I was trying to do a Fourier transform on a Gaussian, and Wolfram seems to give me a correct answer but with no step on the integral. I want to know how the integral is done by steps, specifically, an integral with second order complex exponentials:
The Fourier Transform: \begin{equation} F_t[A e^{-B(t-L / 2)^2}](x) \end{equation}
A, B, L are just parameters.
And the calculation by wolfram: \begin{equation} \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty}\left(A e^{-B(t-L / 2)^2}\right) e^{i x t} d t=\frac{A e^{-x^2 /(4 B)+(i L x) / 2}}{\sqrt{2} \sqrt{B}} \end{equation}
How was this done?
If $\text{Re}(B) > 0$, then:
$$ \begin{aligned} \mathcal{F}_t\left[A\,e^{-B(t-L/2)^2}\right](x) & := \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty} A\,e^{-B(t-L/2)^2}e^{\text{i}\,x\,t}\,\text{d}t \\ & = \frac{A}{\sqrt{2\pi B}}\int_{-\infty}^{+\infty} e^{-u^2}e^{\text{i}\,x\left(\frac{u}{\sqrt{B}}+\frac{L}{2}\right)}\,\text{d}u \\ & = \frac{A\,e^{\frac{\text{i}\,L\,x}{2}}}{\sqrt{2\pi B}}\int_{-\infty}^{+\infty} e^{-u^2}e^{\text{i}\frac{x}{\sqrt{B}}u}\,\text{d}u \\ & = \frac{A\,e^{\frac{\text{i}\,L\,x}{2}}}{\sqrt{2\pi B}}\left[\int_{-\infty}^{+\infty} e^{-u^2}\cos\left(\frac{x}{\sqrt{B}}u\right)\text{d}u + \text{i}\int_{-\infty}^{+\infty} e^{-u^2}\sin\left(\frac{x}{\sqrt{B}}u\right)\text{d}u\right] \\ & = \frac{A\,e^{\frac{\text{i}\,L\,x}{2}}}{\sqrt{2\pi B}}\left[\sqrt{\pi}\,e^{-\left(\frac{x}{2\sqrt{B}}\right)^2} + \text{i}\,0\right] \\ & = \frac{A\,e^{\frac{\text{i}\,L\,x}{2}-\frac{x^2}{4B}}}{\sqrt{2B}} \end{aligned} $$
as you can check via Wolfram|Alpha.
If the integral of the sine is trivially zero by symmetry, for that of the cosine:
$$ I(k) := \int_{-\infty}^{+\infty} e^{-u^2}\cos(k\,u)\,\text{d}u $$
differentiating under the integral sign, we have:
$$ I'(k) = \int_{-\infty}^{+\infty} -u\,e^{-u^2}\sin(k\,u)\,\text{d}u $$
so, integrating by parts, we have:
$$ I'(k) = \left[\frac{1}{2}\,e^{-u^2}\sin(k\,u)\right]_{-\infty}^{+\infty} - \frac{k}{2}\int_{-\infty}^{+\infty} e^{-u^2}\cos(k\,u)\,\text{d}u $$
i.e. with great amazement (at least the first time it should be like this):
$$ I'(k) = -\frac{k}{2}\,I(k) \quad \quad \overset{\text{ODE}}{\Rightarrow} \quad \quad I(k) = c_1\,e^{-(k/2)^2}. $$
Having to be $I(0)=\sqrt{\pi}$ (proof follows), we have $c_1=\sqrt{\pi}$, end.
Noting that:
$$ \iint\limits_{\mathbb{R}^2} e^{-x^2-y^2}\text{d}x\,\text{d}y = \lim_{R \to +\infty} \int_0^{2\pi} \text{d}\theta \int_0^R e^{-\rho^2}\rho\,\text{d}\rho = \pi $$
it follows that:
$$ \pi = \int_{-\infty}^{+\infty} e^{-x^2}\text{d}x \int_{-\infty}^{+\infty} e^{-y^2}\text{d}y \quad \quad \Rightarrow \quad \quad \int_{-\infty}^{+\infty} e^{-u^2}\text{d}u = \sqrt{\pi}\,. $$
All these tricks are essential to get around the fact that $e^{-u^2}$ doesn't admit a primitive expressible through elementary functions and they work only on domains that extend to infinity. Bye! ^_^