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
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Unfortunately, WA spits out an ##\operatorname{arctan}## or, likewise, a complex logarithm, so this will take me a while to... -
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}...
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MMeden Agan replied to the thread Prove that the integral is equal to ##\pi^2/8##.How about differentiation under integral sign or integral representation of ##\arcsin##? What do you think?
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MMeden Agan replied to the thread Prove that the integral is equal to ##\pi^2/8##.Mhm... A delightful surprise! Most likely my instructor took that integral from there. However, I would be interested to hear what we...
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MMeden Agan reacted to renormalize's post in the thread Prove that the integral is equal to ##\pi^2/8## with Like.
It turns out that this very integral was posted 5 days ago on Mathematics Stack Exchange by "Dan" (is that you @Meden Agan ?)...
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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.
That's why I haven't tried it so far. ##\sqrt{f(\alpha)f(-\alpha)}## looked like a perfect circle, but this wasn't the case...
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MMeden Agan replied to the thread Prove that the integral is equal to ##\pi^2/8##.Mhm... Do you think that identity can make anything easier? IMO, probably complicates the integrand even more.
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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 gave WA the integrand, and here is what you want to calculate: the area below ... I also fed WA with the integral...