integer) we get:
cot⁡(μπ)=1μπ+2μπ∑n=1∞1μ2−n2\cot (\mu \pi)=\frac{1}{\mu \pi}+\frac{2 \mu}{\pi} \sum_{n=1}^{\infty} \frac{1}{\mu^{2}-n^{2}}
Now rearranging the terms we have
cot⁡(μπ)−1μπ=2μπ∑n=1∞1μ2−n2\cot (\mu \pi)-\frac{1}{\mu \pi}=\frac{2 \mu}{\pi} \sum_{n=1}^{\infty} \frac{1}{\mu^{2}-n^{2}}
After multiplying by π\pi and integrating μ\mu from 00 to xx we have:
π∫0x(cot⁡(μπ)−1μπ)dμ=∫0x∑n=1∞−2μn2−μ2dμ(2.3)\pi \int_{0}^{x}\left(\cot (\mu \pi)-\frac{1}{\mu \pi}\right) d \mu=\int_{0}^{x} \sum_{n=1}^{\infty} \frac{-2 \mu}{n^{2}-\mu^{2}} d \mu \tag{2.3}
Rearranging the left side
π∫0x(cot⁡(μπ)−1μπ)dμ=log⁡sin⁡(πx)πx−lim⁡a→0sin⁡(πa)πa=log⁡sin⁡(πx)πx\begin{aligned} \pi \int_{0}^{x}\left(\cot (\mu \pi)-\frac{1}{\mu \pi}\right) d \mu &=\log \frac{\sin (\pi x)}{\pi x}-\lim _{a \rightarrow 0} \frac{\sin (\pi a)}{\pi a} \\ &=\log \frac{\sin (\pi x)}{\pi x} \end{aligned}
On the right hand side of (2.3)(2.3) we can swap the sum and integral because we showed that equation (2.3)(2.3) was uniformly convergent. Therefore we end up with
∫0x∑n=1∞−2μn2−μ2dμ=∑n=1∞∫0x−2μn2−μ2=∑n=1∞log⁡(1−x2n2)=lim⁡α→∞∑n=1αlog⁡(1−x2n2)=lim⁡α→∞log⁡∏n=1α(1−x2n2)=log⁡lim⁡α→∞∏n=1α(1−x2n2)=log⁡∏n=1∞(1−x2n2)\begin{aligned} \int_{0}^{x} \sum_{n=1}^{\infty} \frac{-2 \mu}{n^{2}-\mu^{2}} d \mu &=\sum_{n=1}^{\infty} \int_{0}^{x} \frac{-2 \mu}{n^{2}-\mu^{2}} \\ &=\sum_{n=1}^{\infty} \log \left(1-\frac{x^{2}}{n^{2}}\right) \\ &=\lim _{\alpha \rightarrow \infty} \sum_{n=1}^{\alpha} \log \left(1-\frac{x^{2}}{n^{2}}\right) \\ &=\lim _{\alpha \rightarrow \infty} \log \prod_{n=1}^{\alpha}\left(1-\frac{x^{2}}{n^{2}}\right) \\ &=\log \lim _{\alpha \rightarrow \infty} \prod_{n=1}^{\alpha}\left(1-\frac{x^{2}}{n^{2}}\right) \\ &=\log \prod_{n=1}^{\infty}\left(1-\frac{x^{2}}{n^{2}}\right) \end{aligned}
This gives us
log⁡sin⁡(πx)πx=log⁡∏n=1∞(1−x2n2)\log \frac{\sin (\pi x)}{\pi x}=\log \prod_{n=1}^{\infty}\left(1-\frac{x^{2}}{n^{2}}\right)
After exponentiation both sides and multiplying the πx\pi x over, we end up with the product formula for sine:
sin⁡(πx)πx=∏n=1∞(1−x2n2)(2.4) \frac{\sin (\pi x)}{\pi x}= \prod_{n=1}^{\infty}\left(1-\frac{x^{2}}{n^{2}}\right) \tag{2.4}

Evaluation of Zeta Function For Even Integers

Since we know that ζ(1)\zeta(1) diverges the natural question is to ask for the value of ζ(2)\zeta(2). And there is amazing history related with zeta function evaluated at 22 . In 1644 Pietro Mengoli proposed a problem to all mathematicians. Determine the value of the infinite sum
∑k=1∞1k2\sum_{k=1}^{\infty}\frac{1}{k^2}
This problem, which is known as the Basel problem, was eventually solved by Euler in 1735 after multiple renowned mathematicians had failed. Euler’s solution granted him instant fame in the mathematical community, but that was hardly a reason for Euler to stop there. Euler ingeniously derived and proved the general formula for which the Basel problem is a special case . The purpose of this section is to present this formula, which happens to be the value of Euler’s zeta series at the even integers.

