5C Midterm

8/2/2019 5C Midterm

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Math 5C, Summer 2011

Take-Home Midterm

Due August 23, 2011 in Lecture

Name 1

Perm No. 1

Name 2

Perm No. 2

Style

Substance

Total

Directions:

1. Attach this page as a coversheet to your paper.

2. You may work in groups of 2 and use course materials—including the TA and myself, but no other resourcesare allowed. Especially the internet!

3. There are 30 points on this exam.

4. Substance: Each part of each problem is worth 1 point for correctness—20 points possible in total.

5. Style: There are 10 points for overall presentation; see below.

6. First and foremost your writing should be in fully developed sentences; a mathematical formula should betreated as part of a sentence.

Bad:

calculus fact:

f g = f g −

gf → f = 0

Good: From the product rule (f g) = f g + gf we can deduce that f g = f g −

gf,

which is known as ‘integration by parts.’ Our remarks from before imply that f is identically 0.

7. It is encouraged that you craft your midterm into a coherent piece of prose, though you may simply answerparts (a), (b), etc. if you wish.

8. Finally, I’m unashamedly a sucker for mathematical aesthetic. Your final product should be on crispbeautiful paper with nicely drawn symbols and handwriting (or computer typesetting if you’re super-cool).Remember the old saying: If momma [the grader] ain’t happy, ain’t nobody happy.

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Problem 1

In this problem we’re going to understand ∆ = · = ∂ 2x + ∂ 2y a bit better. Let D be the unit disk in R2

centered at the origin.

1. Use derivative rules to compute ∆(ln(x2 + y2)).

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

∆(ln(x2 + y2)) = · ((ln(x2 + y2)))

=

∂

∂xi +

∂

∂yˆ j

·

2x

x2 + y2i +

2y

x2 + y2ˆ j

=

∂

∂x

2x

x2 + y2

+

∂

∂y

2y

x2 + y2

= 2(x2 + y2)−1 − 4x2(x2 + y2)−2 + 2(x2 + y2)−1 − 4y2(x2 + y2)−2 = 4(x2 + y2)−1 − 4(x2 + y2)−1 = 0

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2. Assuming that ∆f = 0, show that ∂D

∂f

∂nds = 0.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

∆f = 0 ⇒ · (f ) = 0 ⇒ ∂ 2f ∂x2 + ∂ 2f

∂y2 = 0

Using Green’s Theorem: R · F dA =

∂R F · nds

Let f = F , and substitute into Green’s Theorem. D

· (f )dA =

∂D

f · nds

Using the fact: · (f ) = δf and f · n = ∂f ∂n

D

δfdA =

∂D

∂f

∂nds

0 =

∂D

∂f

∂nds

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

3. Ignoring the previous two parts, directly compute ∂D

∂

∂nln(x2 + y2) ds.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

∂D

∂

∂nln(x2 + y2) ds =

∂D

ln(x2 + y2) · nds

For a unit disk, n = r = (x, y) and x2 + y2 = 1 on the boundary.

∂D ln(x

2

+ y

2

) · nds = ∂D 2x

x2 + y2 ,

2y

x2 + y2· (x, y)ds

=

∂D

(2x2 + 2y2)ds = 2

∂D

ds = 2(circumference) = 4π

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Something should look wrong; the last three questions seem to contradict each other. The problem is thatln(x2 + y2) has a singularity at the origin, so our goal is to understand what’s going on. For this we’regoing to need a smooth ‘test function’ φ. Assume that φ is zero on and near ∂D; in particular, φ and allof its derivatives are zero on the boundary of D.

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4. Evaluate ∂D

φ

∂

∂nln(x2 + y2)− ln(x2 + y2)

∂

∂nφ

ds.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Since we are on boundary of D, we know φ and all derivatives of φ = 0, thus both the first and second termgo to zero.

