I am reading a set of physics notes, and I have come across the following claim:
$f(\vec{k})=e^{-\frac{1}{2}\big(\frac{k}{\Lambda}\big)^2}\ \ \Rightarrow\ \ f(\vec{x})=\bigg(\dfrac{\Lambda}{2\pi}\bigg)^3e^{-\frac{1}{2}\Lambda^2x^2}$
The convention used to define the Fourier transform in these notes is:
$$f(\vec{k})=\int d^3xe^{i\vec{k}\cdot\vec{x}}f(\vec{x})\ \ \ \ \text{and}\ \ \ \ f(\vec{x})=\int\dfrac{d^3k}{(2\pi)^3}e^{-i\vec{k}\cdot\vec{x}}f(\vec{k})$$
I want to prove the original claim by calculating the inverse Fourier transform of the Gaussian function $f(\vec{k})$ given. I am slightly confused by the fact that the function depends only on $k=|\vec{k}|$, so I am unsure of whether I need to do the calculation in three dimensions. My suspicion is that I should do it in 3D and that is the reason why my result is wrong. My attempt (in 1D) is:
$$f(\vec{k})=e^{-\alpha k^2}\ \ \ \text{with}\ \ \ \alpha=\dfrac{1}{2\Lambda^2}$$
$$f(\vec{x})=\dfrac{1}{2\pi}\int dk\ e^{-ikx}e^{-\alpha k^2}=\dfrac{1}{2\pi}\int dk\ e^{-(\alpha k^2+ikx)}=\dfrac{1}{2\pi}\int dk\ e^{-\alpha\big(k+\frac{ix}{2\alpha}\big)^2-\frac{x^2}{4\alpha}}=$$ $$=\dfrac{1}{2\pi}\sqrt{\dfrac{\pi}{\alpha}}\ e^{-\frac{x^2}{4\alpha}}=\dfrac{\Lambda}{\sqrt{2\pi}}e^{-\frac{1}{2}\Lambda^2x^2}$$
where I used that
$$\int dk\ e^{-\alpha k^2}=\int dk\ e^{-\alpha(k^2+C)}=\sqrt{\dfrac{\pi}{\alpha}}$$
My result is almost the one I need to get, but not quite. Any advice on how to proceed? Is there a typo in the notes or am I wrong in my calculation? How could the inverse transform be calculated in 3D if it involves the product $\vec{k}\cdot\vec{x}$ in the integrand?
Indeed, there is a typographical error. We shall assume that the Fourier Transform. $\hat f$, of the function $f$ is given and then find $f$ from application of the Fourier Inversion Theorem.
Proceeding, we are given that $\hat f(\vec k)=e^{-|\vec k|^2/2\Lambda^2}$. Then, we have
$$\begin{align} \mathscr{F^{-1}}\{\hat f\}(x)&=\frac1{(2\pi)^3}\int_{\mathbb{R^3}} e^{-|\vec k|^2/2\Lambda^2}e^{-i\vec k\cdot\vec x}\,d^3k\\\\ &=\prod_{i=1}^3\frac1{2\pi}\int_{-\infty}^\infty e^{-k_i^2/2\Lambda^2}e^{-i\vec k_ix_i}\,dk_i\tag1 \end{align}$$
Let's evaluate the integral on the right-hand side of $(1)$. First, we enforce the substitution $k_i\mapsto k_i/\sqrt2\Lambda$ so that
$$\begin{align} \int_{-\infty}^\infty e^{-k_i^2/2\Lambda^2}e^{-i\vec k_ix_i}\,dk_i&=\sqrt2\Lambda\int_{-\infty}^\infty e^{-(k_i^2+i k_i\sqrt2\Lambda x_i)}\,dk_i\\\\ &=\sqrt2\Lambda e^{\Lambda^2 x_i^2/2}\int_{-\infty}^\infty e^{-(k_i+i \sqrt2\Lambda x_i/2)^2}\,dk_i\\\\ &=\sqrt{2\pi}\Lambda e^{\Lambda^2 x_i^2/2}\tag2 \end{align}$$
Substituting $(2)$ into $(1)$ yields
$$\begin{align} \mathscr{F^{-1}}\{\hat f\}(x)&=\frac1{(2\pi)^{3/2}}\Lambda^3 e^{-\Lambda^2 |\vec x|^2/2} \end{align}$$
We could have also converted the problem to spherical coordinates. Proceeding accordingly, we have
$$\begin{align} \mathscr{F^{-1}}\{\hat f\}(x)&=\frac1{(2\pi)^3}\int_{\mathbb{R^3}} e^{-|\vec k|^2/2\Lambda^2}e^{-i\vec k\cdot\vec x}\,d^3k\\\\ &=\frac1{(2\pi)^3}\int_0^{2\pi}\int_0^\pi \int_0^\infty e^{- k^2/2\Lambda^2}e^{ik |\vec x| \cos(\theta)}\,k^2\,\sin(\theta)\,dk\,d\theta\,d\phi\\\\ &=\frac1{(2\pi)^2} \int_0^\infty k^2 e^{- k^2/2\Lambda^2}\left(\int_0^\pi e^{ik |\vec x| \cos(\theta)}\,\sin(\theta)\,d\theta\right)\,dk\\\\ &\frac1{(2\pi)^2} \int_0^\infty k^2 e^{- k^2/2\Lambda^2} \left(\frac{2\sin(k|\vec x|)}{k|\vec x|}\right)\,dk\\\\ &=\frac1{2\pi ^2|\vec x|} \int_0^\infty k e^{- k^2/2\Lambda^2}\sin(k|\vec x|)\,dk\tag 3 \end{align}$$
We will evaluate the integral on the right-hand side of $(3)$ by solving an ordinary differential equation. We let $I(|\vec x|)$ denote the integral
$$I(|\vec x|)= \int_0^\infty e^{- k^2/2\Lambda^2}\cos(k|\vec x|)\,dk\tag 4$$
Differentiating both sides of $(4)$ reveals
$$I'(|\vec x|)=-\int_0^\infty k e^{- k^2/2\Lambda^2}\sin(k|\vec x|)\,dk\tag5$$
Next, integrating by part the integral in $(5)$ with $u= \sin(k|\vec x|)$ and $v=\Lambda^2 e^{- k^2/2\Lambda^2}$, we find that
$$\begin{align} I'(|\vec x|)&=-\Lambda^2 \int_0^\infty e^{-k^2/2\Lambda^2}\cos(k|\vec x|)\,dk\\\\ &=-\Lambda^2 |\vec x|I(|\vec x|)\tag6 \end{align}$$
Integrating $(6)$, it is easy to see that $I(|\vec x|)=C e^{-\Lambda^2 |\vec x|/2}$. Using $I(0)=\sqrt{\pi/2}\Lambda$, we find $C=\sqrt{\pi/2}\Lambda$ and therefore, $I(|\vec x|)=\sqrt{\pi/2}\Lambda e^{-\Lambda^2 |\vec x|^2/2}$.
Putting it all together, we have
$$\mathscr{F^{-1}}\{\hat f\}(x)=\frac1{(2\pi)^{3/2}}\Lambda^3 e^{-\Lambda^2 |\vec x|^2/2}$$
as expected!