/sci/ - Science & Math » Thread #10891508

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Anonymous Fri 16 Aug 01:49:50 2019 No.10891508 View Reply Original Report

Quoted By: >>10891539 >>10891561 >>10891587 >>10891649 >>10891663 >>10891711 >>10893322

Show your work

Anonymous

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Anonymous Fri 16 Aug 2019 01:56:07 No.10891539 Report

Quoted By:

>>10891508
Take your homework somewhere else.

Anonymous

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Anonymous Fri 16 Aug 2019 01:58:50 No.10891551 Report

Quoted By:

-1
My work: https://www.wolframalpha.com/input/?i=e%5E(integrate(i%2F(sqrt(x)%2Bxsqrt(x)))+0+to+inf)

Anonymous

Anonymous Fri 16 Aug 2019 02:01:14 No.10891561 Report

Quoted By: >>10891587 >>10892667

>>10891508
What's your differential?

Anonymous

Anonymous Fri 16 Aug 2019 02:06:44 No.10891587 Report

Quoted By:

>>10891561
Fuck dx I always forget the littlest things >>10891508

Anonymous

Anonymous Fri 16 Aug 2019 02:23:03 No.10891649 Report

Quoted By:

>>10891508
-1

Really really really easy.

$\int_{0}^{\infty} \frac{1}{\sqrt{x} + x\sqrt{x}}dx = \int_{0}^{\infty} \frac{2u}{u+u^3} du = 2 \int_{0}^{\infty} \frac{1}{1+u^2} du = \limit\limits_{u \to \infty} 2 \tan^{-1}(u) = \pi$

exp(i*pi) = -1.

Anonymous

Anonymous Fri 16 Aug 2019 02:26:29 No.10891663 Report

Quoted By: >>10891692 >>10891729 >>10892811

>>10891508
-1

Really really really easy.

$\int_{0}^{\infty} \frac{1}{\sqrt{x} + x\sqrt{x}}dx = \int_{0}^{\infty} \frac{2u}{u+u^3} du = 2 \int_{0}^{\infty} \frac{1}{1+u^2} du = 2\lim\limits_{u \to \infty} \tan^{-1}(u) = \pi$

exp(i*pi) = -1.

Anonymous

Anonymous Fri 16 Aug 2019 02:34:53 No.10891692 Report

Quoted By: >>10891720 >>10891721 >>10891724

>>10891663
can you explain the arctan part to me. i got the first 3 steps.

Anonymous

Anonymous Fri 16 Aug 2019 02:42:15 No.10891711 Report

Quoted By: >>10891721

>>10891508
The easy way:
just remember that $\int \frac{1}{1+x^2} dx = \tan^{-1}(x) + \text{constant}$ .

Or use the substitution $x = sin(\theta)$ to show that the integral does indeed equal arctan. (trig substation).

Anonymous

Anonymous Fri 16 Aug 2019 02:46:34 No.10891720 Report

Quoted By: >>10891754

>>10891692
Take calculus then come back

Anonymous

Anonymous Fri 16 Aug 2019 02:46:42 No.10891721 Report

Quoted By: >>10891754

>>10891711
>>10891692
I'm sorry. use the substation x=tan(theta).

Anonymous

Anonymous Fri 16 Aug 2019 02:48:06 No.10891724 Report

Quoted By: >>10891754

>>10891692
anon I...

Anonymous

Anonymous Fri 16 Aug 2019 02:50:15 No.10891729 Report

Quoted By:

>>10891663
Correct

Anonymous

Anonymous Fri 16 Aug 2019 02:57:47 No.10891754 Report

Quoted By: >>10892708

>>10891720
>>10891724
>>10891721
I just needed a refresher from not having done calc in so long.

Anonymous

Anonymous Fri 16 Aug 2019 12:44:47 No.10892667 Report

Quoted By: >>10893386

>>10891561
Found the Riemannfag

Anonymous

Anonymous Fri 16 Aug 2019 13:19:24 No.10892707 Report

Quoted By:

missed a dx

Anonymous

Anonymous Fri 16 Aug 2019 13:20:09 No.10892708 Report

Quoted By:

>>10891754
That you do

Anonymous

Anonymous Fri 16 Aug 2019 14:35:37 No.10892811 Report

Quoted By: >>10892815

>>10891663
Why is it exp(i*pi)and not just i*pi?

Anonymous

Anonymous Fri 16 Aug 2019 14:39:07 No.10892815 Report

Quoted By: >>10892833

>>10892811
Because
exp(i*theta) = cos(theta) + i*sin(theta).

If you expand exp(x), cos(x) and sin(x) in the form of a series, you'll see that this is true.

Anonymous

Anonymous Fri 16 Aug 2019 14:57:02 No.10892833 Report

Quoted By: >>10893278

>>10892815
>exp(i*theta) = cos(theta) + i*sin(theta).
I remember that, but I don't really see where we are pulling e from after you get i*pi.

Anonymous

Anonymous Fri 16 Aug 2019 19:22:42 No.10893278 Report

Quoted By:

>>10892833
Jesus christ I'm fuckign stupid.

Anonymous

Anonymous Fri 16 Aug 2019 19:39:09 No.10893322 Report

Quoted By: >>10893350

>>10891508
-1

Anonymous

Anonymous Fri 16 Aug 2019 19:44:44 No.10893350 Report

Quoted By:

>>10893322
Congrats, you get the prestigious honor of being the fourth person to answer the OP!

Anonymous

Anonymous Fri 16 Aug 2019 19:57:17 No.10893386 Report

Quoted By:

>>10892667
shut up, Lebesgue

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