Proving Improper Integral with Complex Analysis

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SUMMARY

The discussion focuses on proving the improper integral \(\int_{-\infty}^{\infty} e^{\imath (k - k') x} dx = 2\pi \delta(k-k')\) using complex analysis techniques. Participants suggest evaluating the limit \(\lim_{L\rightarrow \infty} \int_{-L}^{L} e^{\imath (k - k') x} dx\) and relate the Dirac delta function to the limit of \(\delta(x) = \lim_{g\rightarrow \infty} \frac{\sin(gx)}{\pi x}\). The conversation also explores the potential advantages of a piecewise approach by considering cases where \(k \neq k'\) and then \(k = k'\).

PREREQUISITES
  • Complex analysis fundamentals
  • Understanding of the Dirac delta function
  • Knowledge of improper integrals
  • Familiarity with limits and piecewise functions
NEXT STEPS
  • Study the application of contour integration in complex analysis
  • Learn about the properties and applications of the Dirac delta function
  • Explore techniques for evaluating improper integrals
  • Research piecewise function analysis in mathematical proofs
USEFUL FOR

Mathematicians, physics students, and anyone interested in advanced calculus and complex analysis techniques for evaluating integrals and understanding distributions.

jusy1
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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?
 
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Wouldn't it be easier to prove this considering piecewise, i.e. consider k≠k' and then consider k=k'?
 

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