How does answer of this integral equal to kronecker delta?

Question

I can't figure out how does the result of this integral equal to $\delta_{mn}$ ?! $$\frac{1}{T_0}\int_{\alpha}^{\alpha+T_0}e^{-\frac{j2\pi(n-m)}{T_0}}\,dt$$ PS: Is there any easy or intuitive approach to this ?

Answer 1

Just integrate directly. If $n=m$, the exponent is zero \begin{equation} \frac{1}{T_0}\int_\alpha^{\alpha+T_0}\,\mathrm{d}t = \frac{t}{T_0}\,\bigg|^{\alpha+T_0}_\alpha = 1. \end{equation} Otherwise, \begin{equation} \frac{1}{T_0}\int_\alpha^{\alpha+T_0}e^{-\frac{2\pi i(n-m)}{T_0}t}\,\mathrm{d}t = -\frac{e^{-\frac{2\pi i(n-m)}{T_0}t}}{2\pi i(n-m)}\,\bigg|^{\alpha+T_0}_\alpha = \frac{e^{-\frac{2\pi i(n-m)}{T_0}\alpha}-e^{-\frac{2\pi i(n-m)}{T_0}(\alpha+T_0)}}{2\pi i(n-m)} = \frac{e^{-\frac{2\pi i(n-m)}{T_0}\alpha} - e^{-\frac{2\pi i(n-m)}{T_0}\alpha}e^{-2\pi i(n-m)}}{2\pi i(n-m)} = \frac{e^{-\frac{2\pi i(n-m)}{T_0}\alpha}-e^{-\frac{2\pi i(n-m)}{T_0}\alpha}}{2\pi i(n-m)} = 0. \end{equation}