= Solution
For the two-sided exponential, split the integral at zero:
$$
\boxed{\widehat f(k)=\int_0^\infty e^{-(1+ik)x}dx+\int_{-\infty}^0e^{(1-ik)x}dx=\frac1{1+ik}+\frac1{1-ik}=\frac2{1+k^2}.}
$$
For the interval indicator, the <Fourier transform of an interval indicator> gives
$$
\boxed{\widehat g(k)=\int_{-2\pi}^{2\pi}e^{-ikx}dx=\frac{2\sin(2\pi k)}k.}
$$
At $k=0$, the continuous limiting value is $4\pi$, the length of the interval. Changing the values at its two endpoints has no effect on the transform.
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