Fourier transform of e^|x|... |e^(iy-1)t}|=e^-t, how?

Question

I am hopelessly confused. Somebody please help? :(

The red underlined is the problem.

Answer 1

Your confusion appears to be that you missed the fact that they want the Fourier transform of $f(x)= e^{-|x|}$, which by definition is $$ \int_{-\infty}^\infty e^{i\xi t} f(t)\, dt= \int_{-\infty}^\infty e^{i\xi t} e^{-|t|}|\, dt $$ That is not the same as saying that $e^{i\xi t} e^{-|t|}$ is the same as $e^{-|t|}$ or that for $t>$ that $e^{(i\xi-1) t} = e^{-t}$.

When taking the transform, you are multiplying by the oscillating factor $e^{i\xi t}$ and integrating; it is no surprise that the thing you are integrating is not identical to the original function.

Answer 2

Solved!

The underlined can be re-written as

...and e^i(anything), is just a line of magnitude 1 with angle (anything).

Thanks anyway to anyone who tried!