An interesting integral $\cos(x)\cos(x^2)\cos(x^3)...$ I became interested in functions involving powers of cosines after seeing this post. In particular, some fiddling on Desmos leaves me with this question.

What is the value of $$\int_{-1}^1\cos(x)\cos(x^2)\cos(x^3)...\,dx\quad?$$

Of course, we can concisely write it as $\int_{-1}^1\prod_{i=1}^\infty\cos(x^i)\,dx$, and can easily prove that it is above the $x$-axis there. 
We can see that it is quite like the unit semicircle; thus we expect the integral to be bounded above by $\pi/2$.

Any improvements on this bound are welcome.
 A: The upper bound is not horribly difficult to prove. By the Weierstrass product for the cosine function we have 
$$ -\log\cos x = \sum_{n\geq 1}\frac{(4^n-1)\zeta(2n)}{n \pi^{2n}}x^{2n} \tag{1}$$
hence
$$ \sum_{m\geq 1} -\log\cos x^m = \sum_{n\geq 1}x^{2n}\sum_{d\mid n}\frac{(4^d-1)\zeta(2d)}{d\pi^{2d}}=\sum_{n\geq 1} c_n x^{2n}\tag{2}$$
while
$$ -\log\sqrt{1-x^2} = \sum_{n\geq 1}\frac{1}{2n} x^{2n}\tag{3} $$
hence it is enough to show $c_n\geq \frac{1}{2n}$, which is trivial since
$$ c_n\geq \frac{(4^1-1)\zeta(2\cdot 1)}{1\cdot \pi^{2\cdot 1}}=\frac{1}{2}.\tag{4}$$
Indeed, this inequality proves
$$\forall x\in(-1,1),\qquad \prod_{m\geq 1}\cos(x^m)\approx\exp\left(\frac{x^2}{2(x^2-1)}\right) $$
such that the given integral turns out to be pretty close to
$$ 2\int_{0}^{1}\exp\left(\frac{x^2}{2(x^2-1)}\right)\,dx = \int_{0}^{+\infty}\frac{e^{-z/2}\,dz}{(z+1)\sqrt{z^2+z}}=\sqrt{\pi}\,U\left(\tfrac{1}{2},0,\tfrac{1}{2}\right)\approx \sqrt{2}$$
where $U$ is Tricomi's confluent hypergeometric function.
