$\renewcommand{\Re}{\operatorname{Re}}$ $\renewcommand{\Im}{\operatorname{Im}}$ $\newcommand{\erf}{\operatorname{erf}}$ $\newcommand{\dag}{\dagger}$ $\newcommand{\const}{\mathrm{const}}$ $\newcommand{\arcsinh}{\operatorname{arcsinh}}$
In the previous Section 5.1 we introduced Fourier transform and Inverse Fourier transform \begin{align} & \hat{f}( k )= \frac{\kappa}{2\pi}\int_{-\infty}^\infty f(x)e^{-i k x}\,dx \tag{FT}\\ & \check{F}(x)= \frac{1}{\kappa} \int_{-\infty}^\infty F( k ) e^{i k x}\,d k \tag{IFT} \end{align} with $\kappa=1$ (but here we will be a bit more flexible):
Theorem 1. $F= \hat{f} \iff f=\check{F}$. (Already "proved"--formally)
Theorem 2.
Proof. Easy. Preservation of inner product follows from preservation of norm.
Remark 1.
Theorem 3.
Proof. Here for brevity we do not write that all integrals are over $\mathbb{R}$ and set $\kappa=2\pi$.
$\displaystyle{\hat{g}=\int e^{-i k x}g(x)\,dx = \int e^{-i k x}f(x-a)\,dx= \int e^{-i k (x+a)}f(x)\,dx= e^{-i k a}\hat{f}( k )}$.
We replaced $x$ by $(x+a)$ in the integral.
$\displaystyle{\hat{g}=\int e^{-i k x}g(x)\,dx = \int e^{-i k x}e^{ibx}f(x)\,dx= \int e^{-i ( k -b)x}f(x)\,dx= \hat{f}( k -b)}$.
$\displaystyle{\hat{g}=\int e^{-i k x}g(x)\,dx = \int e^{-i k x}f'(x)\,dx \overset{\text{by parts}}= -\int \bigl(e^{-i k x}\bigr)'f(x)\,dx= i k \hat{f}( k )}$.
$\displaystyle{\hat{g}=\int e^{-i k x}g(x)\,dx = \int e^{-i k x}xf(x)\,dx= \int i\partial_ k \bigl(e^{-i k x}\bigr) \, f(x)\,dx= i\hat{f}{}'( k )}$.
$\displaystyle{\hat{g}=\int e^{-i k x}g(x)\,dx = \int e^{-i k x}f(\lambda x )\,dx = \int e^{-i k \lambda^{-1}x}f(x)\,|\lambda|^{-1}dx= |\lambda|^{-1}\hat{f}(\lambda^{-1} k )}$.
Here we replaced $x$ by $|\lambda|^{-1}x$ in the integral and $|\lambda|^{-1}$ is an absolute value of Jacobian.
Remark 2.
Corollary 4. $f$ is even (odd) iff $\hat{f}$ is even (odd).
Definition 1. Convolution of functions $f$ and $g$ is a function $f *g$: \begin{equation} (f*g)(x):=\int f(x-y)g(y)\,dy. \label{eq-5.2.3} \end{equation}
Theorem 4.
Proof. Again for brevity we do not write that all integrals are over $\mathbb{R}$.
Example 1.
Correspondingly, $f(x)=\left\{\begin{aligned} &e^{\alpha x} &&x <0,\\ &0 &&x >0 \end{aligned}\right.\implies \hat{f}(k)= \dfrac{\kappa}{2\pi (\alpha +i k )}$.
Then $f(x)=e^{-\alpha |x|}\implies \hat{f}(k)= \dfrac{\kappa\alpha}{\pi (\alpha^2 + k ^2 )}$.
Example 2.
Let $f(x)= \frac{1}{\alpha ^2+x^2}$ with $\Re (\alpha)>0$. Then
$\hat{f}(k)=\kappa e^{-\alpha |k|}$. Indeed,
using Example~1[3] we conclude that $f(x)$ is an inverse Fourier transform of $\kappa e^{-\alpha |k|}$
since we need only to take into account different factors and replace $i$ by $-i$ (for even/odd functions the latter step could be replaced by multiplication by $\pm 1$).
