$$f(x)=\begin{matrix} 12x-4x^2-8 & x \in[1,2] \\ 12x+4x^2+8 & x \in[-2,-1] \\ 0 & x \notin[1,2], x \notin[-2,-1] \end{matrix} $$
Find $\hat{f(\xi)}$ and $\hat{f(0)}$
Solution:
$$\int_{-\infty}^{+\infty}f(x)dx=0$$ Is there a problem if the function I want to transform's integral is equal to 0? Does it effect the transformation in any way?
Calculating the Fourier transformation $$\hat{f(\xi)} = \int_{-\infty}^{+\infty}e^{-ix\xi}f(x)dx=\hat{f_{1}(\xi)}+\hat{f_{2}(\xi)}$$
Splitting the integral to two parts $$\hat{f_{1}(\xi)}=\int_{-2}^{-1}(12x-4x^{2}-8)e^{-ix\xi}dx=\\=4\frac{\cos\xi}{\xi^{2}}+4\frac{\cos2\xi}{\xi^{2}}+8\frac{\sin\xi}{\xi^3}-8\frac{\sin2\xi}{\xi^3}+i\left(4\frac{\sin2\xi}{\xi^{2}}+4\frac{\sin2+\xi}{\xi^{2}}+8\frac{\cos2\xi}{\xi^3}-8\frac{\cos\xi}{\xi^3}\right)$$
$$\hat{f_{2}(\xi)}=\int_{1}^{2}(12x+4x^{2}+8)e^{-ix\xi}dx=\\=-4\frac{\cos\xi}{\xi^{2}}-4\frac{\cos2\xi}{\xi^{2}}-\frac{\sin\xi}{\xi^3}+8\frac{\sin2\xi}{\xi^3}+i\left(4\frac{\sin2\xi}{\xi^{2}}+4\frac{\sin2+\xi}{\xi^{2}}+8\frac{\cos2\xi}{\xi^3}-8\frac{\cos\xi}{\xi^3}\right)$$
Answer $$\hat{f(\xi)}=\hat{f_{1}(\xi)}+\hat{f_{2}(\xi)}=i\left(8\frac{\sin2\xi}{\xi^{2}}+8\frac{\sin2+\xi}{\xi^{2}}+16\frac{\cos2\xi}{\xi^3}-16\frac{\cos\xi}{\xi^3}\right)$$
Question:
$\hat{f(0)}=\infty$ or $\hat{f(0)}$ is not defined. But this is solution I've never stumbled before, I fear it is wrong. How can I check? Is there anything specific about that I am missing here?
Your function is odd, therefore the transform should be odd too. From this you can find the value of the transform at zero. Also, the function is absolutely integrable and piecewise smooth, so you should certainly expect the inverse transform to exist at least in the sense of the principal value integral.
To compute the direct transform, since the function is odd, discard the real part, which cancels if you fix the missing factor of $8$ in one of the terms, take $2i$ times the imaginary part you obtained from integrating over $[1, 2]$, fix the typos (you have $\sin 2 \xi$ instead of $\sin \xi$ and $\sin 2 + \xi$ instead of $\sin 2 \xi$), and you have your answer.