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##.All I can say here is ##\log \left(\sqrt{1+q(y)^2}+q(y)\right) = \sinh^{-1} \left(q(y)\right)##. I'm not even sure it is helpful. After...
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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
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The minus sign comes from the re-substitution of ##\alpha## to ##y.## No idea. Meanwhile, I confirmed your formula, which looks much... -
MMeden Agan reacted to fresh_42's post in the thread Prove that the integral is equal to ##\pi^2/8## with Like.
Confirmed! It is really too hot here to differentiate ... It means we are now at \begin{align*} I&=2\int_{\sqrt{9-2\sqrt{8}}}^3...
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MMeden Agan replied to the thread Prove that the integral is equal to ##\pi^2/8##.Yes, that's what I get. I fed WolframAlpha with the integral here, and returned the value ##\dfrac{\pi^2}{8} \approx 1.2337...## (even...
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MMeden Agan replied to the thread Prove that the integral is equal to ##\pi^2/8##.Yes, double-check this step:
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MMeden Agan replied to the thread Prove that the integral is equal to ##\pi^2/8##.I fed this site with the integral. I put $$\sqrt{\frac{y^2(y^2+2y-7)}{32(y+1)(9-y^2)-t^2(y^2+2y-7)(y^2+7)(9-y^2)}}$$ into the bar and...
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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.
Unfortunately, WA spits out an ##\operatorname{arctan}## or, likewise, a complex logarithm, so this will take me a while to...
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MMeden Agan replied to the thread Prove that the integral is equal to ##\pi^2/8##.Seems all fine to me. The inner integral can be evaluated easily, can't it? It comes out to be a logarithmic form, if I'm right. As per...
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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'd rather assume that this tiny error is due to the singularity. If you want to check what I've done, here is it: The polynomial...
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MMeden Agan replied to the thread Prove that the integral is equal to ##\pi^2/8##.Still not working, see here. The numerical value should be ##\dfrac{\pi^2}{16} \approx 0.6168{\color{red}{50}}...##, different from...
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MMeden Agan replied to the thread Prove that the integral is equal to ##\pi^2/8##.Mhm, unfortunately it doesn't work. I fed WolframAlpha with that integral here, and returned a completely different value than expected...
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MMeden Agan replied to the thread Prove that the integral is equal to ##\pi^2/8##.Totally agree. I have also tried to approach the integral with complex analysis, but all useless and monstrously meaningless algebra...
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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 think this: The whole part is typical of trig integration problems. We have ##(x-a)(x+a)## or ##x^2-2ax+c## and ##c## is not...
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MMeden Agan replied to the thread Prove that the integral is equal to ##\pi^2/8##.This whole thing is frankly unmanageable. Perhaps changing the upper bound to the inner integral (by giving it as a function 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.
I think that there are nice possibilities to get rid of the arcus sine. I used \begin{align*} I&=2\int_{0}^a \operatorname{arcsin}...