Using Complex Variables one can calculate it directly (residue should be calculated at $-\alpha i\operatorname{sign}(k)$).
Example 3. Let $f(x)=e^{-\frac{\alpha}{2}x^2}$ with $\Re(\alpha)\ge 0$. Here even for $\Re (\alpha)=0$ Fourier transform exists since integrals are converging albeit not absolutely.
Note that $f'=-\alpha x f$. Applying Fourier transform and Theorem 3 [3],[4] to the left and right expressions, we get $i k \hat{f}= -i\alpha \hat{f}'$; solving it we arrive to $\hat{f}=Ce^{-\frac{1}{2\alpha} k ^2}$.
To find $C$ note that $C=\hat{f}(0)= \frac{\kappa}{2\pi}\int e^{-\frac{\alpha}{2}x^2}\,dx$ and for real $\alpha >0$ we make a change of variables $x=\alpha^{-\frac{1}{2}}z$ and arrive to $C=\frac{\kappa}{\sqrt{2\pi \alpha}}$ because $\int e^{-z^2/2}\,dz=\sqrt{2\pi}$. Therefore \begin{equation*} \hat{f}( k )= \frac{\kappa}{\sqrt{2\pi\alpha}}e^{-\frac{1}{2\alpha} k ^2}. \end{equation*}
Remark 3. Knowing Complex Variables one can justify it for complex $\alpha $ with $\Re(\alpha)\ge 0$; we take a correct branch of $\sqrt{\alpha}$ (condition $\Re(\alpha)\ge 0$ prevents going around origin).
In particular, $(\pm i )^{\frac{1}{2}}=e^{\pm \frac{i\pi}{4}}=\frac{1}{\sqrt{2}}(1\pm i)$ and therefore for $\alpha=\pm i\beta $ with for $\beta >0$ we get $f=e^{\mp\frac{i\beta}{2 }x^2}$ and \begin{equation*} \hat{f}( k )=\frac{\kappa}{\sqrt{2\pi\beta}}e^{\pm \frac{\pi i}{4}} e^{\pm\frac{i}{2\beta} k ^2} =\frac{\kappa}{2\sqrt{\pi\beta}} (1\mp i)e^{\pm\frac{i}{2\beta} k ^2}. \end{equation*}
Theorem 5. Let $f(x)$ be a continuous function on the line $(-\infty,\infty)$ which vanishes for large $|x|$. Then for any $a>0$ \begin{equation} \sum_{n=-\infty}^\infty f(an) = \sum_{n=-\infty}^\infty \frac{2\pi }{a}\hat{f}(\frac{2\pi }{a}n) . \label{eq-5.2.4} \end{equation}
Proof. Observe that function \begin{equation*} g(x)= \sum_{n=-\infty}^\infty f(x+an) \end{equation*} is periodic with period $a$. Note that the Fourier coefficients of $g(x)$ on the interval $(-\frac{a}{2}, \frac{a}{2})$ are $b_m=\frac{2\pi}{a}\hat{f}(\frac{2\pi}{a})$, where $\hat{f}(k)$ is the Fourier transform of $f(x)$.
Finally, in the Fourier series of $g(x)$ on $(-\frac{a}{2}, \frac{a}{2})$ plug $x = 0$ to obtain $g(0)=\sum_m b_m$ which coincides with (\ref{eq-5.2.4}).
Remark 4. 1. Properties of Multidimensional Fourier transform and Fourier integral are discussed in Subsection 5.2.A. 2. It is very important to do all problems from Subsection 5.2.P: instead of calculating Fourier transforms directly you use Theorem 3 to expand the "library'' of Fourier transforms obtained in Examples 1--3.
$\Leftarrow$ $\Uparrow$ $\Downarrow$ $\downarrow$ $\Rightarrow$