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 by parts

  1. Jan 6, 2008 #1


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    [SOLVED] Integration by parts

    1. The problem statement, all variables and given/known data
    I've been starring at this for 30 minutes and can't figure out what's wrong. I end up with 1/xi instead of xi. The book says,

    [tex]\int_{-\infty}^{+\infty}te^{-t^2/2}\sin(\xi t)dt=\int_{-\infty}^{+\infty}e^{-t^2/2}\xi \cos(\xi t)dt[/tex]

    3. The attempt at a solution

    I set [itex]u=te^{-t^2/2}[/itex] and [itex]dv = \sin(\xi t)[/itex]. So I get [itex]du=e^{-t^2/2}-t^2e^{-t^2/2}[/itex] and unless I'm completely crazy, [itex]v=-\xi^{-1}\cos(\xi t)[/tex], so that

    [tex]\int_{-\infty}^{+\infty}te^{-t^2/2}\sin(\xi t)dt=te^{-t^2/2}\xi^{-1}\cos(\xi t)|_{-\infty}^{+\infty} + \int_{-\infty}^{+\infty}e^{-t^2/2}\xi^{-1}\cos(\xi t)dt - \int_{-\infty}^{+\infty}t^2e^{-t^2/2}\xi^{-1}\cos(\xi t)dt = \int_{-\infty}^{+\infty}e^{-t^2/2}\xi^{-1}\cos(\xi t)dt[/tex]

  2. jcsd
  3. Jan 6, 2008 #2
    What is [itex] v[/itex]?

    [tex]I=-\int_{-\infty}^{+\infty}(e^{-t^2/2})'\sin(\xi t)dt=-te^{-t^2/2}\sin(\xi t)|_{-\infty}^{+\infty} + \int_{-\infty}^{+\infty}e^{-t^2/2}(\sin(\xi t))'dt = \int_{-\infty}^{+\infty}e^{-t^2/2}\xi\cos(\xi t)dt[/tex]
  4. Jan 6, 2008 #3


    User Avatar
    Science Advisor
    Homework Helper

    I can't entirely follow what you're doing with the u and dv. I learned partial integration as
    [tex]\int_a^b f'(x) g(x) dx = - \left. f(x) g(x) \right|_a^b + \int_a^b f(x) g'(x) dx[/tex]
    Applying this to
    [tex]f'(t) = t e^{-t^2/2}, \qquad \implies \qquad f(t) = - e^{-t^2/2}[/tex]
    [tex]g(t) = \sin(\xi t), \qquad \implies \qquad g'(t) = \xi \cos(\xi t)[/tex]
    then gives me
    [tex]\int_{-\infty}^{+\infty}te^{-t^2/2}\sin(\xi t)dt =
    \left. e^{-t^2/2} \sin(\xi t) \right|_{-\infty}^{+\infty} +
    \int_{-\infty}^{+\infty} -e^{-t^2/2}\xi \cos(\xi t)dt.
    The boundary term vanishes so that gives exactly minus the result you'd like (probably a minus error on my side) but definitely a [itex]\xi[/itex] and not [itex]\xi^{-1}[/itex]. As it should be (otherwise you'd get strange results for [itex]\xi = 0[/itex]).

    [edit]I should learn to type LaTeX even faster, Rainbow Child beat me by 5 whole minutes :smile:[/edit]
    Last edited: Jan 6, 2008
  5. Jan 6, 2008 #4


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I see my mistake: the last term does not vanish. Thx!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook