# Help solving $\frac{1}{{2\pi}}\int_{-\infty}^{+\infty}{{e^{-{{\left({\frac{t}{2}} \right)}^2}}}{e^{-i\omega t}}dt}$

I need help with what seems like a pretty simple integral for a Fourier Transformation. I need to transform $\psi \left( {0,t} \right) = {\exp^{ - {{\left( {\frac{t}{2}} \right)}^2}}}$ into $\psi(0,\omega)$ by solving:

$$\frac{1}{{2\pi }}\int_{ - \infty }^{ + \infty } {\psi \left( {0,t} \right){e^{ - i\omega t}}dt}$$

So far I've written (using Euler's formula):

$$\psi \left( {0,\omega } \right) = \frac{1}{{2\pi }}\int_{ - \infty }^{ + \infty } {\psi \left( {0,t} \right){e^{ - i\omega t}}dt} = \frac{1}{{2\pi }}\left( {\int_{ - \infty }^{ + \infty } {{e^{ - {{\left( {\frac{t}{2}} \right)}^2}}}\cos \omega tdt - i\int_{ - \infty }^{ + \infty } {{e^{ - {{\left( {\frac{t}{2}} \right)}^2}}}\sin \omega tdt} } } \right)$$

$$\begin{array}{l} = \frac{1}{{2\pi }}\left( {{I_1} - i{I_2}} \right)\\ \end{array}$$

I just don't recall a way to solve this integrals by hand. Wolfram Alpha tells me that the result of the first integral is ${I_1} = 2\sqrt \pi {e^{ - {\omega ^2}}}$ and for the second $I_{2}=0$. But on my notes I have ${I_1} = 2\sqrt \pi {e^{ - {{\left( {{\omega ^2}/2} \right)}^2}}}$.

Can anybody tell me how one can solve this type of integrals and if the result from Wolfram Alpha is accurate? Any help will be appreciated.

The second one is trivially $I_2=0$ because $e^{-k\cdot t^2}$ is an even function for any $k\in\mathbb{R}^+$ whereas $\sin (\omega t)$ is an odd function, thus their product is odd and its integral on $[-\infty,\infty]$ is $0$, as it cancels out.

The first one calculated by Mathematica gives me also $2\sqrt{\pi}e^{-\omega^2}$, so maybe there is a typo in your notes. Notice that those types of integrals are expressible in terms of the error function, and are by no means elementary. You should use the Fourier transform operation directly, using its properties, instead of calculating explicitly the integrals, just as when applying Laplace transformations to solve differential equations.

There are some elementary properties of the Fourier transform for $f(x) = e^{-x^2/2}$ that will save you a lot of grunt work. The proofs are left to you, but essentially only require some basic manipulations and the fact that $f$ decays very rapidly. Let $\hat{ \cdot }$ denote the Fourier transform operator.

(1) $\hat{ [xf(x)]} = i \frac{d}{dx} (\hat{f}) (x)$

(2) $\hat{[ \frac{d}{dx} f(x)] } = i x \hat{f} (x)$

Note also that $f$ arises naturally as a solution to the differential equation $\frac{df}{dx} + xf =0$ so that $\hat{f}$ necessarily satisfies this equation as well by properties (1) and (2). Thus $\hat{f} = C f$ - the normalizing constant is recovered by evaluating the standard Gaussian integral. Now for the Fourier transform of $e^{-x^2 / 4}$ you simply need to make a change of variables.

If you complete the square in the argument of the exponentials, $$-\frac{1}{4}(t^2 + 4i \omega t) \to -\frac{1}{4}(t^2+4i\omega t -4\omega^2) -\omega^2 = -\frac{1}{4}(t+i2\omega)^2-\omega^2.$$ After a change of variables $u=\frac{t}{2}+i\omega$, the integral becomes $$2e^{-\omega^2}\int_{-\infty -i\omega}^{\infty -i\omega} e^{-u^2}\ du,$$ which, apart from the pesky factors of $-i\omega$ in the bounds, is a standard Gaussian integral equal to $\sqrt{\pi}$. As a physicist, I'm inclined to just sweep these factors under the rug, but we can do better: if we form a contour in the complex plane along the paths $(-\infty,i0)\to\ (\infty,i0)\to(\infty,-i\omega)\to(-\infty,-i\omega)\to(-\infty,i0),$ the integrand kills the contributions moving in the imaginary direction, and the overall integral is zero since $e^x$ has no poles. So, our integral is equal to the one along the real axis, and we can discard the $-i\omega$ terms.