We start with the product formula for sin⁡(x)\sin(x), which was produced in (2.4)(2.4)
sin⁡(x)=x∏k=1∞(1−x2π2k2){\sin (x)}= x\prod_{k=1}^{\infty}\left(1-\frac{x^{2}}{\pi^2 k^{2}}\right)
After taking the natural logarithm of both sides and substituting x=−iu/2x = −iu/2 we have
log⁡(sin⁡(−iu2))=log⁡(−iu2)+∑k=1∞log⁡(1−(−iu2)2k2π2)\log \left(\sin \left(-\frac{i u}{2}\right)\right)=\log \left(-\frac{i u}{2}\right)+\sum_{k=1}^{\infty} \log \left(1-\frac{\left(-\frac{i u}{2}\right)^{2}}{k^{2} \pi^{2}}\right)and after some simplification we get
log⁡(sin⁡(−iu2))=log⁡(−iu2)+∑k=1∞log⁡(1+u24k2π2)\log \left(\sin \left(-\frac{i u}{2}\right)\right)=\log \left(-\frac{i u}{2}\right)+\sum_{k=1}^{\infty} \log \left(1+\frac{u^{2}}{4 k^{2} \pi^{2}}\right)
We can then differentiate both sides with respect to uu.
(−i2)cos⁡(−iu2)sin⁡(−iu2)=1u+∑k=1∞2u4k2π2+u2(2.5)\left(-\frac{i}{2}\right) \frac{\cos \left(-\frac{i u}{2}\right)}{\sin \left(-\frac{i u}{2}\right)}=\frac{1}{u}+\sum_{k=1}^{\infty} \frac{2 u}{4 k^{2} \pi^{2}+u^{2}} \tag{2.5}Focusing on the left hand side and using the following identities
cos⁡(z)=12(eiz+e−iz) and sin⁡(z)=12i(eiz−e−iz)\cos (z)=\frac{1}{2}\left(e^{i z}+e^{-i z}\right) \quad \text { and } \quad \sin (z)=\frac{1}{2 i}\left(e^{i z}-e^{-i z}\right)gives

(−i2)cos⁡(−iu2)sin⁡(−iu2)=(−i2)(e−u/2+eu/22)(−e−u/2+ea/22i)=12(e−u/2+eu/2)(−e−u/2+eu/2)=12(e−u/2)(e−u/2)(eu−1+2)(−1+eu)=12+1eu−1\begin{aligned}\left(-\frac{i}{2}\right) \frac{\cos \left(-\frac{i u}{2}\right)}{\sin \left(-\frac{i u}{2}\right)} &=\left(-\frac{i}{2}\right) \frac{\left(\frac{e^{-u / 2}+e^{u / 2}}{2}\right)}{\left(\frac{-e^{-u / 2}+e^{a / 2}}{2 i}\right)} \\ &=\frac{1}{2} \frac{\left(e^{-u / 2}+e^{u / 2}\right)}{\left(-e^{-u / 2}+e^{u / 2}\right)} \\ &=\frac{1}{2} \frac{\left(e^{-u / 2}\right)}{\left(e^{-u / 2}\right)} \frac{\left(e^{u}-1+2\right)}{\left(-1+e^{u}\right)} \\ &=\frac{1}{2}+\frac{1}{e^{u}-1} \end{aligned}
Then, after substituting this back into equation (2.5)(2.5), multiplying by u and moving a term over we end up
ueu−1+u2−1=∑k=1∞2u24k2π2+u2(2.6)\frac{u}{e^{u}-1}+\frac{u}{2}-1=\sum_{k=1}^{\infty} \frac{2 u^{2}}{4 k^{2} \pi^{2}+u^{2}} \tag{2.6}
Now for the homestretch, focusing on the left hand side we
