Some complex Fourier-like infinite integral

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SUMMARY

The integral \(\int_{0}^{\infty} e^{ikx} dx\) is evaluated to yield \(\mathcal{I}=\pi \delta(k) + i \frac{1}{k}\). The real part is proven using the known result \(\int_{-\infty}^{\infty} e^{ikx} dx=2\pi \delta(k)\). However, the derivation of the imaginary part, \(\frac{1}{k}\), remains unclear to the participants. The discussion highlights the confusion surrounding the limit of \(\cos(kM)\) as \(M\) approaches infinity, which complicates the understanding of the imaginary component.

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Ahmes
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Homework Statement


The following integral is given:
\int_{0}^{\infty} e^{ikx} dx
k & x are real.

Homework Equations


We know that:
\int_{-\infty}^{\infty} e^{ikx} dx=2\pi \delta(k)

The Attempt at a Solution


The answer is:
\mathcal{I}=\pi \delta(k) + i \frac{1}{k}
I could easily prove the real part with the formula in 2, but couldn't be persuaded why the imaginary part is so (it barely makes sense). I tried:
\Im \int_{0}^{\infty} e^{ikx} dx = \int_{0}^{\infty} =\sin(kx) dx = \lim_{M\rightarrow \infty} \frac{1}{k}(-\cos(kM)+1)
But \lim_{M\rightarrow \infty} \cos(kM) is quite meaningless and I can't see why the whole expression should be k^{-1}...

I'll appreciate any help,
Thanks!
 
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