During some research, I obtained the following convergent product $$ P_a(x) := \prod_{j = 1}^{\infty} \, \cos\left(\frac{x}{j^{a}}\right) \quad (x \in \mathbb{R}, a > 1).$$
Considering how I got it, I know it's convergent and continuous at $0$ (for any fixed $a$), but if I look at $P_a$ now, it doesn't seem so obvious for me.
Try : I showed that $P_a \in L^1(\mathbb{R})$, i.e. it is absolutely integrable on $\mathbb{R}$. Indeed, by using the linearization of the cosine function and the inequalities $\ln(1-y)\leqslant -y$ (for any $y<1$) and $1-\cos\,z \geqslant z^2/2$ (for any real $z$), we obtain \begin{eqnarray*} P_a(x)^2 &=& \prod_{j = 1}^{\infty} \left(1-\frac{1-\cos(x\,j^{-a})}{2}\right) \leqslant \prod_{j > |x|^{1/a}} \left(1-\frac{1-\cos(x\,j^{-a})}{2}\right) \leqslant \exp\left(-C |x|^{1/a} \right), \end{eqnarray*} for some absolute constant $C>0$. However, I have not been able to use this upper bound to prove continuity at $0$ (via uniform convergence for example, if we can).
Question : I wanted to know how to study $P_a$ (e.g. its convergence and continuity at $0$), if you think it has an other form "without product" (or other nice properties) and finally if anyone has already seen this type of product (in some references/articles), please. Thank you in advance.
May be, what you could do is $$P_a(x) = \prod_{j = 1}^{\infty} \, \cos\left(\frac{x}{j^{a}}\right)\implies \log\big[P_a(x)\big]= \sum_{j = 1}^{\infty}\log\Big[\cos\left(\frac{x}{j^{a}}\right) \Big] $$
Now, using Euler polynomials $$\log\big[\cos(t)\big]=\sum_{n = 1}^{\infty}(-1)^n\frac{2^{2 n-3} (E_{2 n-1}(1)-E_{2 n-1}(0)) }{n (2 n-1)!}t^{2 n}$$ $$\log\big[\cos(t)\big]=-\sum_{n = 1}^{\infty}\frac{2^{2 n-3} |E_{2 n-1}(1)-E_{2 n-1}(0)| }{n (2 n-1)!}t^{2 n}$$ $$\log\Big[\cos\left(\frac{x}{j^{a}}\right) \Big]=-\sum_{n = 1}^{\infty}\frac{2^{2 n-3} |E_{2 n-1}(1)-E_{2 n-1}(0)| }{n (2 n-1)!} x^{2n} j^{-2an} $$ Since all coefficients have the same sign, we can switch the order of summations $$\sum_{j=1}^\infty j^{-2an}=\zeta(2an)$$ $$\log\big[P_a(x)\big]=-\frac 18\sum_{n = 1}^{\infty}\frac{|E_{2 n-1}(1)-E_{2 n-1}(0)| \,\zeta(2an)}{n (2 n-1)!} (2x)^{2n} $$ is an infinite sum of negative numbers. Now, it remains to prove the convergence (it is and it should be quite fast).