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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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# Improper integral

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