Let :
$$I_k=\int_{0}^{1}\left(\prod_{n=1}^{k}\left(1+\arctan\left(\left(\frac{y}{4n^{2}}\right)\right)\right)\right)dy$$
And :
$$h\left(x\right)=\int_{0}^{1}\left(\prod_{n=1}^{\operatorname{floor}\left(x\right)}\left(1+\arctan\left(\left(\frac{y}{n^{2}}\right)\right)\right)\right)dy$$
$$m\left(x\right)=x^{2}\left(h\left(x\right)-h\left(x-1\right)\right)$$
Now if we have :
$$\lim_{x\to \infty}m(x)=A,\lim_{k \to \infty}I_k=B$$
Which number is bigger meaning : $A\leq B$ or $B\leq A$? A numerical approach is accepted.
As attempt we have using Fatou's lemma :
$$\lim_{k \to \infty}I_k\geq \int_{0}^{1}\lim_{k \to \infty}\left(\prod_{n=1}^{k}\left(1+\arctan\left(\left(\frac{y}{4n^{2}}\right)\right)\right)\right)dy$$
Or as the product is decreasing :
$$ \int_{0}^{1}\lim_{k \to \infty}\left(\prod_{n=1}^{k}\left(1+\arctan\left(\left(\frac{y}{4n^{2}}\right)\right)\right)\right)dy\geq \int_{0}^{1}\lim_{k \to \infty}\left(\left(1+\arctan\left(\left(\frac{y}{4k^{2}}\right)\right)\right)\right)^{k^2}dy$$
Which leads us to as lower bound :
$$\int_{0}^{1}e^{\frac{y}{4}}dy\simeq 1.1360...$$
Where we have used for $x\in[0,1]$and $k\geq 6$ :
$$\left(1+\arctan\left(\frac{x}{k^{2}}\right)\right)^{k^{2}}-\prod_{n=1}^{k}\left(1+\arctan\left(\frac{x}{n^{2}}\right)\right)\leq 0$$
For the other limit $m(x)$ see my answer which use Am-gm here What is $\lim\limits_{x\to \infty}[g(x)-g(x-1)]\overset?=$