ueu−1+u2−1=∑k=0∞Bkukk!+u2−1=1−u2+∑k=2∞Bkukk!+u2−1=∑k=2∞Bkukk!=∑k=1∞B2ku2k(2k)!(2.7)\begin{aligned} \frac{u}{e^{u}-1}+\frac{u}{2}-1 &=\sum_{k=0}^{\infty} \frac{B_{k} u^{k}}{k !}+\frac{u}{2}-1 \\ &=1-\frac{u}{2}+\sum_{k=2}^{\infty} \frac{B_{k} u^{k}}{k !}+\frac{u}{2}-1 \\ &=\sum_{k=2}^{\infty} \frac{B_{k} u^{k}}{k !} \\ &=\sum_{k=1}^{\infty} \frac{B_{2 k} u^{2 k}}{(2 k) !} \tag{2.7}\end{aligned}
In the last step we were able to relabel the indices because all the odd terms were zero since Bk=0B_k = 0 for odd k>1k > 1 . The right hand side of equation (2.6)(2.6) becomes
∑k=1∞2u24k2π2+u2=∑k=1∞2u2(2kπ)2(11+(u2π)2)=∑k=1∞2u2(2kπ)2∑j=0∞(−1)j(u2πk)2j=∑k=1∞∑k=1∞2(−1)j−1(u2πk)2j+2=∑j=1∞∑k=1∞2(−1)j−1(2π)2j∑k=1∞(1k2j)u2j=∑j=1∞2(−1)j−1(2π)j−1ζ(2j)u2j=∑j=1∞2(−1)j−1(2π)2jζ(2j)uij(2.8)\begin{aligned} \sum_{k=1}^{\infty} \frac{2 u^{2}}{4 k^{2} \pi^{2}+u^{2}} &=\sum_{k=1}^{\infty} \frac{2 u^{2}}{(2 k \pi)^{2}}\left(\frac{1}{1+\left(\frac{u}{2 \pi}\right)^{2}}\right) \\ &=\sum_{k=1}^{\infty} \frac{2 u^{2}}{(2 k \pi)^{2}} \sum_{j=0}^{\infty}(-1)^{j}\left(\frac{u}{2 \pi k}\right)^{2 j} \\ &=\sum_{k=1}^{\infty} \sum_{k=1}^{\infty} 2(-1)^{j-1}\left(\frac{u}{2 \pi k}\right)^{2 j+2} \\ &=\sum_{j=1}^{\infty} \sum_{k=1}^{\infty} \frac{2(-1)^{j-1}}{(2 \pi)^{2 j}} \sum_{k=1}^{\infty}\left(\frac{1}{k^{2 j}}\right) u^{2 j} \\ &=\sum_{j=1}^{\infty} \frac{2(-1)^{j-1}}{(2 \pi)^{j-1}} \zeta(2 j) u^{2 j} \tag{2.8} \\ &=\sum_{j=1}^{\infty} \frac{2(-1)^{j-1}}{(2 \pi)^{2 j}} \zeta(2 j) u^{i j} \end{aligned}
Finally, changing the index from j to k and substituting equations (2.7)(2.7) and (2.8)(2.8) back into equation (2.6)(2.6) yields:
∑k=1∞B2k(2k)!u2k=∑k=1∞2(−1)k−1(2π)2kζ(2k)u2k\sum_{k=1}^{\infty} \frac{B_{2 k}}{(2 k) !} u^{2 k}=\sum_{k=1}^{\infty} \frac{2(-1)^{k-1}}{(2 \pi)^{2 k}} \zeta(2 k) u^{2 k}
Finally, we equate the corresponding coefficients in the sums and rearrange to arrive at our goal
(−1)k−1B2k(2π)2k2(2k)!=ζ(2k)\boxed{ {\frac{(-1)^{k-1} B_{2 k}(2 \pi)^{2 k}}{2(2 k) !}=\zeta(2 k)}}
This is a rather profound formula and it generators some of the most surprising and beautiful relations in all of mathematics. And hence for k=1k=1 we have ζ(2)=B2.(2π)22(2!)=π26\zeta(2) = \frac{B_2.(2 \pi)^2}{2(2!)} = \frac{\pi^2}{6}
In its full glory , to appreciate the shear beauty of the result we see it in its fullest form :
112+122+132+142+⋯=π26\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+ \cdots = \frac{\pi^2}{6}