1
$\begingroup$

To compute the inverse Fourier transform I need to evaluate the following integral $$\int_{-\infty}^{\infty}\exp\left(-{(\omega-\omega_0)^2\over2\sigma_\omega}\right)\cdot\exp\left(-i\phi\right)\cdot\exp(-i\omega t)\;\mathrm{d}\omega$$

Can I ask you guys for a hint how I should proceed? What are the common methods for tackling such problems?

$\endgroup$
2
$\begingroup$

You may just recall the gaussian integral result : $$ \int_{-\infty}^{+\infty}\exp\left(-a{(x-x_0)^2}\right)\mathrm{d}x= \sqrt{\frac{\pi}{a}},\quad \Re a>0,\,x_0 \in \mathbb{C}, $$ and rewrite your initial integral $$I=\int_{-\infty}^{+\infty}\exp\left(-{(\omega-\omega_0)^2\over2\sigma_\omega}\right)\cdot\exp\left(-i\phi\right)\cdot\exp(-i\omega t)\;\mathrm{d}\omega$$ as $$I=\exp\left(-i\phi\right)\cdot\exp(\frac{t^2\sigma_\omega}{2}-i\omega_0 t)\cdot\int_{-\infty}^{+\infty}\exp\left(-{(\omega+it\sigma_\omega-\omega_0)^2\over2\sigma_\omega}\right)\mathrm{d}\omega$$ leading to

$$ I=\sqrt{2\pi\sigma_\omega}\cdot\exp\left(-i\phi\right)\cdot\exp(\frac{t^2\sigma_\omega}{2}-i\omega_0 t). $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.