∂D0

∂

∂n

ln(x2 + y2)− ln(x2 + y2)0 ds

∂D

(0− 0) ds = 0

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

5. Assuming that we can make sense of ∆ ln(x2 + y2), show that it must satisfy D

φ∆ln(x2 + y2) dA =

D

ln(x2 + y2)∆φdA.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Working with only the left side of the equation, we add a form of zero: (−

D ln(x2 + y2)∆φ dA +

D ln(x2 +

y2)∆φ dA). So on the left, we have: D

φ∆ln(x2 + y2) dA−

D

ln(x2 + y2)∆φ dA +

D

ln(x2 + y2)∆φ dA

=

D

(φ∆ln(x2 + y2)− ln(x2 + y2)∆φ) dA +

D

ln(x2 + y2)∆φ dA

From Homework 1, we know the identity: · (f g − gf ) = f ∆g − g∆f . We can then say:

=

D

· (φ ln(x2 + y2)− ln(x2 + y2)φ) dA +

D

ln(x2 + y2)∆φ dA

Using Green’s Theorem:

= ∂D

(φ ln(x2 + y2)− ln(x2 + y2)φ) · nds + D

ln(x2 + y2)∆φ dA

On the boundary of D, we know φ and all derivatives of φ = 0. Thus:

=

∂D

(0 ln(x2 + y2)− ln(x2 + y2)0) · nds +

D

ln(x2 + y2)∆φ dA =

∂D

(0) · nds +

D

ln(x2 + y2)∆φ dA

⇒

D

φ∆ln(x2 + y2) dA =

D

ln(x2 + y2)∆φ dA

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

In the next few parts we’ll try to evaluate the integral on the right.

6. Given a small positive number a let B(a) be the annulus {(x, y) : a2 ≤ x2 + y2 ≤ 1} and C (a) be the circle

{(x, y) : x2 + y2 = a2}. Given a vector field F, rewrite B(a)

· F dA

in terms of line integrals on C (a) and ∂D. Be very careful when considering boundaries and orientation!

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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As seen in the picture, B(a) is the annulus with outer radius 1. We can think of the annulus as a unit diskD with a smaller disc A(a), who’s boundary is C(a), taken out of it. Thus B(a) = D - A(a). Thus, if wethink about the boundary of B(a), we can say: ∂ B(a) = ∂ D - ∂ A(a) = ∂ D - C(a).

The subtraction of C(a) comes from the orientation of the inner circle and the normal pointing in theopposite direction of the outer boundary.

Using this information, we can write: B(a)

· F dA =

∂B(a)

F · nds =

∂D

F · nds−

C (a)

F · nds

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

7. Show that B(a)

ln(x2 + y2)∆φ dA = C (a)

φ ∂

∂nln(x2 + y2)− ln(x2 + y2) ∂

∂nφ

ds.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

From Homework 1, we know the identity: ·(f g−gf ) = f ∆g−g∆f . If we set g = φ and f = ln(x2+y2),and take the integral over B(a) on both sides we get:

B(a)

[ln(x2 + y2)∆φ− φ∆ln(x2 + y2)]dA =

B(a)

· [ln(x2 + y2)φ− φ ln(x2 + y2)]dA

On the right side, we use the identity derived in problem 6. On the left side, we know ∆ ln(x2 + y2) = 0.Thus we get:

B(a)

[ln(x2+y2)∆φ−φ(0)]dA = ∂D

[ln(x2+y2)φ−φ ln(x2+y2)]·nds− C (a)

[ln(x2+y2)φ−φ ln(x2+y2)]·nds

Over the boundary of D, φ and all derivatives of φ = 0, thus the entire integral over ∂D goes to zero. So,we are left with:

B(a)

ln(x2 + y2)∆φdA =

C (a)

[− ln(x2 + y2)φ + φ ln(x2 + y2)] · nds

The outward normal derivative of f, ∂f ∂n = f · n. We can use this to change the form of our last equation:

B(a)

ln(x2 + y2)∆φdA =

C (a)

[φ∂

∂nln(x2 + y2)− ln(x2 + y2)

∂

∂nφ]ds

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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8. As a → 0 the continuity of φ means that φ ≈ φ(0, 0) (a constant) all along C (a). On the other hand ∂φ/∂nchanges values wildly along C (a), but for some constant M we have |∂φ/∂n| < M . Use these facts toevaluate

lima→0

C (a)

φ

∂

∂nln(x2 + y2)− ln(x2 + y2)

∂

∂nφ

ds.

To make the computations easier, notice that for a circle centered at the origin ∂/∂n = ∂/∂r, the derivativewith respect to the polar coordinate r.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

lima→0

C (a)

φ

∂

∂nln(x2 + y2)− ln(x2 + y2)

∂

∂nφ

ds

= lima→0

C (a)

φ

∂

∂nln(x2 + y2)

ds−

C (a)

ln(x2 + y2)

∂

∂nφ

ds

We take φ ≈ φ(0, 0) which we will notate as φ0, which is constant.

