What's coming in the second part of this course?

• Fourier transform. Given $f(x)$, we want to express $f(x) = \int e^{ikx} g(k) dk$, since $e^{ikx}$ is easy to deal with.
• Laplace transform. Given $f(t)$, on $t \in [0, \infty)$, we want to write $f(t) = \int_{c + i \R} e^{pt} F(p) dp$.
• Gamma function, Stirling Formula

When we solve equation of the form $$ \frac{df(x)}{dx} = \lambda f(x), $$ the general soluition is $$ f(x) = c e^{\lambda x}. $$ We say $e^{\lambda x}$ is an eigenfunction for the operator $d/dx$ with eigenvalue $\lambda$, here $\lambda$ can be any complex number.

• For $\lambda$ purely imaginary, $\lambda = i k$, the function $e^{i k x}$, which has constant size $|e^{ikx}| = 1$.
• For $\lambda$ purely real, $\lambda \in \R$, the function $e^{\lambda x}$, which is real, and has exponential growth (if $\lambda > 0$) or exponential decay.
• For $\lambda$ general complex number, say $\lambda = a + i b$ for $a,b\in\R$, then $e^{\lambda x} = e^{ax} e^{ibx} $ has both oscillation factor $e^{ibx}$ and exponential growth/decay $e^{ax}$.

If you give me a function $f(x)$, we can try to write it as a linear combination of these easy to understand pieces $e^{\lambda x}$. Say $$ f(x) = \int_{C} c(\lambda) e^{\lambda x} d\lambda $$ The benefit for doing this? If a differential operator $d/dx$ walk to it, we would have $$ (d/dx) f(x) = \int_{\lambda} c(\lambda) (d/dx) e^{\lambda x} d\lambda = \int_{\lambda} c(\lambda) \lambda e^{\lambda x} d\lambda. $$ Of course, you would immediately complain:

• what is this integration contour $C$? Is it along the real axis of $\lambda$? Is it along the imaginary axis of $\lambda$?
• Why we can switch the order of integration and differentiation?

Let's keep these questions in our head, we will return to those.

So, what is Fourier transformation? If you go on wiki, or open some textbook, you see the following definition: given a function $f(x)$ (with some condition on it), we can define $$ F(p) := \int_{x \in \R} f(x) e^{-ipx} dx. $$ This is called the Fourier transformation.

Let's throw in some input and see what we get.

$$ f(x) = \cosh(x) = (e^{x} + e^{-x} )/2. $$ This function grows to infinity when $x \to \pm \infty$, indeed, when $x \to +\infty$, $e^x$ dominate, and $x \to -\infty$ we have $e^{-x}$ dominate. Either way, you blow up. Good luck when integrating $\int_\R f(x) dx$.

But, let's try to do the Fourier transformation anyway. We get $$ F(p) = \int_x f(x) e^{-ipx} dx = (1/2) \int_\R e^{(1-ip)x} dx + (1/2) \int_\R e^{(-1-ip)x} dx. $$ OK, we cannot continue.

$f(x)$ need to have sufficient 'decay' near infinity for the Fourier transformation to make sense. More precisely, a sufficient condition is that $f(x)$ is 'absolutely integrable', which means $$ \int_\R |f(x)|dx < \infty. $$ Given this condition, we know $$ |F(p)| = | \int_\R e^{-ipx} f(x) dx | \leq \int_\R |e^{-ipx} f(x)| dx = \int_\R |f(x)| dx < \infty. $$ so we are safe.

The Gaussian. $f(x) = e^{-x^2}. $ It decays pretty nicely in both directions.

Since we learned our lesson, we need to do a check first $$ \int_\R e^{-x^2} dx = \sqrt{\pi}. $$ Great, it is finite.

Now, let's Fourier transform. $$ F(p) = \int e^{-ipx - x^2} dx = \int_\R e^{- (x-ip/2)^2 - p^2/4} dx = e^{-p^2/4} \int_{u \in -ip/2+\R} e^{-u^2} du $$ where we used the new variable $u = -ip/2+x$.

