Can you see why?
Beautiful solution by @ValFadeev.
As you can see, the trick is showing that the integral *does not* depend on N.pic.twitter.com/vOJFq5BCNQ
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Just to elucidate how counterintuitive this result is, I've plotted the integrand for different values of N.
All of these curves must admit the EXACT same area!pic.twitter.com/03EsDf4QD9Show this thread
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I don't get how 1/z for z=0 is taken as -> ∞ . It could be - ∞ with equal validity.
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The integral bounds are "really" 0+ to infinity, so when we make the substitution 1/x we take lim_{x--> 0+} 1/x
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That’s one of them!
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I just wanted to pop by and say that I’ve learned a ton from your twitter feed.
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thanks friend! means a lot
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