= lima→0

φ0

C (a)

∂

∂nln(x2 + y2)

ds−

C (a)

ln(x2 + y2)

∂

∂nφ

ds

Converting to spherical coordinates:

= lima→0

φ0

C (a)

∂

∂rln(r2)

rdθ −

C (a)

ln(r2)

∂

∂rφ

rdθ

= lima→0

φ0

C (a)

2

r

rdθ −

C (a)

2r ln(r)

∂

∂nφ

dθ

Since we defined r = a, lima→0 2r ln(r) = 0; and since ∂ ∂nφ is bounded by a constant M, we can say that

lima→0 2r ln(r) ∂ ∂nφ = 0.

=

2φ0

C (a)

dθ −

C (a)

(0) dθ

= 2φ0 θ|2π0 = 4πφ0

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

9. Evaluate the following integrals using the previous part. By definition, D

φ∆ln(x2 + y2) dA =

D

ln(x2 + y2)∆φ dA = lima→0

B(a)

ln(x2 + y2)∆φdA.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

From problem 8, we know:

lima→0

C (a)

φ

∂

∂nln(x2 + y2)− ln(x2 + y2)

∂

∂nφ

ds = 4πφ0

From problem 7, we have the equality that: B(a)

ln(x2 + y2)∆φ dA =

C (a)

φ

∂

∂nln(x2 + y2)− ln(x2 + y2)

∂

∂nφ

ds

Thus substituting into the equation in problem 8, we get:

lima→0

B(a)

ln(x2 + y2)∆φ dA = 4πφ0

Thus, by definition: D

φ∆ln(x2 + y2) dA =

D

ln(x2 + y2)∆φ dA = lima→0

B(a)

ln(x2 + y2)∆φ dA = 4πφ0

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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10. The Dirac delta function δ is not really a function; nonetheless we pretend like it is a function and defineit by the property that for any function f ,

D

fδdA = f (0, 0).

Write ∆ ln(x2 + y2) in terms of δ.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

From problem 9, we see that: D

φ∆ln(x2 + y2) dA = 4πφ0

This shows that ∆ ln(x2 + y2) acts similarly to the Dirac delta function: when taking the integral over Dof a function multiplied by ∆ ln(x2 + y2), we get that function evaluated at (0,0) multiplied by 4 π. We canthen relate ∆ ln(x2 + y2) and δ as:

∆ln(x2 + y2) = 4πδ

Checking that this makes sense, we reevaluate the previous integral using the properties of the Dirac deltafunction:

D

φ∆ln(x2 + y2) dA =

D

φ4πδdA = 4π

D

φδdA = 4πφ0

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

11. In some appropriate sense the test functions φ can approximate nearly any function you like. Therefore wecan evaluate D

φ∆ln(x2 + y2) dA

for arbitrary φ, regardless of what it does on ∂D. In a single step evaluate D

e2x+3y∆ln(x2 + y2) dA.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

We take e2x+3y as our function. Knowing how ∆ ln(x2 + y2) acts like δ, we can easily find a solution in asingle step:

D

(e2x+3y)∆ln(x2 + y2) dA = 4π

D

(e2x+3y)δ dA = 4πe2x+3y(0,0)

= 4π

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

12. In R3 it turns out that ∆(1/

x2 + y2 + z2) = Cδ for some constant C . Assuming this, integrate againsta very simple function to find C .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Using the property of the Dirac delta function described in problem 10, we know: B

∆(1/

x2 + y2 + z2)dV =

B

CδdV = C (0, 0) = C

Since C is a constant function, finding C at any point will tell us what C is at all points. Thus we solve theintegral to find C. We use divergence theorem to change the integral from B to the boundary.

B

∆(1/ x2 + y2 + z2)dV = B

· ((1/ x2 + y2 + z2))dV = ∂B

(1/ x2 + y2 + z2) · d A

=

∂B

−x

(x2 + y2 + z2)3/2,

−y

(x2 + y2 + z2)3/2,

−z

(x2 + y2 + z2)3/2

· (x, y, z)dσ

=

∂B

−x2 + y2 + z2

(x2 + y2 + z2)3/2dσ

Converting to polar coordinates, and knowing on the boundary of B, r = 1, we solve the integral.

