A question on the Dirac delta distribution

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SUMMARY

The integral $$\int e^{-i(k+k’)x}\,\mathrm{d}x$$ is definitively proportional to the Dirac delta distribution ##\delta(k+k’)## in the distributional sense. This conclusion is supported by manipulating the integral with a test function ##\hat \phi(k)##, which is the Fourier transform of ##\phi(x)##. The transformation leads to the result that confirms the relationship between the integral and the delta function, specifically yielding $$2\pi \hat \phi(-k') = \int 2\pi \delta(k+k') \hat\phi(k) dk$$.

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Homework Statement
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Relevant Equations
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Is it correct to say that

$$\int e^{-i(k+k’)x}\,\mathrm{d}x$$

is proportional to ##\delta(k+k’)##?
 
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In the distributional sense, yes.

Consider a test function ##\hat \phi(k)## being the Fourier transform of ##\phi(x)##. It holds that
$$
\int \hat\phi(k) \int e^{-i(k+k')x} dx\, dk = \int e^{-ik'x} \int \hat\phi(k) e^{-ikx} dk\, dx
= \int e^{-ik'x} \phi(-x) dx = \int e^{ik'x} \phi(x) dx = 2\pi \hat \phi(-k') = \int 2\pi \delta(k+k') \hat\phi(k) dk.
$$
 

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