/sci/ - Science & Math » Thread #12695418

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I hear you guys like math

Anonymous Thu 11 Feb 23:26:19 2021 No.12695418 View Reply Original Report

Quoted By: >>12695440 >>12696067 >>12696073 >>12696930

But can you prove this? Hint: They're both equal to $\pi / sin \pi p$ .

Anonymous

Anonymous Thu 11 Feb 2021 23:38:05 No.12695440 Report

Quoted By:

>>12695418
pissing pip?

Anonymous

Anonymous Fri 12 Feb 2021 03:20:02 No.12696027 Report

Quoted By:

Hypergeometric series

Anonymous

Anonymous Fri 12 Feb 2021 03:25:53 No.12696037 Report

Quoted By: >>12696039

im gay lol

Anonymous

Anonymous Fri 12 Feb 2021 03:26:24 No.12696039 Report

Quoted By:

>>12696037
lol

Anonymous

Anonymous Fri 12 Feb 2021 03:46:29 No.12696067 Report

Quoted By:

>>12695418
smells like I'd stary by using t=e^x, dt=dx

Anonymous

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Anonymous Fri 12 Feb 2021 03:48:27 No.12696073 Report

Quoted By: >>12696080

>>12695418 (OP)
smells like I'd start by using t=e^x, dt=dx

The lower bound would be 1 instead of 0, so some refactoring is required to get from t^p to t^-p.
Maybe the anti-symmetric part cancels out and that's the trick

Anonymous

Anonymous Fri 12 Feb 2021 03:51:32 No.12696080 Report

Quoted By: >>12696086

>>12696073
>smells like I'd start by using t=e^x, dt=dx
Anon... If t = e^x, then dt = e^x dx

Anonymous

Anonymous Fri 12 Feb 2021 03:53:20 No.12696086 Report

Quoted By:

>>12696080
right, I had just noticed
So
dt = t dx
dx = t^-1 dt

But again some symmetric-anti-symmetric games may be involved

I think to remember that x/(1+x) has some nice properties in that regard

Anonymous

Anonymous Fri 12 Feb 2021 10:43:51 No.12696930 Report

Quoted By:

>>12695418
set x = e^{-u}

Anonymous

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Anonymous Fri 12 Feb 2021 13:25:07 No.12697220 Report

Quoted By: >>12697221

Hm, well, substituting by y=e^x will get you the right limits and almost the right integrand: e^((1-p)z)/(1+e^z) dz. It doesn't seem true that the parameter p in both integrals can be identical, but that they are related by the transform p -> 1-p.
I leave the proof of convergence to someone else, that shit is boring. Explicit expression of the first integral seems to be possible after PFD or the second through trigonometric identities, but I can't be arsed.

Anonymous

Anonymous Fri 12 Feb 2021 13:26:55 No.12697221 Report

Quoted By:

>>12697220
oop, p->-1-p*

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