Solve the following integral:

$$I = \int_{-\infty}^{+\infty} e^{-at^2} \cos (\omega_0 t ) e^{-i \omega t} dt$$ where $a, \omega_0 > 0$; $a, t, \omega_0, \omega \in \mathbb{R}$ and $i$ is the imaginary unit.

In other words, the Fourier transform of $e^{-at^2} \cos (\omega_0 t )$ is to be determined. As a first attempt, I tried to use Euler's formula:

$$ I = \int_{-\infty}^{+\infty} e^{-at^2} \cos (\omega_0 t ) \cos ( \omega t ) dt - i \int_{-\infty}^{+\infty} e^{-at^2} \cos (\omega_0 t ) \sin ( \omega t ) dt$$

Observing that the integrand function in the second integral is odd with respect to $t$, the integral reduces to

$$ I = \int_{-\infty}^{+\infty} e^{-at^2} \cos (\omega_0 t ) \cos ( \omega t ) dt$$

which can be split with Werner's formulas:

$$ I = \frac{1}{2} \int_{-\infty}^{+\infty} e^{-at^2} \cos [(\omega_0 + \omega) t ] dt + \frac{1}{2} \int_{-\infty}^{+\infty} e^{-at^2} \cos [(\omega_0 - \omega) t ] dt$$

I tried to solve the first integral by parts, but it doesn't help (I don't know an antiderivative for $e^{-at^2}$; using the antiderivative of cosine leads again to an integral with $t e^{-at^2} \sin [(\omega_0 + \omega)t]$.

A direct solution for the two integrals in the RHS is provided in Abramowitz and Stegun (10th printing, 1972, page 302, formula 7.4.6): but the intermediate steps are not showed there.

So, is there any way to proceed? Or another approach to follow?

  • $\begingroup$ It looks like something that should be Fourier transformed, but I haven't checked it. $\endgroup$ Mar 9 '17 at 12:12
  • $\begingroup$ @Lovsovs Yes, it is the Fourier transform of the function $e^{-at^2} \cos (\omega_0 t)$. $\endgroup$
    – BowPark
    Mar 9 '17 at 12:17
  • $\begingroup$ Have you tried www.symbolab.com ? $\endgroup$
    – toliveira
    Mar 9 '17 at 12:23
  • $\begingroup$ Okay, you were of course already aware of that! Btw, you write "I don't know an antiderivative for $e^{-at^2}$," but this is just the standard Gaussian integral (without limits it is the error function, with limits over $\mathbb{R}$ it is $\sqrt{\pi/a}$), but I'd expect you to already know this as well. $\endgroup$ Mar 9 '17 at 12:23
  • $\begingroup$ I suppose that the antiderivative can be expressed using the imaginary error function. $\endgroup$ Mar 9 '17 at 12:36

Use Euler's formula to replace $\cos \omega_0t$ by $\displaystyle \frac{e^{i\omega_0t}+e^{-i\omega_0t}}{2}$ to get

$$I = \frac 12\int_{-\infty}^\infty e^{-at^2}e^{-i(\omega-\omega_0)t} \, \mathrm dt + \frac 12\int_{-\infty}^\infty e^{-at^2}e^{-i(\omega+\omega_0)t}\, \mathrm dt = \frac{\mathcal F[e^{-at^2}](\omega-\omega_0) + \mathcal F[e^{-at^2}](\omega+\omega_0)}{2}$$ where $\mathcal F[g(t)](\omega) = G(\omega)$ denotes the Fourier transform of $g(t)$.


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