Is this solution correct in order to find the Fourier transform of $\mathrm{g}(\sigma, \mathrm{t})=\mathrm{e}^{-\lambda|t|} \mathrm{e}^{\mathrm{i} \sigma \mathrm{t}} ?$
If it seems wrong, please let me know and if you have time I would really appreciate if you could show me the right solution
\begin{aligned} \mathrm{G}(\sigma, \omega)=\mathcal{F}\{\mathrm{g}(\sigma, \mathrm{t})\} & =\int_{-\infty}^{\infty} \mathrm{e}^{-\mathrm{i} \omega \mathrm{t}} \mathrm{e}^{\mathrm{i} \sigma t-\lambda|t|} \mathrm{dt} \\ & =\int_0^{\infty} \mathrm{e}^{-\mathrm{i} \omega \mathrm{t}} \mathrm{e}^{\mathrm{i} \sigma t-\lambda \mathrm{t}} \mathrm{dt}+\int_{-\infty}^0 \mathrm{e}^{-\mathrm{i} \omega t} \mathrm{e}^{\mathrm{i} \sigma t+\lambda t} \mathrm{dt} \\ & =\int_0^{\infty} \mathrm{e}^{-\mathrm{i} \omega t+\mathrm{i} \sigma t-\lambda \mathrm{t}} \mathrm{dt}-\int_0^{-\infty} \mathrm{e}^{-\mathrm{i} \omega t+\mathrm{i} \sigma t+\lambda t} \mathrm{dt} \\ & =\int_0^{\infty} \mathrm{e}^{-\mathrm{i} \omega t+\mathrm{i} \sigma t-\lambda \mathrm{t}} \mathrm{dt}+\int_0^{\infty} \mathrm{e}^{\mathrm{i} \omega t-\mathrm{i} \sigma t-\lambda \mathrm{t}} \mathrm{dt} \\ & =\frac{\left[\mathrm{e}^{-\mathrm{i} \omega t+\mathrm{i} \sigma t-\lambda t}\right]_0^{\infty}}{-\mathrm{i} \omega+\mathrm{i} \sigma-\lambda}+\frac{\left[\mathrm{e}^{\mathrm{i} \omega \mathrm{t}-\mathrm{i} \sigma t-\lambda t}\right]_0^{\infty}}{\mathrm{i} \omega-\mathrm{i} \sigma-\lambda} \\ & =\frac{0-1}{-\mathrm{i} \omega+\mathrm{i} \sigma-\lambda}+\frac{0-1}{\mathrm{i} \omega-\mathrm{i} \sigma-\lambda} \\ & =\frac{1}{\lambda+\mathrm{i}(\omega-\sigma)}+\frac{1}{\lambda-\mathrm{i}(\omega-\sigma)} \\ & =\frac{\mathrm{i}(\omega+\sigma)+\lambda-\mathrm{i}(\omega-\sigma)+\lambda}{\lambda^2+(\omega+\sigma)^2} \\ & =\frac{2 \lambda}{\lambda^2+(\omega-\sigma)^2} \end{aligned}
Here in the 4th line of integration $t$ is replaced by $-\mathrm{t}$. and although $\mathrm{e}^{\pm \mathrm{i}(\omega-\sigma) \mathrm{t}} \sin$ is not defined at $\mathrm{t}>$ $\infty$ but for exponential decay part $\lim _{\mathrm{t} \rightarrow \infty} \mathrm{e}^{-\lambda \mathrm{t}} \mathrm{e}^{\pm \mathrm{i}(\omega-\sigma) \mathrm{t}}=0$.
Similarly for, $\mathrm{g}(-\sigma, \mathrm{t})=\mathrm{e}^{-\lambda|t|} \mathrm{e}^{-\mathrm{i} \sigma \mathrm{t}}$ the fourier transformation will be, $$ \mathrm{G}(-\sigma, \omega)=\mathcal{F}\{\mathrm{g}(-\sigma, \mathrm{t})\}=\frac{2 \lambda}{\lambda^2+(\omega+\sigma)^2} $$
$\mathrm{f}(\mathrm{t})=\mathrm{e}^{-\lambda|t|} \cos (\sigma \mathrm{t})$ $$ \begin{aligned} \mathrm{F}(\omega)=\mathcal{F}\{\mathrm{f}(\mathrm{t})\} & =\int_{-\infty}^{\infty} \mathrm{e}^{-\mathrm{i} \omega \mathrm{t}} \mathrm{e}^{-\lambda|\mathrm{t}|} \cos (\sigma \mathrm{t}) \mathrm{dt} \\ & =\frac{1}{2}\left[\int_{-\infty}^{\infty} \mathrm{e}^{-\mathrm{i} \omega t} \mathrm{e}^{\mathrm{i} \sigma t-\lambda|t|} \mathrm{dt}+\int_{-\infty}^{\infty} \mathrm{e}^{-\mathrm{i} \omega t} \mathrm{e}^{-\mathrm{i} \sigma t-\lambda|t|} \mathrm{dt}\right] \\ & =\frac{1}{2}[\mathrm{G}(\sigma, \omega)+\mathrm{G}(-\sigma, \omega)] \\ & =\frac{1}{2}\left[\frac{2 \lambda}{\lambda^2+(\omega-\sigma)^2}+\frac{2 \lambda}{\lambda^2+(\omega+\sigma)^2}\right] \\ & =\frac{\lambda}{\lambda^2+(\omega+\sigma)^2}+\frac{\lambda}{\lambda^2+(\omega-\sigma)^2} \end{aligned} $$