I need to calculate the integral (that is, an inverse Fourier transform of a certain function): $$f(t)=\displaystyle\frac{1}{2\pi}\int_{-\infty}^\infty\frac{1}{1-\omega^2}e^{i\omega t}d\omega.$$
My attempt is to write the integral as $$\displaystyle\frac{1}{2\pi}.\frac{1}{2}\int_{-\infty}^\infty\bigg[\frac{1}{1-\omega}+\frac{1}{1+\omega}\bigg]e^{i\omega t}dt$$ and use residue theorem. Since each singularity in the integrands is a simple pole, I have found $$\displaystyle\frac{1}{2\pi}.\frac{1}{2}\int_{-\infty}^\infty\bigg[\frac{1}{1-\omega}+\frac{1}{1+\omega}\bigg]e^{i\omega t}d\omega=2\pi i\frac{1}{2\pi}.\frac{e^{it}+e^{-it}}{2}=i\cos t.$$ I didnt make sure with the above result because the Fourier transform of $\cos t$ is $$\frac{\delta(\omega-1)+\delta(\omega+1)}{2},$$ where $\delta$ is Dirac delta. Do these two results contradict? Another attempt after differentiating $f(t)$ with respect to $t$ twice and adding this to $f$ I got a second order ODE:$$f''+f=\delta(t).$$ Can I use the Laplace transform to solve this equation with the initial conditions $f(0)=f'(0)=0$?I appriciate if someone help clarifying these.
Can you please check whether this is a correct solution by using $\mathcal{F}[\operatorname{sgn}(t)]=\frac{2}{i\omega}$? From the FT of sign function, we get $\mathcal{F}^{-1}[\frac1\omega]=\frac{i\operatorname{sgn}(t)}{2}$. Now we are close to the target but still need some shifting. We will continue
\begin{align} f(t)&=\frac{1}{2\pi}\int_{-\infty}^\infty\frac{1}{1-\omega^2}e^{i\omega t}d\omega\\ &=\frac12\mathcal{F}^{-1}\bigg[\frac{1}{1-\omega}+\frac{1}{1+\omega}\bigg]\\ &=\frac12\mathcal{F}^{-1}\bigg[\frac{1}{1-\omega}\bigg]+\frac12\mathcal{F}^{-1}\bigg[\frac{1}{1+\omega}\bigg]\\ &=-\frac12\mathcal{F}^{-1}\bigg[\frac{1}{\omega}\bigg]e^{it}+\frac12\mathcal{F}^{-1}\bigg[\frac{1}{\omega}\bigg]e^{-it}\\ &=\frac12\mathcal{F}^{-1}\bigg[\frac{1}{\omega}\bigg]\left(e^{-it}-e^{it}\right)\\ &=\frac12\frac{i\operatorname{sgn}(t)}{2}\left(e^{-it}-e^{it}\right)\\ &=\frac12\operatorname{sgn}(t)\sin{t}. \end{align}