Meden Agan's latest activity
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MMeden Agan replied to the thread 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...
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MMeden Agan reacted to fresh_42's post in the thread Prove that the integral is equal to ##\pi^2/8## with
Like.
If we look at the graphic, then we see an almost linear behavior with slope around ##-1## (or so) when we approach ##\alpha=0## and a... -
MMeden Agan replied to the thread Prove that the integral is equal to ##\pi^2/8##.Sad. I can't believe that is the only possible solution to the integral.
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MMeden Agan reacted to fresh_42's post in the thread Prove that the integral is equal to ##\pi^2/8## with Like.
I want to do the Weierstraß substitution now just to see where I end up. However, it doesn't look as if this would help. E.g. $$...
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MMeden Agan replied to the thread Prove that the integral is equal to ##\pi^2/8##.Any significant developments? I've come up with a heuristic argument, but I'm becoming more and more convinced that is not correct either.
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MMeden Agan reacted to fresh_42's post in the thread Prove that the integral is equal to ##\pi^2/8## with Like.
The problem is that we have ##\alpha## and ##p(\operatorname{trig}(\alpha))## in one equation - a dimension conflict. The trick that...
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MMeden Agan replied to the thread Prove that the integral is equal to ##\pi^2/8##.IMO, the expression $$2 \int\limits_{0}^{1} \frac{\alpha(z)}{\sqrt{1-z^2}} \, \mathrm dz = \frac{\pi^2}{8}$$ is the best way of...
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MMeden Agan reacted to fresh_42's post in the thread Prove that the integral is equal to ##\pi^2/8## with Like.
Sure. Ultimately, there is no way to cheat. The problem has to be addressed somewhere. As long as we operate with ##f(\alpha)## and...
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MMeden Agan replied to the thread Prove that the integral is equal to ##\pi^2/8##.Mhm, but unfortunately that derivative is frankly unmanageable. We seem to have complicated the integral again.
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MMeden Agan reacted to fresh_42's post in the thread Prove that the integral is equal to ##\pi^2/8## with Like.
I made another integration by parts since the numerator ##\alpha\cdot z'## (cp. post #77) was so inviting. That resulted in (##v## is...
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MMeden Agan replied to the thread Prove that the integral is equal to ##\pi^2/8##.All of that is puzzling me. We have to prove that $$2 \int\limits_{0}^{1} \frac{\alpha(z)}{\sqrt{1-z^2}} \, \mathrm dz =...
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MMeden Agan reacted to fresh_42's post in the thread Prove that the integral is equal to ##\pi^2/8## with Like.
You can see from the many versions that I listed for my integral that I'm still searching for a good key. At least the integral limits...
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MMeden Agan replied to the thread 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...
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MMeden Agan replied to the thread Prove that the integral is equal to ##\pi^2/8##.Mhm... Could you show that?
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MMeden Agan reacted to fresh_42's post in the thread Prove that the integral is equal to ##\pi^2/8## with Like.
Yes, that's basically my formula for ##f(\alpha)f(-\alpha).## It reveals the symmetry of the curve. If we ignore the inverse sine, then...
