Math 360: A primitive integral and elementary functions

D. DeTurck

University of Pennsylvania

October 16, 2017

D. DeTurck Math 360 001 2017C: Integral/functions 1 / 32 Setup for the integral – partitions

Definition: (partitions of an interval) Let a < b, then a sequence of numbers a = x0 < x1 < x2 < ··· < xn = b is called a partition of the interval [a, b]. There is no requirement that the xi ’s be evenly spaced, only that they be strictly increasing and that x0 = a and xn = b. Sometimes we refer to partition by P. For instance, we say P0 is a refinement of the partition P (and 0 0 write P > P) if all of the xi ’s in P are also in P . The norm of the partition P (denoted |P|) is the maximum of xi − xi−1 for i = 1,..., n. (The textbook calls this the “gap” of P).

Given two partitions P1 and P2 of [a, b] it is easy to see that they have a “common refinement” — just use all the x’s from both partitions in a new P3.

D. DeTurck Math 360 001 2017C: Integral/functions 2 / 32 Setup for the integral – monotonic functions

At first, we are only going to consider the integrals of monotonic functions.

Let f :[a, b] → R be a monotonic function. For the moment, we’ll assume that f is increasing (i.e., f (x) ≤ f (y) if x ≤ y), but everything will work for decreasing functions with the obvious adjustments. f does not have to be continuous (and in fact it can be allowed to be discontinuous at [countably ] infinitely many places), as long as it is defined for all x ∈ [a, b] and is monotonic.

D. DeTurck Math 360 001 2017C: Integral/functions 3 / 32 Setup for the integral – upper and lower sums

For f monotonically increasing on [a, b], and P a partition of [a, b].

Definition: (upper and lower sums) The upper sum of f corresponding to the partition P is

n X U(f , P) = f (xi )(xi − xi−1) i=1 and the lower sum of f corresponding to the partition P is

n X L(f , P) = f (xi−1)(xi − xi−1) i=1

Since f is increasing, we have f (xi ) is the maximum value of f on [xi−1, xi ] and f (xi−1) is the minimum. Therefore U(f , P) ≥ L(f , P)

D. DeTurck Math 360 001 2017C: Integral/functions 4 / 32 Setup for the integral – definition

Since there’s always a common refinement P for any partitions P1 and P2, we have

L(f , P1) ≤ L(f , P) ≤ U(f , P) ≤ U(f , P2)

where P is a common refinement for P1 and P2. So for any pair of partitions P1 and P2 we have

L(f , P1) ≤ U(f , P2).

Therefore sup L(f , P) over all partitions P is less than or equal to inf U(f , P). If these are equal then their common value is called the integral: b f (x) dx ˆa

D. DeTurck Math 360 001 2017C: Integral/functions 5 / 32 Proving existence – regular partitions

To show that the integral exists, it is sufficient to find, for any ε > 0, a partition P such that

U(f , P) < L(f , P) + ε.

For monotonic functions, we can do this by using sufficiently fine regular partitions – these are partitions having the xi ’s evenly b − a spaced (so x − x = for all i = 1,..., n). i i−1 n Theorem b If f :[a, b] → R is monotonic, then f (x) dx exists. ˆa Proof: Given ε > 0, choose n so large that the regular partition P f (b) − f (a) with n steps, will have U(f , P) − L(f , P) = < ε. n

D. DeTurck Math 360 001 2017C: Integral/functions 6 / 32 Functions of bounded variation

Proposition (Greater generality) If f (x) can be written as the sum of two monotonic functions p(x) and q(x) with p increasing and q decreasing on [a, b] (such a function is called a function of bounded variation), then b b b f (x) dx exists and is equal to p(x) dx + q(x) dx. ˆa ˆa ˆa The set of functions of bounded variation on a closed bounded interval is actually quite general, so even though we started out with a somewhat restrictive definition we have created a pretty powerful form of the integral.

