Fourier transform of trigonometric function

Question

I would like to ask for some help on the Fourier transform of the following function.

$F(t)=\frac{cos(\Omega t)}{(\lambda^2+t^2)}$

I can do the Fourier transformation with the cosine function.

Thanks in advance.

Answer 1

I will here define the Fourier transform by $$ \hat{f}(\omega) = \int_{-\infty}^{\infty} f(t) \, e^{-i\omega t} \, dt $$

Rewriting $\cos \Omega t$ as $\frac12(e^{i\Omega t}+e^{-i\Omega t})$ gives $$ \hat{F}(\omega) = \int_{-\infty}^{\infty} \frac{\cos(\Omega t)}{\lambda^2+t^2} \, e^{-i\omega t} \, dt = \int_{-\infty}^{\infty} \frac{\frac12(e^{i\Omega t}+e^{-i\Omega t})}{\lambda^2+t^2} \, e^{-i\omega t} \, dt = \frac12 \left( \int_{-\infty}^{\infty} \frac{1}{\lambda^2+t^2} \, e^{-i(\omega-\Omega)t} + \int_{-\infty}^{\infty} \frac{1}{\lambda^2+t^2} \, e^{-i(\omega+\Omega)t} \right) \\ = \frac12 \left( \hat{G}(\omega-\Omega) + \hat{G}(\omega+\Omega) \right) , $$ where $$ G(t) = \frac{1}{\lambda^2+t^2} . $$

Can you calculate the Fourier transform of $G$?