We can move the contour for $u$ as long as the integral is convergent. Recall from last Wednesday's lecture, $e^{-u^2}$ decays fast to $0$ when $|u| \to \infty$ in the sectors $\arg(u) \in (-\pi/4, \pi/4)$ and $\arg(u) \in \pi + (-\pi/4, \pi/4)$. So, we are going to move the integration contour of $u$ from the shifted real line $-ip/2+\R$ back to $\R$. We get $$ \int_{u \in -ip/2+\R} e^{-u^2} du = \int_{u \in \R} e^{-u^2} du = \sqrt{\pi}.$$ Thus, we get $$ F(p) = \sqrt{\pi} e^{-p^2/4}.$$

Success!

Let $f(x) = 1/(1+x^2)$. Let's check that $f(x)$ is absolutely integrable first $$ \int_\R f(x) dx = \int_{C_R} f(z) dz = 2\pi i Res_{z=i} f(z) = 2\pi i \frac{1}{2i } = \pi. $$ where $C_{+,R}$ is the contour consist of two part

• $[-R, R]$
• a semi-circle in the upper half plane of radius $R$.

Equivalently, we can choose a different contour $C_{-,R}$, which also consist of two part

• $[-R, R]$
• a semi-circle in the lower half plane of radius $R$.

Note that when we go around $C_{-,R}$ it goes around the singularity of $f(z)$ at $z=-i$ in the 'negative' (clockwise) direction, hence we get $$ \int_{C_{-,R}} f(z) dz = (-2\pi i) Res_{z=-i} f(z) = -2\pi i \frac{1}{-2i} = \pi. $$ Whew! the two minus sign cancels, we get the same answer.

Now, let's compute the little variation $$ F(p) = \int_\R e^{-ipx} f(x) dx. $$ We first take the 'truncation' approximation, which recovers the original limit under the limit $R \to \infty$. $$ F(p) = \lim_{R \to \infty} \int_{-R}^R e^{-ipx} f(x) dx. $$ Then, we try to 'close the contour', by connecting the point $R$ to $-R$ along some way. Due to this exponential factor $e^{-ipz}$, it matters whether we close the contour in the upper half-plane, or the lower one. Indeed, we have $$ |e^{-ipz}| = e^{Re(-ip(x+iy))} = e^{Re(-ipx + py)} = e^{py} = e^{p Im(z)}. $$ So,

• if $p \geq 0$, to avoid $|e^{-ipz}|$ blow up to infinity, we need to keep $Im(z)$ bounded from above on the contour.
• if $p \leq 0$, to avoid $|e^{-ipz}|$ blow up to infinity, we need to keep $Im(z)$ bounded from below on the contour.

Thus, for $p \geq 0$, we can complete the by the lower-half-semicircle $C_{-,R}$. We get $$ (p\geq 0) F(p) = \int_{C_{-,R}} \frac{e^{-ipz}}{1+z^2} dz = (-2\pi i) Res_{z=-i} \frac{e^{-ipz}}{1+z^2} = \pi e^{-p}. $$ Similarly, for $p \geq 0$, we get $$ (p\leq 0) F(p) = \int_{C_{+,R}} \frac{e^{-ipz}}{1+z^2} dz = (2\pi i) Res_{z=i} \frac{e^{-ipz}}{1+z^2} = \pi e^{p}. $$ We can write the result in a cool way as $$ F(p) = \pi e^{-|p|}. $$ So when $|p| \to \infty$, we see $F(p) \to 0$ very quickly.

warm up:Does the function $f(x)=1$ admits Fourier transformation? Why? How about $f(x) = 1 / (1+|x|)$?

Find the Fourier transformation of the following functions. $$ f(x) = \begin{cases} 1 & 0<x<1 \cr 0 & \text{else} \end{cases} $$

$$ f(x) = \begin{cases} 1+x & -1<x<0 \cr 1-x & 0<x<1 \cr 0 & \text{else} \end{cases} $$