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!

Difficult integral

  1. Oct 24, 2013 #1
    Mod note: Moved from the math technical sections.
    I need to show that
    [tex]\int_{-\infty}^\infty \frac{\sin^2 (pa/\hbar)}{p^2} \, dp = \frac{\pi a}{\hbar}.[/tex]
    I haven't got a clue how to integrate this function! Any help would be much appreciated thanks.
    Last edited by a moderator: Oct 24, 2013
  2. jcsd
  3. Oct 24, 2013 #2
    I've found through the transformation [itex]u=pa/\hbar[/itex] that it is equivalent to showing
    [tex] \int_{-\infty}^{\infty} \frac{\sin^2 u}{u^2}\,du = \pi,[/tex] if that helps anyone.
  4. Oct 24, 2013 #3


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    You've got your substitution and your integral.

    All you have to do is figure out p and dp in terms of u and du.

    The algebra is really simple.
  5. Oct 24, 2013 #4
    Yeah I did that substitution myself to make it easier... now how do I show that the integral is equal to pi??
  6. Oct 24, 2013 #5
    Using the cosine of double angle formula, and then integration by parts, this can be reduced to integrals of $$ \frac {\sin x} {x} $$
  7. Oct 24, 2013 #6
    Under a change of variable x = pa/h:

    [tex]\frac{\hbar}{a}\int_{-\infty}^\infty \frac{sin^2 x}{x^2}dx[/tex]

    As suggested you can apply integration by parts and then sine double angle rule to obtain

    [tex]\frac{\hbar}{a}\int_{-\infty}^\infty \frac{sin(2x)}{x}dx[/tex]

    The integral below has many different proofs, few of them are here
    [tex]\int_{-\infty}^\infty \frac{sin(ωx)}{x}dx=\pi[/tex] for any ω > 0.
  8. Oct 24, 2013 #7

    Can someone tell me what is the difference between a question asked here and a question asked on the homework forums?

    This is an honest question. For example, shouldn't this question ("Difficult integral") be in the homework section?

    Sorry for the off topic perishingtardi
  9. Oct 24, 2013 #8


    Staff: Mentor

    You are correct. As the sticky says at the top of this forum section, "This forum is not for homework or any textbook-style questions."

    I am moving this thread to the Homework section.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted

Similar Discussions: Difficult integral
  1. Difficult integral (Replies: 4)

  2. Difficult Integral (Replies: 10)

  3. Difficult Integral (Replies: 31)

  4. Difficult integral. (Replies: 5)

  5. Difficult Integration (Replies: 22)