If not, could you please show me how to do it? I would really appreciate it!
Quesiton
Consider the function $$ f(t)=\mathrm{e}^{-\lambda|t|} \cos (\sigma t) $$ with positive parameters $\lambda$ and $\sigma$, Calculate the Fourier transformation $\tilde{f}(\omega)$.
Solution
Use the definition of the Fourier transform: $$\tilde{f}(\omega)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} f(t) e^{-i\omega t} dt$$
Substitute the function f(t) into the equation: $$\tilde{f}(\omega)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} (e^{\lambda |t|})cos(\sigma t) e^{-i\omega t} dt$$
Split the integral into two parts, one for t >= 0 and one for t < 0: $$\tilde{f}(\omega)=\frac{1}{\sqrt{2\pi}}\left(\int_{0}^{\infty} e^{\lambda t}cos(\sigma t) e^{-i\omega t} dt + \int_{-\infty}^{0} e^{-\lambda t}cos(\sigma t) e^{-i\omega t} dt\right)$$
Use the substitution u = $e^{\lambda t}$ in the first integral: $$\tilde{f}(\omega)=\frac{1}{\sqrt{2\pi}\lambda}\left(\int_{1}^{\infty} \frac{cos(\sigma \ln u)}{u} e^{-i\omega \ln u} du + \int_{-\infty}^{0} e^{-\lambda t}cos(\sigma t) e^{-i\omega t} dt\right)$$
Use the substitution $v = \sigma t$ in the second integral: $$\tilde{f}(\omega)=\frac{1}{\sqrt{2\pi}\lambda}\left(\int_{1}^{\infty} \frac{cos(\sigma \ln u)}{u} e^{-i\omega \ln u} du + \frac{1}{\sigma}\int_{-\infty}^{0} cos(v) e^{-i(\omega/\sigma) v} dv\right)$$
Use the definition of the cosine transform and the definition of the inverse cosine transform: $$\tilde{f}(\omega)=\frac{1}{\sqrt{2\pi}\lambda}\left(\int_{1}^{\infty} \frac{cos(\sigma \ln u)}{u} e^{-i\omega \ln u} du + \frac{1}{\sigma} \frac{\sqrt{2\pi}}{2}(\delta(\omega + \sigma) + \delta(\omega - \sigma))\right)$$
Simplify the first integral by completing the square in the exponent and applying a change of variables:
$$\tilde{f}(\omega)=\left(\frac{\lambda}{\sqrt{\lambda^2 + (\omega + \sigma)^2}} + \frac{\lambda}{\sqrt{\lambda^2 + (\omega - \sigma)^2}}\right) + \frac{1}{\sigma} \frac{\sqrt{2\pi}}{2}(\delta(\omega + \sigma) + \delta(\omega - \sigma))$$
Simplify the expression by combining the two terms: $$\tilde{f}(\omega)=\frac{\lambda}{\lambda^2 + (\omega + \sigma)^2} + \frac{\lambda}{\lambda^2 + (\omega - \sigma)^2}$$
NEW SOLUTION