D. DeTurck Math 360 001 2017C: Integral/functions 7 / 32 Basic properties

Because the upper and lower sums have these properties, it follows that the integral does: 1 Linearity: b b b αf (x) + βg(x) dx = α f (x) dx + β g(x) dx for ˆa ˆa ˆa constants α, β and f and g are functions of bounded variation on [a, b]. 2 Monotonicity: If f (x) ≥ g(x) for all x ∈ [a, b] then b b a f (x) dx ≥ a g(x) dx. 3 If´ f (x) is of bounded´ variation on [a, b], then so is |f (x)| and b b

f (x) dx ≤ |f (x)| dx. ˆa ˆa

D. DeTurck Math 360 001 2017C: Integral/functions 8 / 32 More basic properties

4 If a < b < c then

c b c f (x) dx = f (x) dx + f (x) dx. ˆa ˆa ˆb

5 If m ≤ inf{f (x) | x ∈ [a, b]} and M ≥ sup{f (x) | x ∈ [a, b]} then b (b − a)m ≤ f (x) dx ≤ (b − a)M. ˆa 6 Mean value theorem for integrals: If f is continuous and of bounded variation on [a, b] then there is a c with a < c < b such that 1 b f (c) = f (x) dx. b − a ˆa

D. DeTurck Math 360 001 2017C: Integral/functions 9 / 32 Fundamental Theorem – integrals of derivatives

Fundamental theorem of calculus I: Integrals of derivatives Let F (x) be a differentiable function on [a, b] with derivative F 0(x), and suppose F 0 is a function of bounded variation on [a, b]. Then

b F 0(x) dx = F (b) − F (a). ˆa We can prove this for functions with monotonic derivatives and then use linearity. If F 0(x) is monotonically increasing, then for any interval (xi−1, xi ) partition P we have

0 F (xi ) − F (xi−1) 0 F (xi−1) ≤ ≤ F (xi ) xi − xi−1 by the mean value theorem (for derivatives).

D. DeTurck Math 360 001 2017C: Integral/functions 10 / 32 Fundamental Theorem – proof conclusion

But then

n X 0 L(f , P) = F (xi−1)(xi − xi−1) i=1 n X ≤ F (xi ) − F (xi−1) i=1 n X 0 ≤ F (xi )(xi − xi−1) = U(f , P). i=1

But the middle sum telescopes to F (xn) − F (x0) = F (b) − F (a). Now F (b) − F (a) is trapped between sup L(f , P) and inf U(f , P), P P b both of which are equal to the integral f (x) dx. ˆa

D. DeTurck Math 360 001 2017C: Integral/functions 11 / 32 Second fundamental theorem – derivatives of integrals

Fundamental theorem of calculus II: Derivatives of integrals Let f (x) be a continuous function of bounded variation on [a, b]. Define the function F (x) via

x F (x) = f (t) dt. ˆa Then F is differentiable and F 0(x) = f (x). First, if h > 0 we have

x+h F (x + h) − F (x) = f (t) dt = hf (c) ˆx for some c between x and x + h by properties 4 and 6 (mean value theorem for integrals).

D. DeTurck Math 360 001 2017C: Integral/functions 12 / 32 Second fundamental theorem – proof conclusion

Therefore the difference quotient

F (x + h) − F (x) = f (c) where x < c < x + h h But since f is continuous, f (c) → f (x) as h → 0+.

x For h < 0 we use that F (x + h) − F (x) = − f (t) dt = hf (c) ˆx+h for some c between x + h and x, and the proof goes through as for the h > 0 case.

D. DeTurck Math 360 001 2017C: Integral/functions 13 / 32 Using FTC I to compute integrals

Of course, we can use the first FTC to calculate integrals, once we have anti-derivative formulas obtained by turning around derivative formulas. Right now, though, we don’t have much more than derivatives of rational powers of x. So we have that since d (xr ) = rxr−1 dx if r is a rational number, then

b 1 xr−1 dx = (br − ar ) provided r 6= 0. ˆa r Of course we usually replace r by r + 1 and say

b 1 xr dx = (br+1 − ar+1) provided r 6= −1. ˆa r + 1

D. DeTurck Math 360 001 2017C: Integral/functions 14 / 32 What about r = −1?

b 1 So now we have a question: what is dx? Because ˆa x f (x) = 1/x is monotonically decreasing on any interval [a, b] where a > 0 the integral should exist.

So, provisionally, let’s define the function L(x) via:

x 1 L(x) = dt ˆ1 t and see if we can uncover some of its properties. To begin, we know only that 1 L(1) = 0 and L0(x) = . x

D. DeTurck Math 360 001 2017C: Integral/functions 15 / 32 Multiplicative property of L(x)

Consider the function L(ax) for a positive constant a. By what we know about L and the chain rule we have

d dL a 1 L(ax) = a = = . dx dx ax ax x So the derivative of L(ax) is the same as the derivative of L(x), therefore the two differ by a constant.

