Show from direct integration that the Fourier transform of the function
$ f(t) = \begin{cases} 1, & \text{-1 < t < 0} \\ -1, & \text{0 < t < 1} \\ 0, & \text{otherwise} \\ \end{cases}$
is $f(k)= \frac{2i(1-cos(k))}{k}$ \begin{align} \int_{-1}^0 (1)e^{-ikt} & = [\frac{-ie^{-ikt}}{k}]_1^0 dt- \\ & = \frac{-ie^{-ik(0)}}{k} + \frac{ie^{-ik(-1)}}{k} \\ & = \frac{-i}{k} + \frac{ie^{ik}}{k}\\ \end{align}
Second part
\begin{align} \int_0^1 (-1)e^{-ikt} & = [\frac{ie^{-ikt}}{k}]_0^1 \\ & = \frac{ie^{-ik(1)}}{k} - \frac{ie^{-ik(0)}}{k} \\ & = \frac{ie^{-ik}}{k} - \frac{i}{k} \\ \end{align}
adding them together
\begin{align} \frac{-i}{k} + \frac{ie^{ik}}{k} + \frac{-ie^{ik}}{k} - \frac{i}{k} & = \frac{-2i}{k} + \frac{ie^{ik}+ie^{-ik}}{k} \\ & = \frac{-2i + ie^{ik} +ie^{-ik}}{k} \\ & = \frac{-2i+2i(\frac{e^{ik}+e^{-ik}}{2})}{k} \\ & = \frac{-2i+2i({cos(k)})}{k} \\ & = \frac{2i(-1+cos(k))}{k} \end{align}
I can't see where I have made an error here.
The second part asks to use complex integration to find the inverse Fourier transform, show that \begin{align} f(k)= \frac{1}{2\pi}\int_{-\infty}^\infty\frac{2i(1-cos(k))}{k}e^{ikt} \end{align}
I am struggling finding my residue now that I can't use my p-over-q, so far I have:
\begin{align} \frac{i}{\pi}\int_{-\infty}^\infty\frac{(1-cos(k))}{k}e^{ikt} \end{align}
\begin{align} p(k) & = e^{ikt}-e^{ikt}cosk \\ q(k) & = k \\ q'(k) & = 1 \end{align}
I run into trouble because \begin{align} p(0) & = e^0-e^0cos(0) \\ & = 1 - 1(1)\\ & = 0 \end{align}
So I can't use the p-over-q rule but I don't know how to find my residue to show f(t)=1 at -1<t<0 (I haven't tried to use Jordan's Lemma yet because I can't find the residue)