Lets calculate the Fourier transform of the function, $\mathrm{g}(\sigma, \mathrm{t})=\mathrm{e}^{-\lambda|t|} \mathrm{e}^{\mathrm{i} \sigma \mathrm{t}}$ $$ \begin{aligned} \mathrm{G}(\sigma, \omega)=\mathcal{F}\{\mathrm{g}(\sigma, \mathrm{t})\} & =\int_{-\infty}^{\infty} \mathrm{e}^{-\mathrm{i} \omega \mathrm{t}} \mathrm{e}^{\mathrm{i} \sigma t-\lambda|t|} \mathrm{dt} \\ & =\int_0^{\infty} \mathrm{e}^{-\mathrm{i} \omega \mathrm{t}} \mathrm{e}^{\mathrm{i} \sigma t-\lambda \mathrm{t}} \mathrm{dt}+\int_{-\infty}^0 \mathrm{e}^{-\mathrm{i} \omega t} \mathrm{e}^{\mathrm{i} \sigma t+\lambda t} \mathrm{dt} \\ & =\int_0^{\infty} \mathrm{e}^{-\mathrm{i} \omega t+\mathrm{i} \sigma t-\lambda \mathrm{t}} \mathrm{dt}-\int_0^{-\infty} \mathrm{e}^{-\mathrm{i} \omega t+\mathrm{i} \sigma t+\lambda t} \mathrm{dt} \\ & =\int_0^{\infty} \mathrm{e}^{-\mathrm{i} \omega t+\mathrm{i} \sigma t-\lambda \mathrm{t}} \mathrm{dt}+\int_0^{\infty} \mathrm{e}^{\mathrm{i} \omega t-\mathrm{i} \sigma t-\lambda \mathrm{t}} \mathrm{dt} \\ & =\frac{\left[\mathrm{e}^{-\mathrm{i} \omega t+\mathrm{i} \sigma t-\lambda t}\right]_0^{\infty}}{-\mathrm{i} \omega+\mathrm{i} \sigma-\lambda}+\frac{\left[\mathrm{e}^{\mathrm{i} \omega \mathrm{t}-\mathrm{i} \sigma t-\lambda t}\right]_0^{\infty}}{\mathrm{i} \omega-\mathrm{i} \sigma-\lambda} \\ & =\frac{0-1}{-\mathrm{i} \omega+\mathrm{i} \sigma-\lambda}+\frac{0-1}{\mathrm{i} \omega-\mathrm{i} \sigma-\lambda} \\ & =\frac{1}{\lambda+\mathrm{i}(\omega-\sigma)}+\frac{1}{\lambda-\mathrm{i}(\omega-\sigma)} \\ & =\frac{\mathrm{i}(\omega+\sigma)+\lambda-\mathrm{i}(\omega-\sigma)+\lambda}{\lambda^2+(\omega+\sigma)^2} \\ & =\frac{2 \lambda}{\lambda^2+(\omega-\sigma)^2} \end{aligned} $$
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, $$ \mathrm{G}(-\sigma, \omega)=\mathcal{F}\{\mathrm{g}(-\sigma, \mathrm{t})\}=\frac{2 \lambda}{\lambda^2+(\omega+\sigma)^2} $$
$\mathrm{f}(\mathrm{t})=\mathrm{e}^{-\lambda|t|} \cos (\sigma \mathrm{t})$ $$ \begin{aligned} \mathrm{F}(\omega)=\mathcal{F}\{\mathrm{f}(\mathrm{t})\} & =\int_{-\infty}^{\infty} \mathrm{e}^{-\mathrm{i} \omega \mathrm{t}} \mathrm{e}^{-\lambda|\mathrm{t}|} \cos (\sigma \mathrm{t}) \mathrm{dt} \\ & =\frac{1}{2}\left[\int_{-\infty}^{\infty} \mathrm{e}^{-\mathrm{i} \omega t} \mathrm{e}^{\mathrm{i} \sigma t-\lambda|t|} \mathrm{dt}+\int_{-\infty}^{\infty} \mathrm{e}^{-\mathrm{i} \omega t} \mathrm{e}^{-\mathrm{i} \sigma t-\lambda|t|} \mathrm{dt}\right] \\ & =\frac{1}{2}[\mathrm{G}(\sigma, \omega)+\mathrm{G}(-\sigma, \omega)] \\ & =\frac{1}{2}\left[\frac{2 \lambda}{\lambda^2+(\omega-\sigma)^2}+\frac{2 \lambda}{\lambda^2+(\omega+\sigma)^2}\right] \\ & =\frac{\lambda}{\lambda^2+(\omega+\sigma)^2}+\frac{\lambda}{\lambda^2+(\omega-\sigma)^2} \end{aligned} $$