What constant? Well, putting x = 1 gives that L(ax) at 1 is L(a). So L(ax) = L(a) + L(x). Look familiar?

D. DeTurck Math 360 001 2017C: Integral/functions 16 / 32 Power property of L

r How about L(x ) if r ∈ Q? we have d 1 d rxr−1 r dL(x) L(xr ) = xr = = = r . dx xr dx xr x dx So the derivative of L(xr ) is r times the derivative of L(x). r Moreover, at x = 1, we have L(x )|x=1 = 0. This shows that L(xr ) = rL(x). Again, look familiar? So far, we have defined L(x) only for x ≥ 1. But we could define it for 0 < x < 1 either by saying that L(x) = −L(1/x) or else by 1 1 saying that L(x) = − dx. It’s an easy exercise to show that ˆx x these definitions agree. Likewise, if we define L(x) = L(−x) for x < 0 we will have L0(x) = 1/x there as well.

D. DeTurck Math 360 001 2017C: Integral/functions 17 / 32 The range of ln(x)

We have defined the function L(x) as the integral of 1/x with L(1) = 0 (and L(−1) = 0). So L(x) agrees with the natural logarithm function ln(x).

1 1 Since 1 ≥ ≥ for 1 < x < 2, we have that 1 > ln(2) > 1 . x 2 2 Therefore n ln(2n) > 2 which shows that ln(x) → ∞ as x → ∞. Likewise, ln(1/2) < −1/2, so ln(1/2n) < −n/2, showing that ln(x) → −∞ as x → 0+.

Therefore the range of ln(x) is all of R, and ln(x) is a strictly monotonically increasing function, since its derivative is positive. Therefore it has an inverse function that will map R to (0, ∞). Let’s call the inverse function E(x) and study its properties.

D. DeTurck Math 360 001 2017C: Integral/functions 18 / 32 The inverse of ln x and arbitrary powers

The function E(x) is defined by E(ln(x)) = x for x > 0 and ln(E(x)) = x for all x ∈ R. Because ln(1) = 0 we have E(0) = 1. And since

ln(E(a)E(b)) = ln(E(a)) + ln(E(b)) = a + b,

we have E(a + b) = E(a)E(b). Likewise, since for a ∈ R and r ∈ Q, ln(E(a)r ) = r ln(E(a)) = ra,

we have E(ra) = E(a)r . Therefore E(x) behaves like raising some number to the xth power, and it gives us a way to define xy for all real y, namely

xy = E(y ln x).

D. DeTurck Math 360 001 2017C: Integral/functions 19 / 32 The number e

We have E(x) is like raising some number to the xth power. What number? Well, that would be E(1), which we’ll call e. So e is the number such that ln e = 1. Because 2 1 4 1 1 1 1 ln 2 = dt < 1 and ln 4 = dt > + + > 1 ˆ1 t ˆ1 t 2 3 4 we know that 2 < e < 4. We’ll get better estimates later. But for now, we’ll write ex for E(x). 1 dy dy Derivative: If y = ex then x = ln y so 1 = , and so = y, y dx dx in other words d ex = ex . dx

D. DeTurck Math 360 001 2017C: Integral/functions 20 / 32 An important differential equation

So ex is a solution of the differential equation y 0 = y with initial value y(0) = 1. More generally Important! The unique solution of the initial-value problem

y 0 = ky, y(0) = C is y = Cekx . It is easy to check that Cekx satisfies the differential equation and the initial condition. If f (x) were another solution, then f 0 = kf and f (0) = C and we could calculate:

d  f (x)  1 e−kx = (e−kx f 0(x) − ke−kx f (x)) = (kf − kf ) = 0 dx Cekx C C

so e−kx f (x) is a constant. What constant? Evaluate at zero to get that it’s 1, so f (x) = Cekx after all.

D. DeTurck Math 360 001 2017C: Integral/functions 21 / 32 Better estimates of e

Now that we know that ex satisfies y 0 = y, we know that all the derivatives of ex are ex . In particular, all of them are equal to 1 at x = 0. Therefore, by the Taylor estimate: 1 1 1 1 M e = e1 = 1 + + + + ··· + + 1! 2! 3! n! (n + 1)!

where M is the value of the (n + 1)st derivative of ex evaluated somewhere between 0 and 1. From our earlier primitive estimate we know that 1 < M < 4. So for any n > 0 we can write

1  1 1 1 1  4 < e − 1 + + + + ··· + < (n + 1)! 1! 2! 3! n! (n + 1)!

