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BARTON ZWIEBACH: Today's
subject is momentum space.

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We're going to kind of discover
the relevance of momentum

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space.

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We've been working
with wave functions

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that tell you the
probabilities for finding

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a particle in a given
position and that's sometimes

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called coordinate
space or position space

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representations of
quantum mechanics,

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and we just talked about
wave functions that

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tell you about probabilities
to find a particle in a given

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position.

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But as we've been
seeing with momentum,

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there's a very intimate relation
between momentum and position,

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and today we're going to
develop the ideas that

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lead you to think
about momentum space

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in a way that is quite
complimentary to coordinate

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space.

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Then we will be able to
talk about expectation

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values of operators
and we're going

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to be moved some steps into
what is called interpretation

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of quantum mechanics.

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So operators have expectations
values in quantum mechanism--

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they are defined
in a particular way

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and that will be
something we're going

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to be doing in the second
part of the lecture.

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In the final part
of the lecture,

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we will consider
the time dependence

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of those expectation
values, which is again,

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the idea of dynamics--

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if you want to understand how
your system evolves in time,

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the expectation value--

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the things that you measure--

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may change in time
and that's part

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of the physics of the
problem, so this will tie

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into the Schrodinger equation.

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So we're going to begin
with momentum space.

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So we'll call this
uncovering momentum space.

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And for much of what
I will be talking

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about in the first
half of the lecture,

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time will not be relevant--
so time will become relevant

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later.

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So I will be writing
wave functions that

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don't show the
time, but the time

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could be put there everywhere
and it would make no difference

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whatsoever.

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So you remember these
Fourier transform statements

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that we had that a wave
function of x we could put time,

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but they said let's
suppress time.

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It's a superposition
of plane waves.

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So there is a superposition
of plane waves

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and we used it last time to
evolve wave packets and things

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like that.

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And the other side of Fourier's
theorem is that phi of k

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can be written by a pretty
similar integral, in which you

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put psi of x, you change
the sign in the exponential,

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and of course,
integrate now over x.

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Now this is the wave
function you've always

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learned about first, and
then this wave function,

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as you can see, is encoded
by phi of k as well.

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If you know phi of
k, you know psi of x.

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And so this phi of k has the
same information in principle

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as psi of x--

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it tells you everything
that you need to know.

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We think of it as
saying, well, phi of k

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has the same
information as psi of x.

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And the other thing
we've said about phi of k

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is that it's the weight with
which you're superposing plane

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waves to reconstruct psi of x.

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The Fourier transform
theorem is a representation

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of the wave function in terms of
a superposition of plane waves,

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and here, it's the
coefficient of the wave that

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accompanies each exponential.

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So phi of k is the
weight of the plane

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waves in the superposition.

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So one thing we want to do is
to understand even deeper what

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phi of k can mean.

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And so in order to do that,
we need a technical tool.

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Based on these
equations, one can

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derive a way of representing
this object that we

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call the delta function.

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Delta functions are pretty
useful for manipulating objects

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and Fourier transforms,
so we need them.

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So let's try to obtain what
is called the delta function

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statement.

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And this is done by
trying to apply these two

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equations simultaneously.

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Like, you start
with psi of x, it's

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written in terms of
phi of k, but then what

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would happen if you
would substitute

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the value of phi of k in here?

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What kind of equation you get?

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What you get is an equation
for a delta function.

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So you begin with psi of x
over square root of 2 pi,

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and I'll write it
dk e to the ikx,

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and now, I want
to write phi of k.

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So I put phi of k--

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1 over square root
of 2 pi integral psi,

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and now I have to
be a little careful.

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In here in the second integral,
x is a variable of integration,

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it's a dummy variable.

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It doesn't have any
physical meaning, per se--

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it disappears after
the integration.

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Here, x represents a point
where we're evaluating the wave

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function, so I just cannot
copy the same formula here

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because it would be confusing,
it should be written with

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a different x and x-prime.

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Because that x certainly
has nothing to do with the x

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we're writing here.

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So here it is, we've
written now this integral.

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And let's rewrite it
still differently.

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We'll write as
integral dx-prime.

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We'll change the orders of
integration with impunity.

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If you're trying to be very
rigorous mathematically,

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this is something
you worry about.

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In physics problems
that we deal with,

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it doesn't make a difference.

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So here we have dx-prime,
psi of x-prime, then

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a 1 over 2 pi integral dk e
to the ik x minus x-prime.

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So this is what the
integral became.

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And you look at it and you
say, well, here is psi of x,

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and it's equal to the
integral over x-prime

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of psi of x-prime times
some other function

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of x and x-prime.

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This is a function of x minus
x-prime, or x and x-prime

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if you wish.

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It doesn't depend on
k, k is integrated.

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And this function,
if you recognize it,

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it's what we call
a delta function.

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It's a function that,
multiplied with an integral,

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evaluates the integrand
at a particular point.

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So this is a delta function,
delta of x-prime minus x.

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That is a way these
integrals then would work.

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That is, when you
integrate over x-prime--

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when you have x-prime
minus something,

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then the whole integral
is the integrand

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evaluated at the point where we
say the delta function fires.

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So the consistency of
these two equations

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means that for all
intents and purposes,

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this strange integral
is a representation

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of a delta function.

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So we will write it down.

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I can do one thing here, but--

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it's kind of here you see
the psi of x-prime minus x,

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but here you see
x minus x-prime.

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But that's sign, in
fact, doesn't matter.

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It's not there.

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You can get rid of it.

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Because if in this integral,
you can let K goes to minus k,

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and then the dk changes sign,
the order of integration

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changes sign, and they
cancel each other.

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And the effect is
that you change

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the sign in the exponents, so
if you let k goes to minus k,

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the integral just becomes
1 over 2 pi integral dk e

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to the ik x-prime minus x.

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So we'll say that
this delta function

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is equal to this thing,
exploiting this sign ambiguity

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that you can always have.

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So I'll write it
again in the way

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most people write the formula,
which is at this moment,

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switching x and x-prime.

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So you will write this
delta of x minus x-prime

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is 1 over 2 pi integral from
infinity to minus infinity dk

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e to the ik x minus x-prime.

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So this is a pretty
useful formula

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and we need it all the time
that we do Fourier transforms

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as you will see very soon.

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It's a strange integral, though.

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If you have x equal to x-prime,
this is 0 and you get infinity.

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So it's a function,
it's sort of 0--

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when x minus x-prime
is different from 0,

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somehow all these
waves superimpose to 0.

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But when x is equal to
x-prime, it blows up.

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So it's does the right thing,
it morally does the right thing,

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but it's a singular
kind of expression

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and we therefore
manipulate it with care

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and typically, we use
it inside of integrals.

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So it's a very nice formula,
we're going to need it,

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and it brings here for the
first time in our course,

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I guess, the delta functions.

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And this is something if
should-- if you have not

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ever play without
the functions, this

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may be something interesting
to ask in recitation

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or you can try to prove, for
example, just like we show

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that delta of minus x
is the same as delta

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of x, the delta of
a being a number

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times x is 1 over absolute
value of a times delta of x.

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Those are two simple
properties of delta functions

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and you could practice by
just trying to prove them--

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for example, use this integral
representation to show them.