=

∂B

−1

rdσ = −

∂B

dσ

= −4π = C

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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What we’ve done in this problem is shown that (1/2π) ln

x2 + y2 is the electrostatic potential of a point chargein R2. In general if the density of charge in space is ρ, then the electrostatic potential V solves ∆V = ρ. TheDirac delta is simply the charge density of a point charge. In R

3 the potential of a point charge is proportionalto 1/

x2 + y2 + z2; if we write that instead as 1/r we get the (probably familiar) formula

V =kQ

r,

where k and Q are just a constant and unit of charge respectively. This is the beginning of electrostatics as wellas the mathematical subject of potential theory.

Problem 2

In this problem we’re going to use series to solve a problem in statistical physics and thermodynamics. Anyobject whose temperature is above absolute zero will emit thermal radiation—that is, electromagnetic radiationgenerated by the internal thermal energy of the object. This phenomenon is probably familiar; any object whichis raised to a high enough temperature will glow ‘red-hot’ (in fact, other colors are possible too). Around theturn of the 20th century, understanding the underlying physics of thermal radiation from known principles wasa big problem. When Lord Rayleigh attempted to derive the amount of electromagnetic radiation a ‘blackbody’would emit at a certain frequency, a so-called ultraviolet catastrophe occurred: the rate at which the body emits

energy would beP ∝

∞

0

dλ

λ4= ∞

(a nice instance of a divergent integral!). By looking at experimental data, Max Planck was able to deduce amore accurate law. Though he amazingly guessed a complicated formula from the data, he later derived it byassuming energy came in discrete increments—this was the first time energy quantization and Planck’s constanth appeared in physics. Accepting something as crazy as energy quantization requires some convincing. We’d liketo compute the same rate of energy flow that caused Rayleigh’s law to fall apart. In doing so, we’ll derive theso-called Stefan-Boltzmann law of blackbody radiation.

1. Let’s cut to the chase. After a derivation starting from Planck’s law, the rate of energy flow (or ‘power’)via thermal radiation for a blackbody is given by the formula

P = 2πhAc2

∞

0

ν 3

exp(hν/kT )− 1dν,

where A is the surface area of the object, c is the speed of light, k is Boltzmann’s constant, and T isthe temperature of the object (recall the notation exp t = et). Let’s remove all physical content from theintegral by substituting x = hν/kT . From here deduce the Stefan-Boltzmann law: how does P depend onT ? Your expression for P will still contain the dimensionless integral

∞

0

x3

ex − 1dx.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

We multiply by a form of the number 1: k3T 3

h2h2

k3T 3

P =k3T 3

h2h2

k3T 32πhA

c2

∞

0

ν 3

exp(hν/kT )− 1dν

=k3T 3

h2

2πA

c2

∞

0

x3

ex − 1dν

Since x = hνkT , dx = h

kT dν ⇒ dν = kT h dx.

P =k4T 4

h32πA

c2

∞

0

x3

ex − 1dx

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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At this point we could say that we’re done. However, a great test of Planck’s theory comes from evaluatingthe integral and comparing numbers against experimental results. We’ll stop the story of the physicshere and focus on evaluating the integral. Finding an antiderivative is out of the question and a numerictreatment would be arduous, but we can use series to evaluate the integral explicitly!

2. For x > 0 and s > 1 show thatxs−1

ex − 1 =

∞n=1

xs−1

e−nx

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

The original form of the equation looks familiar to the standard geometric series:∞

k=0 rk = 11−r , so we

will adjust it.xs−1

ex − 1=

e−1

e−1xs−1

ex − 1=

e−xxs−1

1− e−x= e−xxs−1

1

1− e−x

We look at the geometric series for 11−e−x =

∞

n=0 e−nx. Thus:

e−xxs−11

1− e−x= e−xxs−1

∞n=0

e−nx

Since there is no dependency on n, we can bring all the terms into the summation.

=

∞n=0

e−xxs−1e−nx =

∞n=0

xs−1e−(n+1)x

We can then change the summation to start at 1, and change n+1 to n.

