College Math Teaching

August 25, 2014

Fourier Transform of the “almost Gaussian” function with a residue integral

This is based on the lectures on the Fourier Transform by Brad Osgood from Stanford:

And here, F(f)(s) = \int^{\infty}_{-\infty} e^{-2 \pi i st} f(t) dt provided the integral converges.

The “almost Gaussian” integrand is f(t) = e^{-\pi t^2} ; one can check that \int^{\infty}_{-\infty} e^{-\pi t^2} dt = 1 . One way is to use the fact that \int^{\infty}_{-\infty} e^{-x^2} dx = \sqrt{\pi} and do the substitution x = \sqrt{\pi} t; of course one should be able to demonstrate the fact to begin with. (side note: a non-standard way involving symmetries and volumes of revolution discovered by Alberto Delgado can be found here)

So, during this lecture, Osgood shows that F(e^{-\pi t^2}) = e^{-\pi s^2} ; that is, this modified Gaussian function is “its own Fourier transform”.

I’ll sketch out what he did in the lecture at the end of this post. But just for fun (and to make a point) I’ll give a method that uses an elementary residue integral.

Both methods start by using the definition: F(s) = \int^{\infty}_{-\infty} e^{-2 \pi i ts} e^{-\pi t^2} dt

Method 1: combine the exponential functions in the integrand:

\int^{\infty}_{-\infty} e^{-\pi(t^2 +2  i ts}  dt . Now complete the square to get: \int^{\infty}_{-\infty} e^{-\pi(t^2 +2  i ts-s^2)-\pi s^2}  dt

Now factor out the factor involving s alone and write as a square: e^{-\pi s^2}\int^{\infty}_{-\infty} e^{-\pi(t+is)^2}  dt

Now, make the substitution x = t+is, dx = dt to obtain:

e^{-\pi s^2}\int^{\infty+is}_{-\infty+is} e^{-\pi x^2}  dx

Now we show that the above integral is really equal to e^{-\pi s^2}\int^{\infty}_{-\infty} e^{-\pi x^2}  dx = e^{\pi s^2} (1) = e^{-\pi s^2}

To show this, we perform \int_{\gamma} e^{z^2} dz along the retangular path \gamma : -x, x, x+is, -x+is and let x \rightarrow \infty

countour
Now the integral around the contour is 0 because e^{-z^2} is analytic.

We wish to calculate the negative of the integral along the top boundary of the contour. Integrating along the bottom gives 1.
As far as the sides: if we fix s we note that e^{-z^2} = e^{(s^2-x^2)+2si} and the magnitude goes to zero as x \rightarrow \infty So the integral along the vertical paths approaches zero, therefore the integrals along the top and bottom contours agree in the limit and the result follows.

Method 2: The method in the video
This uses “differentiation under the integral sign”, which we talk about here.

Stat with F(s) = \int^{\infty}_{-\infty} e^{-2 \pi i ts} e^{-\pi t^2} dt and note \frac{dF}{ds} = \int^{\infty}_{-\infty} (-2 \pi i t) e^{-2 \pi i ts} e^{-\pi t^2} dt

Now we do integration by parts: u = e^{-2 \pi i ts}, dv = (-2 \pi i t)e^{-\pi t^2} \rightarrow v = i e^{-\pi t^2}, du = (-2 \pi i s)e^{-2 \pi i ts} and the integral becomes:

(i e^{-\pi t^2} e^{-2 \pi i ts}|^{\infty}_{-\infty} - (i)(-2 \pi i s) \int^{\infty}_{-\infty} e^{-2 \pi i ts} e^{-\pi t^2} dt

Now the first term is zero for all values of s as t \rightarrow \infty . The second term is merely:

-(2 \pi s) \int^{\infty}_{-\infty} e^{-2 \pi i ts} e^{-\pi t^2} dt = -(2 \pi s) F(s) .

So we have shown that \frac{d F}{ds} = (-2 \pi s)F which is a differential equation in s which has solution F = F_0 e^{- \pi s^2} (a simple separation of variables calculation will verify this). Now to solve for the constant F_0 note that F(0) = \int^{\infty}_{-\infty} e^{0} e^{-\pi t^2} dt = 1 .

The result follows.

