My question is related to part c of this question.
I ended up trying to plug in the solution to part b into the second equation given as the inverse Fourier Transform and combining it with the value of f(t) given in part ii)a.
Therefore i deduce that
$$e^{−\alpha|t|}=\int_{-\infty}^{\infty}\frac{1}{\pi}\frac{a}{\omega^2+a^2}e^{-i\omega t} dt$$
I tried replacing $\alpha$ with 1 and tried making some substitutions to get the RHS go the equation I deduced above to look similar to that in part c but I currently am unable to do it.
Any help on doing part c will be appreciated.
Just plug in $t=1$, $\alpha=1$, and $\omega=x$.
$$\frac{1}{e}=f(1)=\\=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}g(x)e^{-ix1}dx=\\=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}\frac{2\cdot1}{1+x^2}e^{-ix}dx$$ This yields $$\int_{-\infty}^{\infty}\frac{1}{1+x^2}e^{-ix}dx=\frac{\pi}{e}$$