Question 3 of 7: Laplace-Kernel Function — Area, Fourier Transform, and the $a\to0$ Limit

Question 3: Laplace-Kernel Function — Area, Fourier Transform, and the $a\to0$ Limit (5+9+3+3 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. $f(x)=\dfrac{1}{2a}e^{-|x|/a}$, $a\gt0$ — a two-sided (Laplace-kernel) exponential, symmetric about $x=0$ since $f(-x)=f(x)$.

Find. (a) $\int_{-\infty}^{\infty}f(x)\,dx$ and its graph for $a=2,\,0.5$. (b) $F(\omega)$. (c) the graph of $F(\omega)$ for $a=2,\,0.5$. (d) the limiting behaviour as $a\to0$.

Approach. Split every integral at $x=0$ into the two exponential pieces, integrate each half directly, then combine. The transform integral is done the same way, splitting also on the sign of $\omega x$ inside the complex exponential's real/imaginary parts.

1. (a) Area under $f(x)$. $$\int_{-\infty}^{\infty}f(x)\,dx=\int_{-\infty}^{0}\dfrac{1}{2a}e^{x/a}dx+\int_{0}^{\infty}\dfrac{1}{2a}e^{-x/a}dx=\dfrac{1}{2a}(a)+\dfrac{1}{2a}(a)=\dfrac12+\dfrac12$$ $$\boxed{\text{Area}=1\quad\text{(independent of }a\text{)}}$$ $f$ is therefore a valid probability density (the Laplace distribution) for every $a\gt0$.
2. (a) Graphs of $f(x)$ for $a=2$ and $a=0.5$. Peak height is $f(0)=1/(2a)$: $0.25$ for $a=2$, $1.0$ for $a=0.5$; each side decays as $e^{-|x|/a}$, so the larger $a$ gives the wider, shorter peak and the smaller $a$ the narrower, taller one — both enclose the same unit area.
   f(x) = (1/2a)·exp(−|x|/a): wide-short (a=2) vs. narrow-tall (a=0.5), both unit area
3. (b) Fourier transform. Split at $x=0$ and use $\operatorname{Re}(1/a\mp i\omega)\gt0$ so each half converges: $$F(\omega)=\dfrac{1}{\sqrt{2\pi}}\left[\int_{-\infty}^{0}\dfrac{1}{2a}e^{x(1/a-i\omega)}dx+\int_{0}^{\infty}\dfrac{1}{2a}e^{-x(1/a+i\omega)}dx\right]=\dfrac{1}{2a\sqrt{2\pi}}\left[\dfrac{1}{1/a-i\omega}+\dfrac{1}{1/a+i\omega}\right]$$ The bracket combines to $\dfrac{2/a}{1/a^2+\omega^2}$, so $$\boxed{F(\omega)=\dfrac{1}{\sqrt{2\pi}\,(1+a^2\omega^2)}}$$ (real and even in $\omega$, as expected since $f$ is real and even).
4. (c) Graphs of $F(\omega)$ for $a=2$ and $a=0.5$. Both curves peak at the same height $F(0)=1/\sqrt{2\pi}\approx0.3989$ (independent of $a$), but the half-width where $F$ drops to half its peak occurs at $\omega=1/a$: $\omega=0.5$ for $a=2$ (narrow spectrum) and $\omega=2$ for $a=0.5$ (wide spectrum). The wide time-domain pulse ($a=2$) therefore produces the narrow frequency-domain pulse and vice versa — a concrete instance of the time–bandwidth reciprocity that underlies the uncertainty principle.
   F(ω) = 1/[√(2π)·(1+a²ω²)]: narrow (a=2) vs. wide (a=0.5) spectrum
5. (d) Limit $a\to0$. In the time domain, the peak height $1/(2a)\to\infty$ while the width shrinks like $a$, with the area held fixed at 1 by part (a) — this is exactly the defining property of the Dirac delta, so $f(x)\to\delta(x)$. In the frequency domain, $F(\omega)=1/[\sqrt{2\pi}(1+a^2\omega^2)]\to1/\sqrt{2\pi}$ for every fixed $\omega$ as $a\to0$: the spectrum flattens to a constant. $$\boxed{a\to0:\quad f(x)\to\delta(x),\qquad F(\omega)\to\dfrac{1}{\sqrt{2\pi}}\ \text{(flat spectrum)}}$$ This is consistent with the known transform pair $\delta(x)\leftrightarrow1/\sqrt{2\pi}$: an infinitely localized pulse contains equal energy at every frequency.