Does the Integral of sin(x)/x from 0 to Infinity Converge?

[saint_n]

integral of sinx/x exists?

#1
saint_n
Registered User


Join Date: Apr 2004
Posts: 3 integrating sinx/x between (0,infinty)?

--------------------------------------------------------------------------------

hey ppl!

Can you help me by giving me a method or how you would go around to prove that this

\mid\int\frac{sinx}{x}dx\mid

exists.
Thanx
Saint_n

 

[arildno]

Well, here's a rough argument:
1. In a vicinity of zero (0<=x<=e<<1) , the integrand sin(x)/x=1-x^(2)/2+-+-<1 by making e small enough, and using the fact we have an alternating series.
2.Hence, up to any finite value of the upper integration limit, the integral exists.
3. Split your integral f.ex. as follows:
a) 0<=x<=2*(pi)
b) In successive intervals: 2*n*(pi)<=x<=(2n+1)*(pi), (2n+1)*(pi)<=x<=2*(n+1)*(pi), n>=1
The value obtained on an interval is less in absolute value than on the previous interval, and of opposite sign.

There's still some work to be done to make this rigorous, though..


saint n: You have spread this question over way too many threads!
I see from another you've made that it is the actual interval integrations you're having trouble with; not the general procedure:
So:
Note that T(n)=abs(int(sin(x)/x))<=int(abs(sin(x)/x))<=int(1/x)
You should be able to complete the steps now..

 

[saint_n]

thanx for the help!My light bulb just went on!Couldnt av done it without ya

Saint_n