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 made another integration by parts since the numerator ##\alpha\cdot z'## (cp. post #77) was so inviting. That resulted in (##v## is... -
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 $$2 \int\limits_{0}^{1} \frac{\alpha(z)}{\sqrt{1-z^2}} \, \mathrm dz =...
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MMeden Agan reacted to fresh_42's post in the thread Prove that the integral is equal to ##\pi^2/8## with Like.
You can see from the many versions that I listed for my integral that I'm still searching for a good key. At least the integral limits...
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MMeden Agan replied to the thread Prove that the integral is equal to ##\pi^2/8##.Fantastic work. The integral reduces to ##2 \displaystyle \int\limits_{0}^{1} \frac{\alpha(z)}{\sqrt{1-z^2}} \, \mathrm dz##. We have to...
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MMeden Agan replied to the thread Prove that the integral is equal to ##\pi^2/8##.Mhm... Could you show that?
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MMeden Agan reacted to fresh_42's post in the thread Prove that the integral is equal to ##\pi^2/8## with Like.
Yes, that's basically my formula for ##f(\alpha)f(-\alpha).## It reveals the symmetry of the curve. If we ignore the inverse sine, then...
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MMeden Agan reacted to robphy's post in the thread Prove that the integral is equal to ##\pi^2/8## with Like.
Possibly helpful: $$\frac{\left(\left(\cos x\right)^{2}-\left(\sin x\right)^{2}\sqrt{9-16\left(\sin x\right)^{2}}\right)}{1-\left(\tan...
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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 thought about Feynman, but haven't looked deeper into it. My integral is ##\displaystyle{\int_0^a \dfrac{\alpha...
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MMeden Agan replied to the thread 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: $$2\int\limits_{0}^{\arcsin...
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MMeden Agan reacted to fresh_42's post in the thread Prove that the integral is equal to ##\pi^2/8## with Like.
See my post #69. And I already posted it a couple of posts before. The expression under the root is ##1+8\cos(2x)## which, together...
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MMeden Agan reacted to fresh_42's post in the thread Prove that the integral is equal to ##\pi^2/8## with Like.
Not really. I am at $$ I=-2\int_{\alpha=0}^{\alpha=a}\dfrac{\alpha\cdot...
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MMeden Agan replied to the thread 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...
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MMeden Agan replied to the thread 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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MMeden Agan reacted to fresh_42's post in the thread Prove that the integral is equal to ##\pi^2/8## with Like.
I'm afraid I have to agree. Especially, as all the negative roots and singularities haven't even been addressed. I don't think I'll...
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MMeden Agan replied to the thread Prove that the integral is equal to ##\pi^2/8##.Mhm. This looks awful. What shall we do?
