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Could someone explain in detail to the layman, what steps were jumped to reach the second line?.

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  1. all $t' = -t$, then $dt' = -dt$ and then

    $$ \int_{-T}^0 dt\;Ae^{-j\omega t} = -\int_{T}^0 dt'\; Ae^{j\omega t'} = \int_{0}^T dt'\; Ae^{j\omega t'} = \int_{0}^T dt\; Ae^{j\omega t} $$

    where in the last step I just renamed the dummy variable $t'$ as $t$

  2. Now I'm going to use

    $$ e^{jx} - e^{-jx} = (\cos x + j\sin x) - (\cos x - j\sin x) = 2j\sin x $$

    So that

$$ F(j\omega) = \int_{0}^T dt\; Ae^{j\omega t} - \int_{0}^T dt\; Ae^{-j\omega t} = \int_0^T dt\;A (e^{j\omega t} - e^{-j\omega t}) = \int_0^T dt\;2jA\sin \omega t $$

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  • $\begingroup$ I don't understand the last step of (1). Why can you rename the dummy variable t' = t; if you had already defined it as t' = -t ? $\endgroup$ – Numb3rsS4t Nov 19 '16 at 15:42
  • $\begingroup$ @Numb3rsS4t You are allowed to name the integration variable whatever you like, $t'$, $t$, $\tau$, ... it really doesn't matter $\endgroup$ – caverac Nov 19 '16 at 17:04
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  1. Reverse the limits of second integral and multiply the integral by $-1$ to compensate the reversal.
  2. Make transformation in the second integral from $t$ to $-t$.
  3. Use $e^{-ja} - e^{ja} = -2j\sin(a)$
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