/sci/ - Science & Math

[Anonymous No.5861783]

hey /sci/, I need to know how to demonstrate step by step this. When I do it I get as solution: sin(x)·log(x)... Can you help me? Thx

[Anonymous No.5861787]

lrn2 complex analysis

[Anonymous No.5861795]

>>5861783
Download COMPLEX VARIABLE AND APPLICATIONS by brown and churchill, 8th edition, page 279 read the whole chapter if you want, but there it is explained step by step. You have to know complex variable.

[Anonymous No.5861797]

>>5861787
This. It is easy to prove with the residue theorem.

[Anonymous No.5861804]

>>5861795
http://campus.fi.uba.ar/file.php/171/textos/Complex_variable_-_Churchill.pdf

Have a link, from my university so I don't know ifit will work for you. Page 279.

[Anonymous No.5861807]

thank you!

[Anonymous No.5861828]

I try this, and obviously is not the same... Where is the error?

[Anonymous No.5861845]

>>5861828
you integrate from 1, it should be 0.

>those aostanci

[Anonymous No.5861848]

>>5861828
Why the fuck are you writing sin(x) as seux ?

[Anonymous No.5861852]

By parts integration.

[prep !!bRma1Ao62vX No.5861856]

let <span class="math">J_n = \int_0^{\pi/2} \frac{sin((2n+1)x}{sin(x)} dx[/spoiler]

and <span class="math">K_n = \int_0^{\pi/2} \frac{sin((2n+1)x}{x} dx [/spoiler]

show that <span class="math">J_n - K_n \rightarrow 0 [/spoiler], that <span class="math">\forall n, J_n = \frac{\pi}{2}[/spoiler], and that <span class="math">K_n \rightarrow \int_0^{\infty} \frac{sin(x)}[x}dx[/spoiler].

[Anonymous No.5861854]

>>5861848
in spanish it's called "seno". the u is probably a bad n.

[Anonymous No.5861858]

>>5861848
It's Spanish: sen(x)

[prep !!bRma1Ao62vX No.5861860]

let <span class="math">J_n = \int_0^{\pi/2} \frac{sin((2n+1)x)}{sin(x)} dx[/spoiler]

and <span class="math">K_n = \int_0^{\pi/2} \frac{sin((2n+1)x)}{x} dx [/spoiler]

show that <span class="math">J_n - K_n \rightarrow 0 [/spoiler], that <span class="math">\forall n, J_n = \frac{\pi}{2}[/spoiler], and that <span class="math">K_n \rightarrow \int_0^{+\infty} \frac{sin(x)}{x}dx[/spoiler].

[Anonymous No.5861862]

>>5861828
You made a mistake.

[Anonymous No.5861866]

>>5861828
When you integrate by parts the second time, if dv = cos x, then v = sin x.

[Anonymous No.5861886]

>>5861860
Why Kn0+xsin(x)dx ??¿?¿?

[Anonymous No.5861895]

>>5861860
Why K_n ---> That ??

[prep !!bRma1Ao62vX No.5861907]

>>5861895
>>5861886
come on man!

substitute <span class="math">u=(2n+1)x, dx=du/(2n+1)[/spoiler].
<span class="math"> K_n= \int{0}^{(2n+1)\pi/2} \frac{sin(u)}{u/(2n+1)} du/(2n+1) [/spoiler] so <span class="math">K_n = \int_0^{(2n+1)\pi/2} \frac{sin(u)}{u} du [/spoiler]
so when n goes to infinity...

[Anonymous No.5861956]

>>5861907

Dude, don't lagh but why do you multiply (2n+1)*(PI/2)????

[Anonymous No.5861967]

>>5861956
hahahahahahahahahahahhahahahaha...
go back to kindergarten.