∞n=0

xs−1e−(n+1)x =∞n=1

xs−1e−(n)x

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

3. A central result in modern integration theory is the Lebesgue Monotone Convergence Theorem (or MCTfor short).

Theorem (MCT). Let g1(x), g2(x), g3(x), . . . be a sequence of “measurable” functions—this is a technical

term which includes most functions, such as continuous ones. Fix an interval [a, b], which can be either

bounded or infinite. Assume that the sequence is monotone; that is, for each x ∈ [a, b],

g1(x) ≤ g2(x) ≤ g3(x) ≤ · · ·

Then it follows that

limn→∞

ba

gn(x) dx =

ba

limn→∞

gn(x) dx.

Assume this theorem is true and prove this important corollary.

Corollary (MCT for Series). Let f 1(x), f 2(x), f 3(x) . . . be continuous functions defined on an interval [a, b].

Assume that for any x ∈ [a, b], each f n(x) ≥ 0. Then it follows that

∞n=1

ba

f n(x) dx =

ba

∞n=1

f n(x) dx.

Hint: How can an infinite series be thought of as a sequence?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Assuming the monotone convergence theorem, we can say that:

gn(x) = f 1(x) + f 2(x) + ... + f m(x) =mi=1

f i(x)

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Where all f i(x) are continuous and defined on the interval [a,b].We will assume that all f i(x) ≥ 0 for any x ∈ [a, b].Since gn(x) is monotonically increasing:

limn→∞

ba

ni=1

f i(x) dx =

ba

limn→∞

ni=1

f i(x) dx

limn→∞ ba

[f 1(x) + f 2(x) + ... + f n(x)] dx = ba

∞i=1

f i(x) dx

limn→∞

ba

f 1(x) +

ba

f 2(x) + ... +

ba

f n(x)

dx =

ba

∞i=1

f i(x) dx

limn→∞

ni=1

ba

f i(x) dx =

ba

∞i=1

f i(x) dx

∞i=1

ba

f i(x) dx =

ba

∞i=1

f i(x) dx

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

4. There are many examples of sequences gn(x) for which limit and integral cannot be interchanged. Define

gn(x) =

n if 0 < x < 1/n

0 otherwise

Show that these functions are not monotone on [0, 1] and that

limn→∞

ba

gn(x) dx =

ba

limn→∞

gn(x) dx.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

To show that gn(x) is not monotone, we look at gn(x) at values n =1, 2 and 3:

g1(x) =

1 if 0 < x < 10 otherwise

g2(x) =

2 if 0 < x < 1/20 otherwise

g3(x) =

3 if 0 < x < 1/30 otherwise

At x = 2/5, g1(x) = 1, g2(x) = 2, while at g3(x) = 0, since gn(x) is both increasing and decreasing,it is clearly not monotone.

Also we can show:

limn→∞

ba

gn(x) dx =

ba

limn→∞

gn(x) dx

by evaluating both sides. We take the integral over all real numbers.

limn→∞

∞

−∞

gn(x) dx = limn→∞

1 = 1

Also ∞

−∞

limn→∞

gn(x) dx =

∞

−∞

0 = 0

1 = 0

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

5. Define the gamma function for s > 0 as

Γ(s) =

∞

0

ts−1e−t dt.

Use integration by parts to show that Γ(s + 1) = sΓ(s).

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. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Γ(s + 1) =

∞

0

tse−t

From the product rule (f g) = f g + gf we can deduce that

b

a

f g = f gb

a

− b

a

gf

which is known as ‘integration by parts. We set g = ts and f = e−t which implies g = sts−1 and f = −e−t.As a result:

∞

0

tse−t = (−e−tts)

∞

0

+

∞

0

sts−1e−t

−e−tts = 0 @ 0 and ∞, which leaves us with: ∞

0

tse−t =

∞

0

sts−1e−t

∞

0

tse−t = s

∞

0

ts−1e−t

Γ(s + 1) = sΓ(s)

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

6. Evaluate Γ(1) and explain why Γ(n) = (n− 1)! follows for positive integers n.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Γ(1) =

∞

0

t1−1e−t dt =

∞

0

e−t dt = −e−t∞

0

= 1

To show Γ(n) = (n− 1)!, we prove by induction.We assume Γ(n) = (n− 1)!. From problem 5, we know: Γ(n + 1) = nΓ(n) = n(n− 1)! = n! = ((n + 1)− 1)!

Thus for all positive integers n, Γ(n) = (n− 1)!.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Let’s think of the p series as a function of p; define the Riemann zeta function

ζ ( p) =

∞n=1

1

n p.

