Exponential decay Fourier transform

Question

I managed to get part a) but when I try to use the fact that $$\cos(t\omega)=\frac{e^{it\omega}+e^{-it\omega}}{2}$$ and $$\sin(t\omega)=\frac{e^{it\omega}-e^{-it\omega}}{2i}$$ for part b), then I get a result of 0 and not ${\pi}e^{-t\lambda}$

Answer 1

notice how we have: $$g(\omega,\lambda,t)=\frac{\lambda\cos\omega t+\omega\sin\omega t}{\lambda^2+\omega^2}$$ we know that: $$e^{ix}=\cos x+i\sin x$$ lets break it down and see what we have: $$e^{i\omega t}=\cos\omega t+i\sin\omega t$$ now notice if we take: $$(\lambda-i\omega)e^{i\omega t}=(\lambda-i\omega)\cos\omega t+(\lambda-i\omega)i\sin\omega t$$ $$=(\lambda\cos\omega t+\omega\sin\omega t)+i(\lambda\sin\omega t-\omega\cos\omega t)$$ and that is where the hint @md2perpe came from. Obviously we now need to deal with the integral: now we can notice that our integral is just equal to: $$\sqrt{2\pi}\int_0^\infty\bar{f}(t)e^{i\omega t}d\omega$$ and you still need to account for it being the real part but it shouldnt be too hard