$$\delta(t - 1) e^{i\pi t} + \delta(t - 2) e^{-i \pi t}$$

I don't understand how to simplify this. My two guesses were either there was a property with the delta function or the delta function was a red herring and I needed to use Euler's equation. I know the answer is

$$-\delta(t - 1) + \delta(t - 2)$$

but do not see how to get rid of the exponential.

  • $\begingroup$ conceptually consider that $\delta(t-1)$ can be thought of as taking a value of zero outside of $t=1$ so $\delta(t-1) f(t)$ = $\delta(t-1) e^{i \pi t}$ for any $f(t)$ such that $f(1) = e^{i \pi 1}$. $\endgroup$ – jodag Dec 10 '17 at 21:53
  • $\begingroup$ Technically it does not have a value. Delta functions are only defined when you integrate them together with test functions. $\endgroup$ – mathreadler Dec 10 '17 at 21:53
  • $\begingroup$ True, which is why I said conceptually. I edited to make this more clear. $\endgroup$ – jodag Dec 10 '17 at 21:54

$$\delta(x - a)f(x) = \delta(x-a)\ f(a)$$

Remember also that

$$e^{2\pi i} = 1$$

$$e^{\pi i} = -1$$


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