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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

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    Gold Member

    Wouldn't it be easier to prove this considering piecewise, i.e. consider k≠k' and then consider k=k'?
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