Let ##f(x)= \min\limits_{m, n \in \mathbb Z} \left|x- \sqrt{m^2+2 \, n^2}\right|## be the minimum distance between a positive real ##x## and a number of the form ##\sqrt{m^2 + 2 n^2}## with ##m, n## integers.
Let us consider a radius ##R## and let us consider the set ##S_R## of integer points...
I agree. Your ideas are always excellent, but unfortunately there's nothing we can do.
The form in post #86 made me hopeful. But unfortunately nothing came of it.
IMO, the expression $$2 \int\limits_{0}^{1} \frac{\alpha(z)}{\sqrt{1-z^2}} \, \mathrm dz = \frac{\pi^2}{8}$$ is the best way of representing the integral.
Perhaps it is enough to choose ##\alpha(z)## so that the integrand is symmetric in the interval ##(-1, 1)##, like the original function?
The...
All of that is puzzling me.
We have to prove that $$2 \int\limits_{0}^{1} \frac{\alpha(z)}{\sqrt{1-z^2}} \, \mathrm dz = \frac{\pi^2}{8}.$$
Now, some well-known integrals of this type are:
$$2 \int\limits_{0}^{1}\frac{\frac{1}{2} \, \arcsin z}{\sqrt{1-z^{2}}} \, \mathrm dz = \frac{\pi^2}{8}...
Fantastic work. The integral reduces to ##2 \displaystyle \int\limits_{0}^{1} \frac{\alpha(z)}{\sqrt{1-z^2}} \, \mathrm dz##. We have to prove it is equal to ##\pi^2/8##.
Now, there is no chance of separating ##\alpha## and ##z## by inverting ##f(\alpha)##.
So, how can we merge the two in a new...
The most beautiful form into which we can convert the original integral, according to the picture, is:
$$2\int\limits_{0}^{\arcsin \left(1/\sqrt[4]{8}\right)}\arcsin\left(\sqrt{\frac{\cos^{2}\left(x\right)-\sin^{2}\left(x\right)\sqrt{9-16\ \sin^{2}x}}{1-\tan^{2}x}}\right) \, \mathrm dx.$$
Now...
@fresh_42 Anything new on this beast of an integral?
Yesterday I tried to come up with something, but it's only a remark and can't be enough to evaluate the integral.