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How to get from:

$$\frac{1}{2\pi}\frac{e^{jn}-e^{-jn}}{jn}$$

to:

$$\frac{1}{2\pi}\int\limits_{-1}^{1}e^{j\omega n} d \omega$$

as given here: https://dsp.stackexchange.com/a/38854/16003

mavavilj
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1 Answers1

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The antiderivative of $e^{j\omega n}$ is $\frac{e^{j \omega n}}{jn}$ (as long as $n \neq 0$) and so by the fundamental theorem of calculus

$$ \frac{1}{2\pi} \int_{-1}^1 e^{j \omega n} \, d\omega = \frac{1}{2\pi} \left[ \frac{e^{j \omega n}}{jn} \right]_{\omega = -1}^{\omega = 1} = \frac{e^{jn} - e^{-jn}}{2\pi j n}.$$

levap
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