I know the closed form formula for calculating the values of Zeta function at even integers. I was able to derive it from the coefficients of the power series of $-\dfrac{\pi x}{2}\cot(\pi x)$. Now, I am looking at the way with which Euler was able to derive the values of $\zeta(2n)$.
I have seen this "trick" in Brilliant:
So I have:
$$\dfrac{\sin(x)}{x}=1-\dfrac{x^2}{3!}+\dfrac{x^4}{5!}-\dfrac{x^6}{7!}+\dfrac{x^8}{9!}…\tag{1}$$
$$\dfrac{\sin(x)}{x}=\left(1-\dfrac{x^2}{\pi^2}\right)\left(1-\dfrac{x^2}{4\pi^2}\right)\left(1-\dfrac{x^2}{9\pi^2}\right)\left(1-\dfrac{x^2}{16\pi^2}\right)…\tag{2}$$
$$\dfrac{\sin(ix)}{ix}=1-\dfrac{(ix)^2}{3!}+\dfrac{(ix)^4}{5!}-\dfrac{(ix)^6}{7!}…\tag{3}$$
$$\dfrac{\sin(ix)}{ix}=1+\dfrac{x^2}{3!}+\dfrac{x^4}{5!}+\dfrac{x^6}{7!}…\tag{4}$$
$$\dfrac{\sin(ix)}{ix}=\left(1-\dfrac{(ix)^2}{\pi^2}\right)\left(1-\dfrac{(ix)^2}{4\pi^2}\right)\left(1-\dfrac{(ix)^2}{9\pi^2}\right)\left(1-\dfrac{(ix)^2}{16\pi^2}\right)…\tag{5}$$
$$\dfrac{\sin(ix)}{ix}=\left(1+\dfrac{x^2}{\pi^2}\right)\left(1+\dfrac{x^2}{4\pi^2}\right)\left(1+\dfrac{x^2}{9\pi^2}\right)\left(1+\dfrac{x^2}{16\pi^2}\right)…\tag{6}$$
Multiply $(6)$ and $(2)$, I have:
$$\dfrac{\sin(x)\sin(ix)}{ix^2}=1+\left(\dfrac{x^2}{3!}-\dfrac{x^2}{3!}\right)+\left(\dfrac{x^4}{5!}+\dfrac{x^4}{5!}-\dfrac{x^4}{(3!)^2}\right)+…\tag{7}$$
$$\dfrac{\sin(x)\sin(ix)}{ix^2}=\left(1+\dfrac{x^2}{\pi^2}\right)\left(1-\dfrac{x^2}{\pi^2}\right)\left(1+\dfrac{x^2}{4\pi^2}\right)\left(1-\dfrac{x^2}{4\pi^2}\right)\left(1+\dfrac{x^2}{9\pi^2}\right)\left(1-\dfrac{x^2}{9\pi^2}\right)…\tag{8}$$
$$\dfrac{\sin(x)\sin(ix)}{ix^2}=\left(1-\dfrac{x^4}{\pi^4}\right)\left(1-\dfrac{x^4}{16\pi^4}\right)\left(1-\dfrac{x^4}{81\pi^4}\right)…\tag{9}$$
Comparing the coeffcients of $x^4$ in the infinite Maclaurin series and the infinite product, we will arrive at $\zeta(4)=\dfrac{\pi^4}{90}$.
Now that I wish to continue this process to see if I can get $\zeta(6)$, $\zeta(8)$, $\zeta(10)$, I don't know how to continue. The trick above works fine for $\zeta(4)$, but what about higher $\zeta(2n)$.
I look around and find this interesting post of user17762:
In his answer, he wrote the function on the LHS as $\dfrac{i\sin(z)\sin(\dfrac{z}{i})}{z^2}$. A little bit of manipulation shows that it is exactly like $\dfrac{\sin(x)\sin(ix)}{ix^2}$ or $\dfrac{\sin(x)\sinh(x)}{x^2}$
I have tried to multiply again $\dfrac{\sin(x)\sinh(x)\sin(x/i^2)}{x^2(\frac{x}{i^2})}=\dfrac{\sin^2(x)\sinh(x)}{x^3}$, using the hint found in the answer of the member above. Then I use Wolfram to check the series expansion of this function but the third term of $x^6$ is $\dfrac{5}{3024}$. I thought that I should a find a function where its coefficient of $x^6$ should be $\dfrac{1}{945}$. I am not sure if my reasoning is problematic.
Is there an algorithm to find the function on the LHS?
Best Answer
The generalisation you are looking for is given by the functions $$f_n \colon \mathbb{R} \to \mathbb{R}, \, f_n (x) = \prod \limits_{k=0}^{n-1} \operatorname{sinc}\left(\mathrm{e}^{\mathrm{i} \pi \frac{k}{n}} x\right) \, , $$ for $n \in \mathbb{N}$ ($\operatorname{sinc}(z) = \frac{\sin(z)}{z}$ with $\operatorname{sinc}(0) = 1$). Since the product representation of $\operatorname{sinc}$ converges absolutely, we can rearrange factors to obtain $$ f_n(x) = \prod \limits_{k=0}^{n-1} \prod \limits_{l=1}^\infty \left(1 - \mathrm{e}^{2 \pi \mathrm{i} \frac{k}{n}} \frac{x^2}{\pi^2 l^2} \right) = \prod \limits_{l=1}^\infty \prod \limits_{k=0}^{n-1} \left(1 - \mathrm{e}^{2 \pi \mathrm{i} \frac{k}{n}} \frac{x^2}{\pi^2 l^2} \right) = \prod \limits_{l=1}^\infty \left(1 - \left(\frac{x}{\pi l}\right)^{2n} \right)$$ for $x \in \mathbb{R}$ and $n \in \mathbb{N}$ (see this question for a simple proof of the final step). This representation confirms that these functions are real at real arguments. Clearly, the coefficient in front of $x^{2n}$ in the series expansion of $f_n$ is $[x^{2n}] f_n (x) = - \frac{\zeta(2n)}{\pi^{2n}}$ for $n \in \mathbb{N}$. Now simply multiply the power series representations, i.e. $$ f_n (x) = \prod \limits_{k=0}^{n-1} \sum \limits_{l_k = 0}^\infty \frac{(-1)^{l_k} \mathrm{e}^{2 \pi \frac{k l_k}{n} \mathrm{i}} x^{2l_k}}{(2l_k+1)!} \, ,$$ and compute the coefficient this way to obtain the zeta values by comparison. As you can easily check using CAS, this works for every $n \in \mathbb{N}$ ($[x^6]f_3(x) = - \frac{1}{945}$ or $[x^8]f_4(x) = - \frac{1}{9450}$, for example), but I have not yet managed to prove $[x^{2n}] f_n(x) = \frac{(-1)^{n} \mathrm{B}_{2n} 2^{2n-1}}{(2n)!}$ analytically.