Now: which method was easier? The second required differential equations and differentiating under the integral sign; the first required an easy residue integral.

By the way: the video comes from an engineering class. Engineers need to know this stuff!

October 25, 2013

A Laplace Transform of a function of non-exponential order

Many differential equations textbooks (“First course” books) limit themselves to taking Laplace transforms of functions of exponential order. That is a reasonable thing to do. However I’ll present an example of a function NOT of exponential order that has a valid (if not very useful) Laplace transform.

Consider the following function: n \in \{1, 2, 3,...\}

g(t)= \begin{cases}      1,& \text{if } 0 \leq t \leq 1\\      10^n,              & \text{if } n \leq t \leq n+\frac{1}{100^n} \\  0,  & \text{otherwise}  \end{cases}

Now note the following: g is unbounded on [0, \infty) , lim_{t \rightarrow \infty} g(t) does not exist and
\int^{\infty}_0 g(t)dt = 1 + \frac{1}{10} + \frac{1}{100^2} + .... = \frac{1}{1 - \frac{1}{10}} = \frac{10}{9}

One can think of the graph of g as a series of disjoint “rectangles”, each of width \frac{1}{100^n} and height 10^n The rectangles get skinnier and taller as n goes to infinity and there is a LOT of zero height in between the rectangles.

notexponentialorder

Needless to say, the “boxes” would be taller and skinnier.

Note: this is an example can be easily modified to provide an example of a function which is l^2 (square integrable) which is unbounded on [0, \infty) . Hat tip to Ariel who caught the error.

It is easy to compute the Laplace transform of g :

G(s) = \int^{\infty}_0 g(t)e^{-st} dt . The transform exists if, say, s \geq 0 by routine comparison test as |e^{-st}| \leq 1 for that range of s and the calculation is easy:

G(s) = \int^{\infty}_0 g(t)e^{-st} dt = \frac{1}{s} (1-e^{-s}) + \frac{1}{s} \sum^{\infty}_{n=1} (\frac{10}{e^s})^n(1-e^{\frac{-s}{100^n}})

Note: if one wants to, one can see that the given series representation converges for s \geq 0 by using the ratio test and L’Hoptial’s rule.

June 5, 2012

Quantum Mechanics, Hermitian Operators and Square Integrable Functions

In one dimensional quantum mechanics, the state vectors are taken from the Hilbert space of complex valued “square integrable” functions, and the observables correspond to the so-called “Hermitian operators”. That is, if we let the state vectors be represented by \psi(x) = f(x) + ig(x) and we say \psi \cdot \phi = \int^{\infty}_{-\infty} \overline{\psi} \phi dx where the overline decoration denotes complex conjugation.

The state vectors are said to be “square integrable” which means, strictly speaking, that \int^{\infty}_{-\infty} \overline{\psi}\psi dx is finite.
However, there is another hidden assumption beyond the integral existing and being defined and finite. See if you can spot the assumption in the following remarks:

Suppose we wish to show that the operator \frac{d^2}{dx^2} is Hermitian. To do that we’d have to show that:
\int^{\infty}_{-\infty} \overline{\frac{d^2}{dx^2}\phi} \psi dx = \int^{\infty}_{-\infty} \overline{\phi}\frac{d^2}{dx^2}\psi dx . This doesn’t seem too hard to do at first, if we use integration by parts:
\int^{\infty}_{-\infty} \overline{\frac{d^2}{dx^2}\phi} \psi dx = [\overline{\frac{d}{dx}\phi} \psi]^{\infty}_{-\infty} - \int^{\infty}_{-\infty}\overline{\frac{d}{dx}\phi} \frac{d}{dx}\psi dx . Now because the functions are square integrable, the [\overline{\frac{d}{dx}\phi} \psi]^{\infty}_{-\infty} term is zero (the functions must go to zero as x tends to infinity) and so we have: \int^{\infty}_{-\infty} \overline{\frac{d^2}{dx^2}\phi} \psi dx = - \int^{\infty}_{-\infty}\overline{\frac{d}{dx}\phi} \frac{d}{dx}\psi dx . Now we use integration by parts again:
- \int^{\infty}_{-\infty}\overline{\frac{d}{dx}\phi} \frac{d}{dx}\psi dx = -[\overline{\phi} \frac{d}{dx}\psi]^{\infty}_{-\infty} + \int^{\infty}_{-\infty} \overline{\phi}\frac{d^2}{dx^2} \psi dx which is what we wanted to show.

