Meden Agan's latest activity
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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 it can be simplified in a better way by sorting the factor polynomials. But I agree, it doesn't look good. We have a circle... -
MMeden Agan replied to the thread Prove that the integral is equal to ##\pi^2/8##.All correct, but it's almost impossible to integrate all that stuff. Take a look here. How about using the function ##\sinh^{-1}##...
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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 only needed ##z## as an intermediate variable. Expressing terms like ##p(y)/q'(y)## by ##z=q(y)## is almost impossible. So I...
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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.
1) It was your integration that led to the minus sign. That also happens when we substitute the arctangent with a logarithm. I was...
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MMeden Agan replied to the thread Prove that the integral is equal to ##\pi^2/8##.Shouldn't we express this as a function of ##z##? How do we differentiate an expression in ##y## with respect to the variable ##z##?
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MMeden Agan replied to the thread Prove that the integral is equal to ##\pi^2/8##.Mhm, there are some major problems. 1) Do we care that ##p(y)## is complex when ##y \in (-3,3)##? 2) When ##y=3##, ##z= i##. So we'd...
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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.
Here is my plan: \begin{align*} I&=2\int_{\sqrt{9-2\sqrt{8}}}^3 \left(p(y)\log\left|\sqrt{1+q(y)^2}+q(y)\right|\right) \,dy\\...
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MMeden Agan replied to the thread Prove that the integral is equal to ##\pi^2/8##.Hope you'll be able to outline how to go on. For now I have no idea how to do it.
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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 don't think that returning to ##\operatorname{arcsin}## would be an improvement. My formula is awkward if you ask me, even if it...
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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 Like.
The minus sign comes from the re-substitution of ##\alpha## to ##y.## No idea. Meanwhile, I confirmed your formula, which looks much...
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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.
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...