Integral of exponential with complex argument

Question

I'm doing a problem about Fourier transforms and this integral came up.

$\int_{-\infty}^{\infty}e^{-x^2}e^{-ikx}dx$

I'm not really sure how to approach it.

Answer 1

Combining the exponentials give $I=\int_{-\infty}^{\infty}e^{-(x^2+ikx)}dx$

Completing the square of the exponent gives $(x+\frac{ik}{2})^2 -\frac{k^2}{4}$

This leaves the integral as $I=e^{\frac{-k^2}{4}}\int_{-\infty}^{\infty}e^{-(x+\frac{ik}{2})^2}dx$, which can be converted to a Gaussian integral using the substitution $u=x+\frac{ik}{2}$.

Therefore the solution is $I=\sqrt{\pi}e^{\frac{-k^2}{4}}$

Answer 2

One approach is as already hinted: Complete the square for $x^2+ikx$, then apply Cauchy's theorem.

Another fun argument that I saw years ago in Rudin Functional Analysis: Define $$G(k)=\int e^{-x^2}e^{-ikx}\,dx.$$Differentiate under the integral and then integrate by parts:$$\begin{align}G'(k)&=\int e^{-x^2}(-ix)e^{-ikx}\,dx \\&=\frac i2\int\left(\frac d{dx}e^{-x^2}\right)e^{-ikx}\,dx \\&=-\frac i2\int e^{x^2}\left(\frac d{dx}e^{-ikx}\right)\,dx \\&=-\frac k2G(k).\end{align}$$

Hence $$\frac d{dk}\left(e^{k^2/4}G(k)\right)=0,$$so $$G(k)=ce^{-k^2/4}.$$Find $c$ by figuring out what $G(0)$ is.

Answer 3

If you take the Mellin transform of the integrand with respect to $k$, you get \begin{equation} \mathcal{M}_{k \to s}[I](s) = \Gamma(s)\int_{\infty}^{\infty} e^{-x^2}(i x)^{-s} \; dx = \Gamma(s)\cos\left(\frac{\pi s}{2}\right)\Gamma\left(\frac{1}{2}-\frac{s}{2}\right) \end{equation} by Ramanujan's Master theorem the inverse transform is \begin{equation} I(k)=\sum_{s=0}^\infty \frac{\mathcal{M}_{k \to s}[I](-s)}{\Gamma(s)}\frac{k^s}{s!} =\sqrt{\pi}e^{-k^2/4} \end{equation}