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My current problem simplifies to solving $\int_0^T e^{2\pi ix/T} dx$, $T \not= 0$.

I currently have 2 ways of solving this, one of which uses the definition of $e^{i \pi}$, and the other from definition of integrating exponents.

First method: $\int_0^T e^{2\pi ix/T}dx = \int_0^T 1^{x/T}dx = \int_0^T 1dx = T$

Second method: $\int_0^T e^{2\pi ix/T}dx = e^{2 \pi i x/T}/{\frac T {2 \pi i}}|_0^T = {\frac T {2 \pi i}} (e^{2 i \pi} - 1) = 0$

The two different methods give different answers that are contradictory. Any pointers would be helpful, thanks.

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  • $\begingroup$ $e^{2\pi i x/T}\neq 1$ for $0< x< T$ $\endgroup$ – mickep Apr 16 '17 at 7:38
  • $\begingroup$ @mickep does $e^{kx} = (e^k)^x$ not hold for complex numbers? $\endgroup$ – Thunda Apr 16 '17 at 7:39
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You have $$ e^{2\pi i x/T}=\cos(2\pi x/T)+i\sin(2\pi x/T). $$ Thus, clearly $e^{2\pi i x/T}\neq 1$ for $0<x<T$ (for exaple, $x=T/2$ gives $e^{\pi i}=-1$)

You have to be more careful when using complex powers.

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