1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Integration problem

  1. Oct 14, 2006 #1
    Can someone give me a hint on how to solve

    [tex]I = \int_0^\infty e^{-ax} \sin{bx} dx[/tex]

    ?
     
  2. jcsd
  3. Oct 14, 2006 #2
    Use integration by parts, and set the upper limit as t. Then let t--> infinity (i.e. improper integral)
     
    Last edited: Oct 14, 2006
  4. Oct 14, 2006 #3
    But integration by parts will only give me a similar integral to solve...?
     
  5. Oct 14, 2006 #4
    you have to use it multiple times, so that you can solve the integral
     
  6. Oct 14, 2006 #5
    You can write the trig function as a complex exponential:

    [tex]
    \sin(bx) = \frac{1}{2i}\left[e^{ibx}-e^{-ibx}\right]\,.
    [/tex]

    Then you will have two similar integrals to solve, which are straight forward.
     
  7. Oct 16, 2006 #6
    jprO: Of course. That will do it.

    courtrigad: Partial integration multiple times will give me a loop between [tex]e^{-ax} \sin{bx}[/tex] and [tex]e^{-ax} \cos{bx}[/tex].
     
  8. Oct 16, 2006 #7

    HallsofIvy

    User Avatar
    Science Advisor

    Yes, it does:
    [tex]I = \int_0^\infty e^{-ax} \sin{bx} dx[/tex]
    Let [itex]u= e^{-ax}[/itex], [itex]dv= sin(bx)dx[/itex]
    Then [itex]du= -ae^{-ax}dx[/itex], [itex]v= -\frac{1}{b}cos(bx)[/itex]
    so the integral becomes
    [tex]I= -\frac{1}{b}e^{-ax}cos(bx)\left|_0^\inftjy- \frac{a}{b}\int_0^\inty}e^{-ax}cos(bx)dx[/tex]
    Since e-ax goes to 0 as x goes to infinity, while cos(bx) is bounded, that is
    [tex]I= \frac{1}{b}- \frac{a}{b}\int_0^\inty}e^{-ax}cos(bx)dx[/tex]
    Now let [itex]u= e^{-ax}[/itex], [itex]dv= cos(bx)[/itex]. Again we have [itex]du= -a e^{-ax}dx[/itex], [itex]v= \frac{1}{b}sin(bx)[/tex]
    The integral is now:
    [tex]I= \frac{1}{b}- \frac{a^2}{b^2}\int_0^\infty e^{-ax}sin(bx)dx[/tex]
    But remember that [itex]I= \int_0^\infty e^{-ax}sin(bx)dx[/itex]
    so that equation says
    [itex]\int_0^\infty e^{-ax}sin(bx)dx= \frac{1}{b}- \frac{a^2}{b^2}\int_0^\infty e^{-ax}sin(bx)dx[/tex]
    so
    [tex]\left(1+frac{a^2}{b^2}\right)\int_0^\infty e^{-ax}sin(bx)dx= \frac{1}{b}[/tex]
    [tex]\int_0^\infty e^{-ax}sin(bx)dx= \frac{b}{a^2+b^2}[/tex]
    (Modulo any silly litte errors from working too fast!)
     
  9. Oct 16, 2006 #8
    I don't understand why you do the substitutions. Can't you just solve the integral by parts without doing the substitutions?
     
  10. Oct 16, 2006 #9
    do you agree that integration by parts if of the form:

    [tex] \int udv = uv-\int vdu [/tex]? You need substitutions to do integration by parts.
     
    Last edited: Oct 16, 2006
  11. Oct 17, 2006 #10
    I see what you mean. I'm just not used to do it your way. Actually, I'm doing it your way. I just don't think about what I'm really doing. ;)
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook




Loading...
Similar Threads for Integration problem Date
Integration Problem Apr 3, 2018
Integration problem using u substitution Mar 19, 2018
Integration problem of a quotient Feb 20, 2018
Problem integral Nov 26, 2017
Integration by parts problem Jul 18, 2017