Question

I dont get this...
all comes ahead...
its all in the context of evaluating improper integrals in complex analysis..

Answer 1

\[\int\limits_{-\infty}^{\infty}sint dt=\lim_{r \rightarrow \infty}\int\limits_{-r}^{r}sint dt =\lim_{r \rightarrow \infty}[-cost]=\lim_{r \rightarrow \infty}0=0\]
note; the value of [−cost] is between r and -r.

so my question is how have they got that the limit of -cost is 0?

Answer 2

we dont even know the values of r and -r...so how did htey find that -cost has limit 0?

Answer 3

I'm taking this at face-value, for now, but what if we forget about the limit for the moment, and focus on the integral
\[\huge \int\limits_{-r}^{r}\sin \ t \ dt\]instead?

Answer 4

use the fact that cos(-x) = cos(x) ... this integral shouldn't converge (i think). what you are calculating is Cauchy Principal Value.

Answer 5

\[\large = \left[-\cos \ t \right]_{-r}^{r}=(-\cos \ r)-(-\cos(-r))=0\]

Answer 6

Bit too big, terenz :P

But yes this integral should not converge... Area under the sine curve will never converge... but this may have something to do with the "...in complex analysis" :/

Answer 7

Oh, this was complex analysis? That's beyond my knowledge. Take it away, guys :D

Answer 8

lol i have no idea, I don't remember studying integrals of complex numbers/functions in university calculus.

Answer 9

ahaaa
moment of epiphany..thanks terenzreignz

Answer 10

All in a day's work? :D

Answer 11

lol

Answer 12

But that still doesn't change the fact that the area under a sine/cosine curve is unbounded...

Answer 13

Yeah... that's why I was careful and said I'd take it at face value... whatever that means, LOL

Answer 14

yea this doesn't converge. this is called
http://en.wikipedia.org/wiki/Cauchy_principal_value

Answer 15

I thought that's because the total area of under the sine curve on the right is equal to the area on the left and hence the can cancel out\[\int\limits^r_{-r}\]

Answer 16

^yeah, I'm guessing that's what it's getting at. Idk anything about the cauchy principal value thought.

Answer 17

no they don't ... both are indeterminate.

Answer 18

Honestly...wish complex numbers did not exist...they do really make things more.....complex.. :D

Answer 19

I do think complex number makes life easier. :P

Answer 20

^^ agree with agostino

Answer 21

i guess I do agree to an extent but all the studying.... blurgh...

Answer 22

yeah yeah ... currently I am also stuck on complex analysis.

Answer 23

I feel for you..

Answer 24

what book are you following?

Answer 25

I'm studying from a university published book-the open univeristy

Answer 26

i don't know what it is.

Answer 27

I'm studying thorugh distant learning/home-schooling..
In the UK we have a univeristy called the Open University which caters for distant learning students...so i am doing my maths degree with them..and they have provided all the materials of study

Answer 28

Oh .. i know that OU.

Answer 29

:)
So what book do you use?

Answer 30

http://books.google.com/books/about/Complex_Variables.html?id=9zKl4lXEXlsC&redir_esc=y
currently looking this

Answer 31

Anyone uses Springer's books?

Answer 32

@experimentX just had a look through an online version of the book... imho it is better explained and set out than the book im studying from...
thanks for introducing..will definitely be using it.

@agostino never used it.