/sci/ - Science & Math » Thread #10926195

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integrals

Anonymous Wed 28 Aug 21:39:51 2019 No.10926195 View Reply Original Report

Quoted By: >>10926244 >>10926662 >>10926802 >>10926838 >>10926854 >>10926859 >>10926869 >>10926940

What was your integral of the day anons?

Anonymous

Anonymous Wed 28 Aug 2019 21:53:12 No.10926231 Report

Quoted By: >>10926662

>solving integrals after the age of 12

Anonymous

Anonymous Wed 28 Aug 2019 21:59:28 No.10926244 Report

Quoted By: >>10926614 >>10926785

>>10926195
My integral of the day was 1/(x+1)^2 over the real axis. I couldn't solve it, even if it was supposed to be pretty easy, so I decided to come here to lurk

Anonymous

Anonymous Thu 29 Aug 2019 00:57:33 No.10926614 Report

Quoted By: >>10926785

>>10926244
that's just u substitution

Anonymous

Anonymous Thu 29 Aug 2019 01:03:52 No.10926632 Report

Quoted By: >>10926887

>solving integrals after age 5
When I turned 6 my parents gave me a book on real analysis and said if I didn't pass the test within a month they'd kill me. And that's how I became a maths PhD at age 8

Anonymous

Anonymous Thu 29 Aug 2019 01:17:07 No.10926662 Report

Quoted By: >>10926676

>>10926231
kek
>>10926195
mine was e^x and i took like 3 hours and finally figured it out

Anonymous

Anonymous Thu 29 Aug 2019 01:23:14 No.10926676 Report

Quoted By:

>>10926662
you could probably use a transform and the residue theorem to solve that one

Anonymous

Anonymous Thu 29 Aug 2019 02:09:52 No.10926785 Report

Quoted By:

>>10926244
>>10926614
haha wtf at least we can say we have a wide IQ distribution in this board lads

Anonymous

Anonymous Thu 29 Aug 2019 02:13:49 No.10926802 Report

Quoted By:

>>10926195
$\int\limits_{(2-\epsilon)^{s}-1}^{\infty} {\pi ((t+1)^{1/s}) - \pi (t^{1/s}) \over t}dt = \log (\zeta(s))$

Anonymous

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Anonymous Thu 29 Aug 2019 02:27:55 No.10926838 Report

Quoted By:

>>10926195
Can you explain too me this thing. I know it's Cauchy integral formula, but I doesn't look right. Shouldn't it be $\oint \frac{f(z)}{z-z_0} = 2\pi i f(z_0)$ ?

Anonymous

Anonymous Thu 29 Aug 2019 02:34:35 No.10926854 Report

Quoted By: >>10926859

>>10926195
$\int_0^1 \ln\Bigg(\frac{1}{x+1}\Bigg)dx$ uses substitution and IBP

Anonymous

Anonymous Thu 29 Aug 2019 02:36:36 No.10926859 Report

Quoted By:

>>10926854
>>10926195
Sorry here
$\int_0^1\ln\Bigg(\frac{1}{x+1}\Bigg)dx$

Anonymous

Anonymous Thu 29 Aug 2019 02:41:38 No.10926869 Report

Quoted By:

>>10926195
You write the number 1 like a retard.

Anonymous

Anonymous Thu 29 Aug 2019 02:48:47 No.10926887 Report

Quoted By:

>>10926632
That's unironically how most math prodigies are created

Anonymous

Anonymous Thu 29 Aug 2019 03:08:01 No.10926940 Report

Quoted By:

>>10926195
Couldn't you just do partial fraction decomposition?
1/(1+x^4) = [1/(1+ix^2) + 1/(1-ix^2)]/2
= [1/(1+isqrt(i)x) + 1/(1-isqrt(i)x) + 1/(1+sqrt(i)x) + 1/(1-sqrt(i)x)]/4.
Integrate to get
[log(1+isqrt(i)x)/(isqrt(i)) - log(1-isqrt(i)x)/(isqrt(i)) + log(1+sqrt(i)x)/(sqrt(i)) - log(1-sqrt(i)x)/(sqrt(i))]/4
=[log((1+isqrt(i)x)/(1-isqrt(i)x))/(isqrt(i)) + log((1+sqrt(i)x)/(1-sqrt(i)x))/sqrt(i)]/4
Plugging in the boundary gives [pi/sqrt(i) + pi*sqrt(i)]/4 = 2pi*Re(sqrt(i))/4 = pi*sqrt(2)/4

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