For $$u(x) = \begin{cases} (1-|x|)^2, &|x|≤ 1,\\ 0, &|x|>0\end{cases}$$
I have the integral that gives the Fourier transform of $u({\omega})$
$$ \int_{-1}^{1} ((1-|x|)^2)e^{-ix\omega}dx $$ which is equivalent to the integral:
$$ \int_{0}^{1} ((1-|x|)^2)(e^{-ix\omega}+e^{ix\omega})dx $$
I want to understand how to obtain the second integral from the first integral.
I am trying to understand it better by finding a common Fourier transform
$\vert x\vert = -x$ for $x \lt 0$ and $\vert x\vert = x$ for $x\gt 0$: $$ F_\text{trans}= \int_{-\infty} ^{\infty} f(x) e^{-ikx}\,dx = \int_{-\infty} ^{\infty} e^{-a|x|} e^{-ikx}\,dx \\ = \int_{-\infty} ^{0} e^{ax} e^{-ikx}\,dx + \int_{0} ^{\infty} e^{-ax} e^{-ikx}\,dx \\ = \int_{-\infty} ^{0} e^{(a-ik)x}\,dx + \int_{0} ^{\infty} e^{-(a+ik)x}\,dx. $$
$$ \frac{e^{(-a-ik)x}}{-a-ik}\Big|_{0}^{\infty} + \frac{e^{(a-ik)x}}{a-ik}\Big|_{-\infty}^{0} $$
$$ \frac{e^{-a\infty}e^{-ik\infty}}{-a-ik} + \frac{1}{a+ik} + \frac{1}{a-ik} - \frac{e^{-\infty a}e^{-\infty-ik} }{a-ik}$$
Then I understand the first fraction goes to zero since dividing by a very large number ($\infty$ will obviously result in a very small number. Then the second fraction is just straightforward, so is the third. The fourth however knowing the result of this Fourier transform should also go to zero however I fail to understand how since it seems to become like $\frac{\infty}{\infty (a-ik)}$ due to the double - sign. Could someone please clarify this for me..?