Now did you catch the “hidden assumption”?

Here it is: it is possible for a function \psi to be square integrable but to be unbounded!

If you wish to work this out for yourself, here is a hint: imagine a rectangle with height 2^{k} and base of width \frac{1}{2^{3k}} . Let f be a function whose graph is a constant function of height 2^{k} for x \in [k - \frac{1}{2^{3k+1}}, k + \frac{1}{2^{3k+1}}] for all positive integers k and zero elsewhere. Then f^2 has height 2^{2k} over all of those intervals which means that the area enclosed by each rectangle (tall, but thin rectangles) is \frac{1}{2^k} . Hence \int^{\infty}_{-\infty} f^2 dx = \frac{1}{2} + \frac{1}{4} + ...\frac{1}{2^k} +.... = \frac{1}{1-\frac{1}{2}} - 1 = 1 . f is certainly square integrable but is unbounded!

It is easy to make f into a continuous function; merely smooth by a bump function whose graph stays in the tall, thin rectangles. Hence f can be made to be as smooth as desired.

So, mathematically speaking, to make these sorts of results work, we must make the assumption that lim_{x \rightarrow \infty} \psi(x) = 0 and add that to the “square integrable” assumption.

February 19, 2012

Divergent Improper Integrals: change of variables to an unbounded integrand.

Filed under: calculus, change of variable, improper integrals, integrals, integration by substitution, pedagogy — collegemathteaching @ 10:41 pm

This post was motivated by a student question: my student wanted help with the following problem:

\int^{\infty}_{1} \frac{x^2}{\sqrt{x^3 +1}} dx
Of course the idea is to do a substitution: u = x^3 + 1 which transforms the integral into \frac{1}{3} \int^{\infty}_{2} \frac{1}{\sqrt{u}} du which diverges. So far, so good. But then I told him one of my calculus tips: “it is often a good idea to try to guess the answer ahead of time” and then pointed out that for large values of x, \frac{x^2}{\sqrt{x^3 +1}} \approx \frac{x^2}{\sqrt{x^3}} = \sqrt{x} and of course \int^{\infty}_1 \sqrt{x} dx diverges because the integrand does not go to zero (in fact, is unbounded!) as x tends to infinity.

Then I realized that a change of variables had taken an unbounded function to a bounded one…though one which did not produce a convergent improper integral.

That lead to the natural question: if one has an integrand which is positive but monotonically decreasing to zero on [1, \infty ) , is there a change of variables which will change the integrand to either an unbounded function on [1, \infty ) or at least one that does not decrease to zero?

I admit that I have not answered this question yet, nor have I looked it up. But I can answer this question for a certain class of functions:

Theorem

Given \int^{\infty}_1 \frac{1}{x^r} dx
If 0 < r < 1 , let k > \frac{1}{1-r} . Then the change of variable u = x^{\frac{1}{k}} transforms \int^{\infty}_{1} \frac{1}{x^r} dx to k \int^{\infty}_1 u^{k(1-r) -1} du and of course u^{k(1-r) -1} is unbounded on [1, \infty) .

If 1 < r let k < \frac{1}{1-r} < 0 . Then \int^{\infty}_1 \frac{1}{x^r} dx is transformed into |k|\int^{1}_{0} u^{-1+k(1-r)} du which is an integral of a bounded function over a bounded region.

In short, one class of functions whose improper integral diverges can be transformed to functions that tend to infinity and the class of functions whose integrals converge can be transformed into functions which are bounded over a bounded interval.

Here is such an example: We show the equivalent integrals \int^{1.5}_{1} 3x^{\frac{1}{2}} dx and \int^{(1.5)^3}_1 \frac{1}{\sqrt{u}} du . The transformation is accomplished by using u =x^3 . Note how the transformation stretches the interval of integration to account for the function “shrinkage”.

On the other hand, using u^{-2}=x transforms \int^{\infty}_1 \frac{1}{x^2} dx into 2\int^{1}_{0} u du

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