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In the class, my professor said the following

$$\int_{-\infty}^{\infty} e^{-j(\omega-\omega_0)t} dt=\frac{1}{-j(\omega-\omega_0)}e^{-j(\omega-\omega_0)t}\bigg |_{-\infty}^{\infty}=0,$$ by Euler's formula when $\omega \neq \omega_0$.

I am a bit confused about this! How does the Euler's formula come in when dealing with $\infty$.

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    $\begingroup$ It's not a precise statement, it's describing one of the properties of the Dirac delta. $\endgroup$ May 10 at 2:56
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Intuitively, by taking the inverse FT of a dirac delta (times $2\pi$) in the frequency domain we obtain $$\frac{1}{2\pi}\int_{\mathbb{R}} 2\pi\delta(\omega-\omega_0)e^{i\omega t}d\omega=\frac{e^{i\omega_0t}}{2\pi}\cdot 2\pi=e^{i\omega_0 t}$$ So we know that the FT $$\int_{\mathbb{R}}e^{-i(\omega-\omega_0)t}dt=2\pi\delta(\omega-\omega_0)$$ This works because we introduced the Dirac delta (which is not a function). The function $e^{-i(\omega-\omega_0)t}$ is not integrable in the usual sense.

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