Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Improper integral

  1. Dec 3, 2011 #1
    Hi everybody

    I was trying to prove that [tex]\int_{-\infty}^{\infty}e^{\imath (k - k') x}dx = 2\pi\delta(k-k')[/tex] by solving [tex]\lim_{L\rightarrow \infty} \int_{-L}^{L}e^{\imath (k - k') x}dx[/tex]

    knowing that [tex]\delta(x)=\lim_{g\rightarrow \infty}\frac{\sin(gx)}{\pi x}[/tex]

    But is there a way of proving this result using complex analysis?
     
  2. jcsd
  3. Dec 3, 2011 #2

    Char. Limit

    User Avatar
    Gold Member

    Wouldn't it be easier to prove this considering piecewise, i.e. consider k≠k' and then consider k=k'?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Improper integral
  1. Improper Integration (Replies: 5)

  2. Improper integrals (Replies: 4)

  3. Improper Integrals (Replies: 1)

  4. "Improper" Integral (Replies: 3)

Loading...