This function is ubiquitous in number theory and seemingly random areas of science such as quantummechanics. We’ll only need one important fact about it.

7. We already know that ζ (1) diverges; now let’s find two more values of ζ . Euler came up with the following

formula for the sine function:

x−x3

6+

x5

120+ · · · = sin x = x

1−

x2

π2

1−

x2

4π2

1−

x2

9π2

· · ·

Compare the coefficients of x3 and x5 in these expressions to evaluate ζ (2) and ζ (4).

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

We know on the left the coefficient of x3 is −16 . On the right, we can see that x3 terms are only made

when the outermost x is multiplied by an x2 term. Looking at all x3 terms on the left and right, we getthe equality:

−1

6x3 = (−

x3

π2−

x3

4 pi2−

x3

9π2)

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(−x3)1

6= (−x3)(

1

π2)(

1

12+

1

22+

1

32)

1

6= (

1

π2)ζ (2)

ζ (2) =π2

6

(x−x3

π2)(1−

x2

4π2)(1−

x2

9π2) · · · =

= (x−x3

4π2−

x3

π2+

x5

4π4)(1−

x2

9π2) . . .

= x−x3

4π2−

x3

π2+

x5

4π4+

x3

9π2+

x5

36π4+

x5

9π4−

x7

36π6. . .

1

π

4(

1

4

+1

36

+1

9

) =1

π

4(

1

1

2

· 2

2+

1

1

2

· 3

2+

1

2

2

· 3

2+ . . . )

ζ (2)2 = ( 1

n2)2 =

n,m=1

1

n2m2=n<m

+n=m

1

π4+n>m

x5

120=

x5

π4(

1

12 · 22+

1

12 · 32+

1

22 · 32+ . . . ) =

(ζ 2(2)− ζ (4))

2

1

π4

π4

60= ζ 2(2)− ζ (4)

ζ (4) = ζ (22)− π

4

60 = π

4

36 − π

4

60 = π

4

90

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

8. Use the MCT for series to evaluate ∞

0

xs−1

ex − 1dx

in terms of the gamma function and the Riemann zeta function. Be sure to check the hypotheses of thetheorem! What is the value of the integral in the Stefan-Boltzmann law?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

We have shown in problem 2 that xs−1

ex−1 =∞

n=1 xs−1e−nx, and by applying the Monotone ConvergenceTheorem for Series from problem 3 to the initial integral, we obtain the following expression

∞

0

xs−1

ex − 1dx =

∞

0

∞n=1

xs−1e−nxdx =

∞n=1

∞

0

xs−1e−nxdx

We set u = nx, and du = ndx

Furthermore, we know that (nx)s−1 = xs−1ns−1, ⇒ xs−1 = (nx)s−1

ns−1 .We also multiply by a form of 1, n

n , which gives us:

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∞n=1

∞

0

xs−1e−nxdx =

∞n=1

1

n

1

ns−1

∞

0

n(nx)s−1e−nxdx =

∞n=1

1

ns

∞

0

(u)s−1e−udu

and given that ζ ( p) =∞

n=11np , and that Γ(s) =

∞

0 xs−1e−x dx, we have

∞

0

xs−1

ex

− 1

dx =

∞

n=1

1

ns

∞

0

(u)s−1e−udu = ζ (s)Γ(s)

In order to evaluate the integral in the Stefan-Bolzmann law, we simply use the expression we obtainedabove, and set s = 4

∞

0

xs−1

ex − 1dx =

∞

0

x3

ex − 1dx = ζ (4)Γ(4)

We have shown in problem 7 that ζ (4) = π4

90 , and we proved in problem 6 that Γ(n) = (n− 1)!, therefore,the expression becomes

∞0

x

3

ex − 1 dx = ζ (4)Γ(4) = π

4

90 · 3! = π

4

15

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

I am morally obligated to mention the Riemann hypothesis, which is math’s biggest unsolved problem. Usingformulas such as

ζ ( p) = 2(2π) p−1 sin(πp/2)Γ(1− p)ζ (1− p)

ζ can be defined for any complex number—except 1 of course. For technical reasons, people care about the zeroesof ζ . Here is the Riemann hypothesis: for complex numbers a + bi with 0 < a < 1 is it TRUE or FALSE that

ζ (a + bi) = 0 =⇒ a = 1/2

Millions of dollars and eternal glory await the person who can solve this problem.

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