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Integrating sinx/x between (0,infinty)?

  1. Apr 25, 2004 #1
    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

    [tex]\mid\int\frac{sinx}{x}dx\mid[/tex]

    exists.
    Thanx
    Saint_n
     
  2. jcsd
  3. Apr 25, 2004 #2

    pig

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    If a function is continuous on an interval, it is integrable (wrong word?) on that interval. sinx/x is continous on <0, +inf> so an integral exists.

    However, it is not elementary.
     
  4. Apr 25, 2004 #3
    There are many websites on google to break integrals down step by step. Recall that an integral is the reverse operation of a derivative. What is the reverse of sinx isn't it cosx?

    Try this website http://mss.math.vanderbilt.edu/~pscrooke/toolkit.html

    Good luck!
     
  5. Apr 25, 2004 #4
    interesting...
     
  6. Apr 25, 2004 #5
    Geez... there is not "standard" nice neat integral for sinx/x... And nint fails on the calculator... So.. :(

    Maybe you like to try using Eular's method and start approximating for 0 using steps of 0.1 and find f(infinity) :)
     
  7. Apr 25, 2004 #6
    ebola could u tell me whether its lebesgues integrable over that interval and why it is.. ?? U can intergrate its power seires and let it tend to infinity.. etc
     
  8. Apr 25, 2004 #7
    thanx for the help ppl!!!My light bulb just went on!!!Couldnt av done it without yas

    Saint_n
     
  9. Apr 25, 2004 #8

    uart

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    The integral exists and is usually denoted Si(x)

    [tex]Si(x) = \int_{u=0}^{u=x} \frac{\sin(u)}{u} \, du[/tex]

    For the integral from 0 to infinity that you specifically request the value is Pi/2, though I dont remember how to do it. That is [tex]Si(\infty) = \pi/2[/tex].

    If you merely interested in wheter or not the integral from zero to infinty exists (is bounded) rather than needing to know it's presice value then this is a much easier problem.

    For example you can pretty easily verify that the integral is bounded by:

    [tex]Si(\infty) \leq \pi\, \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} [/tex]
     
    Last edited: Apr 25, 2004
  10. Apr 25, 2004 #9
    writing T(n) = [tex]\mid\int\frac{sinx}{x}dx\mid[/tex]
    Can u write sinx as a power series
    then dividing by x
    and integrating it
    then you get T(n)
    so T(n) = x + (x^3)/(3!3) + (x^5)/(5!5) + ... since i made my interval
    [tex]( (n-1)\pi , n\pi)[/tex]
    Why am i getting
    t(n) > t(n-1)
    What am i doing wrong??Can it be done this way???
    Isnt t(n-1) > t(n)
    I proved it another way but just frustrating that i dont know where i went wrong.
    Wish i could put my calculations up but its just tooo long..
     
  11. Apr 25, 2004 #10

    uart

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    Yeah, integrating the power series should work. But you've got the power series for sin wrong, remember that the sign of the terms alternate.


    But I dont think the power series will help you evaluate this integral from 0 to infinity.

    From the title of the thread I thought you were interested in integrating from 0 to infinty, but I notice in the body of the thread you dont mention those limits, what is the case?
     
  12. Apr 25, 2004 #11
    sorry!!!
    i forgot to mention it.
    But looks like everyone got the idea that it was (0,infinity) which it what i wanted.
    Found out that the method i mentioned before....
    i did a stupid thing by turning the minus to plus, DOPE!!!thanx
     
  13. Apr 25, 2004 #12
    i was thinking of approximating
    [tex]\int\frac{sinx}{x}dx\[/tex]

    by changing it into an alternating series where
    T(n) = [tex]\mid\int\frac{sinx}{x}dx\mid[/tex] where n = 1... infinity
    and waz even thinking of using complex analysis to integrate it and we see when i get through it but im also trying to figure out why T(n-1) > T(n) which im just stuck on at the end..

    sorrie!!!
    Having problems proving T(n) < T(n-1)

    T(n)= [tex]\mid\int\frac{sinx}{x}dx\mid[/tex]
    and using the power series of sine and dividing by x and integrating it with the interval for n-1 is [(n-2)*pi,(n-1)*pi]
    so will
    T(n-1) = [tex] ( (n-1)\pi - \frac{((n-1)\pi)^3}{3!3} + \frac{((n-1)\pi)^5}{5!5} - ...
    -((n-1)\pi - \frac{((n-2)\pi)^3}{3!3} + \frac{((n-2)\pi)^5}{5!5} - ...[/tex]

    T(n) = [tex] ((n)/pi - \frac{((n)\pi)^3}{3!3} + \frac{((n)\pi)^5}{5!5} - ...
    - ((n-1)\pi - \frac{((n-1)\pi)^3}{3!3} + \frac{((n-1)\pi)^5}{5!5} - ...) [/tex]

    this is how far i am but how do i carry on from here???Is it correct???
     
    Last edited: Apr 25, 2004
  14. Apr 25, 2004 #13
    Told ya so! Heh. Use Matlab or something to evaluate the integral.
     
  15. Apr 25, 2004 #14
    1) i dont know how to use matlab / mathematica (if u were thinking about it)

    2) Gotta show all the working out hahaha
     
  16. Apr 26, 2004 #15

    uart

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    Science Advisor



    This is a homework or assignment question right. Are you sure you've interpreted it correctly and that you really require an approximation, or is it more a bound that you interested in. The approximation with an alternating series can only be crude anyway because of the "x" quantitization. (Meaning that it only approximates Si(x) at discrete values of x=k Pi).

    Are you sure you're not just looking for the simple alternating series upper bound that I posted earlier
     
  17. Apr 26, 2004 #16

    uart

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    That is,

    [tex]Si(n \pi) \leq \pi\, \sum_{k=1}^{n} \frac{(-1)^{k+1}}{k} [/tex]
     
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