/sci/ - Science & Math » Thread #13024944

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Prove you're not a pseud.

Anonymous Sat 24 Apr 15:03:54 2021 No.13024944 View Reply Original Report

Quoted By: >>13024968 >>13025053 >>13025120 >>13027285 >>13029003 >>13029546 >>13029663

Surely you can prove this, right?

Anonymous

Anonymous Sat 24 Apr 2021 15:05:34 No.13024950 Report

Quoted By:

Onions basedboy basedjak desu

Anonymous

Anonymous Sat 24 Apr 2021 15:10:43 No.13024968 Report

Quoted By: >>13024983

>>13024944
Degenerate fucking idiot thinks this question is hard and dares to ask my genius if I can solve this
kys

Anonymous

Anonymous Sat 24 Apr 2021 15:15:12 No.13024983 Report

Quoted By:

>>13024968
solve it mister talk a lot

Anonymous

Anonymous Sat 24 Apr 2021 15:29:29 No.13025027 Report

Quoted By:

I can't solve it
t. PhD in math

Anonymous

Anonymous Sat 24 Apr 2021 15:32:55 No.13025040 Report

Quoted By:

I can't solve this.

Anonymous

Anonymous Sat 24 Apr 2021 15:37:09 No.13025053 Report

Quoted By: >>13025080

>>13024944
>writes two expressions that evaluate to the same separated by an equality sign
> asks to "solve"
What?

Anonymous

Anonymous Sat 24 Apr 2021 15:42:42 No.13025080 Report

Quoted By:

>>13025053
You have to prove the quality.

Anonymous

Anonymous Sat 24 Apr 2021 15:53:23 No.13025120 Report

Quoted By: >>13025139 >>13025147 >>13025225

>>13024944
Let
$ x = e^{-u} \\ dx = - e^{-u} du $

$ \int_0^\infty \frac{1}{(1+x)x^p} dx = - \int_{\infty}^{-\infty} \frac{e^{pu}}{1 + e^{-u}} e^{-u}du = \int_{-\infty}^\infty \frac{e^{pu}}{e^u + 1} du = \int_{-\infty}^\infty \frac{e^{px}}{1 + e^x} dx $

Anonymous

Anonymous Sat 24 Apr 2021 15:56:25 No.13025139 Report

Quoted By:

>>13025120
you beat me to it. fuck you.

Anonymous

Anonymous Sat 24 Apr 2021 15:58:05 No.13025147 Report

Quoted By: >>13025157 >>13025263

>>13025120
So why does it require 0 < p < 1?

Anonymous

Anonymous Sat 24 Apr 2021 16:00:39 No.13025157 Report

Quoted By:

>>13025147
Because the integral wouldn't converge otherwise.

Anonymous

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Anonymous Sat 24 Apr 2021 16:13:38 No.13025225 Report

Quoted By:

>>13025120
You forgot this term. Nice try though.

Anonymous

Anonymous Sat 24 Apr 2021 16:22:40 No.13025263 Report

Quoted By: >>13025280 >>13025672 >>13027291

>>13025147
It doesn't. Even if the integral doesn't converge, both sides will equal infinity.

Anonymous

Anonymous Sat 24 Apr 2021 16:29:45 No.13025280 Report

Quoted By: >>13025287 >>13027291

>>13025263
Crank city, wrong side of tracks.

Anonymous

Anonymous Sat 24 Apr 2021 16:31:43 No.13025287 Report

Quoted By:

>>13025280
? Go take your meds now, I'm not in the mood to entertain schizos.

Anonymous

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Anonymous Sat 24 Apr 2021 18:18:00 No.13025672 Report

Quoted By: >>13027291 >>13028483

>>13025263
> $\infty=\infty$

Anonymous

Anonymous Sun 25 Apr 2021 01:28:30 No.13027285 Report

Quoted By: >>13029554

>>13024944
just tell me, can this be solved with keyhole and half circle contours? if so, i wont even bother

Anonymous

Anonymous Sun 25 Apr 2021 01:29:33 No.13027291 Report

Quoted By:

>>13025263
>>13025280
>>13025672
both sides would be infinity, but more importantly, same sort of infinity, so it still makes sense

Anonymous

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Anonymous Sun 25 Apr 2021 07:25:56 No.13028411 Report

Quoted By: >>13028483 >>13030104

Anonymous

Anonymous Sun 25 Apr 2021 07:50:16 No.13028483 Report

Quoted By: >>13029546

>>13028411
Use a better software sweetie
https://www.wolframalpha.com/input/?i=integrate+%28e%5E%280.3x%29%2F%281+%2B+e%5Ex%29%29%2C+-inf+to+inf

>>13025672
The audacity to make this post, holy shit. This place is filled with brainlets and high schoolers.
https://en.wikipedia.org/wiki/Extended_real_number_line#Arithmetic_operations

Anonymous

Anonymous Sun 25 Apr 2021 12:33:12 No.13029003 Report

Quoted By:

>>13024944
huh so this is how I can get /sci/ to do my homework for me...

Anonymous

Anonymous Sun 25 Apr 2021 16:07:16 No.13029546 Report

Quoted By: >>13029556

>>13024944
pi/sin(pi*p)
>>13028483
Use pen and paper sweetie
https://www.wolframalpha.com/input/?i=pi%2Fsin%28pi*0.3%29

Anonymous

Anonymous Sun 25 Apr 2021 16:10:06 No.13029554 Report

Quoted By:

>>13027285
make substitution x = z/[1-z]
Beta function

Anonymous

Anonymous Sun 25 Apr 2021 16:10:56 No.13029556 Report

Quoted By:

>>13029546
No one asked you to compute the integral, sweetie.

Anonymous

Anonymous Sun 25 Apr 2021 16:46:11 No.13029663 Report

Quoted By:

>>13024944
Here we go. Noting that $\int_{-a}^a f(x) dx = \int_0^a f(x)+f(-x) dx$ , we can rewrite the full integral using the geometric series to
$ \int_0^\infty \frac{e^{(p-1)x}+e^{-px}}{1+e^{-x}}dx = \sum_{n\ge 0} (-1)^n\int_0^\infty e^{-(n+1-p)x}+e^{-(n+p)x} dx = \sum_{n\ge 0} \frac{(-1)^n}{n+1-p}+\frac{(-1)^n}{n+p}. $
Splitting this series into even and odd parts and using the cotangent pole expansion, we get
$ \sum_{n\in\mathbb{Z}} \frac{(-1)^n}{n+p} = \sum_{n\in\mathbb{Z}} \frac{1}{n+p}-\sum_{n\in\mathbb{Z}} \frac{2}{2n+p} = \pi\cot \pi p-\pi\cot \frac {\pi p}2 = \frac{\pi}{\sin \pi p}. $

Anonymous

Anonymous Sun 25 Apr 2021 18:40:08 No.13030104 Report

Quoted By:

>>13028411
it clearly converges, i dont care about your shitty app

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