Recent content by Meden Agan

  1. M

    Prove that the integral is equal to ##\pi^2/8##

    I don't see how that helps either. Let me know if you get anything useful, I'm stuck.
  2. M

    Prove that the integral is equal to ##\pi^2/8##

    @fresh_42 Did you get something new? If the answer is no, let's go ahead to analyze the solution on MSE.
  3. M

    Prove that ##\lim\limits_{x \to \infty} f(x) = 0##

    Mhm, could you outline what's unclear step by step? Hope I can provide a detailed clarification.
  4. M

    Prove that ##\lim\limits_{x \to \infty} f(x) = 0##

    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...
  5. M

    Prove that the integral is equal to ##\pi^2/8##

    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.
  6. M

    Prove that the integral is equal to ##\pi^2/8##

    Sad. I can't believe that is the only possible solution to the integral.
  7. M

    Prove that the integral is equal to ##\pi^2/8##

    Any significant developments? I've come up with a heuristic argument, but I'm becoming more and more convinced that is not correct either.
  8. M

    Prove that the integral is equal to ##\pi^2/8##

    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...
  9. M

    Prove that the integral is equal to ##\pi^2/8##

    Mhm, but unfortunately that derivative is frankly unmanageable. We seem to have complicated the integral again.
  10. M

    Prove that the integral is equal to ##\pi^2/8##

    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}...
  11. M

    Prove that the integral is equal to ##\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...
  12. M

    Prove that the integral is equal to ##\pi^2/8##

    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...
  13. M

    Prove that the integral is equal to ##\pi^2/8##

    @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.
  14. M

    Prove that the integral is equal to ##\pi^2/8##

    I totally agree. That is actually what I was going to say to you.
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