I'm trying to workout the Fourier transform of $f(x) = x^2 \exp{(-x^2)}$. We know that
\begin{equation}
\tilde f(x) = \int_{-\infty}^{\infty} f(x)\exp{(-ikx)} \ \mathrm{d}x.
\end{equation}
I have managed to simplify this by subbing in the value of $f(x)$ into the integral to:
\begin{equation}
\tilde f(x) = \exp{(-\frac{1}{4} ik^2)} \int_{-\infty}^{\infty} x^2 \exp{(-(x+\frac{1}{2}ik)^2)} \ \mathrm{d}x
\end{equation}
by using the completing the square method. I'm stuck after this point as I'm struggling to integrate the term above. I've tried the substitution method where I sub $s=x+\frac{1}{2}ik$ but am still getting nowhere. I am missing something obvious here?
[Math] Fourier transform of $x^2 \exp{(-x^2)}$
fourier analysis
Related Solutions
Your expression is correct. Further, set $x=-y$ in the first integral and observe that $$ \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{0}Ae^{\alpha x}e^{-ikx}dx = \frac{1}{\sqrt{2\pi}}\int_0^{\infty}Ae^{-\alpha y}e^{iky}dy= \left( \frac{1}{\sqrt{2\pi}}\int_{0}^{\infty}Ae^{-\alpha x}e^{-ikx}dx\right)^*, $$ where $(\cdot)^*$ denotes the complex conjugate.
I write the Fourier transform as
$$\hat{f}(k) = \int_{-\infty}^{\infty} dx \: \frac{\sin{x}}{x} e^{i k x} $$
Consider, rather, the integral
$$ \frac{1}{i 2} \int_{-\infty}^{\infty} dx \: \frac{e^{i x}-e^{-i x}}{x} e^{i k x} $$
$$ = \frac{1}{i 2} \int_{-\infty}^{\infty} dx \: \frac{e^{i (1+k) x}}{x} - \frac{1}{i 2} \int_{-\infty}^{\infty} dx \: \frac{e^{-i (1-k) x}}{x} $$
Consider the following integral corresponding to the first integral:
$$\oint_C dz \: \frac{e^{i (1+k) z}}{z} $$
where $C$ is the contour defined in the illustration below:
This integral is zero because there are no poles contained within the contour. Write the integral over the various pieces of the contour:
$$\int_{C_R} dz \: \frac{e^{i (1+k)z}}{z} + \int_{C_r} dz \: \frac{e^{i (1+k) z}}{z} + \int_{-R}^{-r} dx \: \frac{e^{i (1+k) x}}{x} + \int_{r}^{R} dx \: \frac{e^{i (1+k) x}}{x} $$
Consider the first part of this integral about $C_R$, the large semicircle of radius $R$:
$$\int_{C_R} dz \: \frac{e^{i (1+k)z}}{z} = i \int_0^{\pi} d \theta e^{i (1+k) R (\cos{\theta} + i \sin{\theta})} $$
$$ = i \int_0^{\pi} d \theta e^{i (1+k) R \cos{\theta}} e^{-(1+k) R \sin{\theta}} $$
By Jordan's lemma, this integral vanishes as $R \rightarrow \infty$ when $1+k > 0$. On the other hand,
$$ \int_{C_r} dz \: \frac{e^{i (1+k) z}}{z} = i \int_{\pi}^0 d \phi \: e^{i (1+k) r e^{i \phi}} $$
This integral takes the value $-i \pi$ as $r \rightarrow 0$. We may then say that
$$\begin{align} & \int_{-\infty}^{\infty} dx \: \frac{e^{i (1+k) x}}{x} = i \pi & 1+k > 0\\ \end{align}$$
When $1+k < 0$, Jordan's lemma does not apply, and we need to use another contour. A contour for which Jordan's lemma does apply is one flipped about the $\Re{z}=x$ axis. By using similar steps as above, it is straightforward to show that
$$\begin{align} & \int_{-\infty}^{\infty} dx \: \frac{e^{i (1+k) x}}{x} = -i \pi & 1+k < 0\\ \end{align}$$
Using a similar analysis as above, we find that
$$\int_{-\infty}^{\infty} dx \: \frac{e^{-i (1-k) x}}{x} = \begin{cases} -i \pi & 1-k < 0 \\ i \pi & 1-k >0 \\ \end{cases} $$
We may now say that
$$\hat{f}(k) = \int_{-\infty}^{\infty} dx \: \frac{\sin{x}}{x} e^{i k x} = \begin{cases} \pi & |k| < 1 \\ 0 & |k| > 1 \\ \end{cases} $$
To translate to your definition of the FT, divide the RHS by $\sqrt{2 \pi}$.
Best Answer
Observe that $$ \int x^2 e^{-x^2} e^{-ikx} = - \frac{\partial^2}{\partial k^2} \int e^{-x^2} e^{-ikx} $$ and complete the square in the latter integral.