# Inverse Fourier Transform of $\frac{1}{1-\omega^2}$

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.

• The given integral only makes sense in principal value. Commented Mar 13, 2018 at 16:52
• @JackD'Aurizio: Thanks for editing and comment. Do you mean the above approaches are not correct? Can you please elaborate a little bit? Commented Mar 13, 2018 at 17:03
• I simply mean that $$f(t)=\frac{1}{2\pi}\int_{-\infty}^{+\infty}\frac{e^{i\omega t}}{1-\omega^2}\,d\omega$$ is not a definition since the RHS does not make sense in the usual Riemann or Lebesgue's interpretation of $\int$. $$f(t)=\frac{1}{2\pi}\color{red}{\text{PV}}\int_{-\infty}^{+\infty}\frac{e^{i\omega t}}{1-\omega^2}\,d\omega$$ is much better. Commented Mar 13, 2018 at 17:34

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}