D. DeTurck Math 360 001 2017C: Integral/functions 22 / 32 e is irrational

Use the last inequality to show that e is not a rational number. If it were, then e = p/q for some positive integers p and q and we would have 1 p  1 1 1 1  4 < − 1 + + + + ··· + < (n + 1)! q 1! 2! 3! n! (n + 1)!

Multiply by n! and get

1 n!p  1 1 1 1  4 < − n! 1 + + + + ··· + < (n + 1) q 1! 2! 3! n! (n + 1)

If n ≥ q then the number in the middle is an integer, but if n > 4 then we’ve trapped an integer between two positive rational numbers both of which are less than 1. A contradiction!

D. DeTurck Math 360 001 2017C: Integral/functions 23 / 32 Decimal approximation of e

Start again from the Taylor estimate 1 1 1 1 M e = e1 = 1 + + + + ··· + + 1! 2! 3! n! (n + 1)!

where we know that 1 < M < 4.

If we evaluate this for n = 19 we get that

2.7182818284590452353 < e < 2.7182818284590452366.

Probably close enough for most purposes.

D. DeTurck Math 360 001 2017C: Integral/functions 24 / 32 Trig?

How should we approach trigonometric functions? Let’s start with the function x 1 A(x) = 2 dt ˆ0 1 + t Clearly A(x) is an increasing function and since x 1 1 A(x) < 1 + 2 dt = 2 − , ˆ1 t x

we have that A(x) is defined for all x ∈ R (we have A(x) is an odd function of x) and |A(x)| < 2 for all x. Since A(x) is increasing and bounded above, it must approach a limit as x → ∞. Let’s call this limit π/2: Definition of π π = 2 lim A(x). x→∞

D. DeTurck Math 360 001 2017C: Integral/functions 25 / 32 Inverse functions

Since A(x) is a monotonically increasing function from R to the 1 1 interval (− 2 π, 2 π), it has an inverse function that we will call 1 1 T (x) from (− 2 π, 2 π) to R. d 1 Since A(x) = , if x = A(y) (or y = T (x)) we have dx 1 + x2 1 dy 1 = 1 + y 2 dx Therefore d T (x) = 1 + T (x)2. dx From our definition of π/2, we see that lim T (x) = ∞ x→π/2−

D. DeTurck Math 360 001 2017C: Integral/functions 26 / 32 S(x) and C(x)

Next. let S(x) and C(x) be the functions satisfying

T (x) 1 S(x) = and C(x) = . p1 + T (x)2 p1 + T (x)2

It’s immediately apparent that S(x)2 + C(x)2 = 1. Moreover

√ 0 2 0 √TT 2 2 2 2 d 1 + T T − T 2 (1 + T ) − T (1 + T ) S(x) = 1+T = dx 1 + T 2 (1 + T 2)3/2 1 = = C(x) (1 + T 2)1/2

d and likewise C(x) = −S(x). dx

D. DeTurck Math 360 001 2017C: Integral/functions 27 / 32 The differential equation for S(x) and C(x)

Both S(x) and C(x) satisfy the second-order DE: y 00 + y = 0. Additionally S(0) = 0 and S0(0) = 1 and C(0) = 1 and C 0(0) = 0. A new kind of initial-value problem! Proposition There is at most one solution of the initial-value problem

y 00 + y = 0 y(0) = a y 0(0) = b. If there were two, call them y and z, then their difference u = y − z would satisfy u00 + u = 0 with u(0) = u0(0) = 0. But then

d (u2 + u02) = 2uu0 + 2u0u00 = 2u0(u + u00) = 0, dx so this quantity is constant, and in fact is zero because it is zero when x = 0. But that means u2 = 0, so u = 0 and so y = z.

D. DeTurck Math 360 001 2017C: Integral/functions 28 / 32 Extensions, periodicity

1 1 At this point, because T (x) is defined only for − 2 π < x < 2 π, the functions S(x) and C(x) have these same restrictions. However, 1 − since T (x) → ∞ as x → 2 π , you can see that S(x) → 1 and 1 − 0 0 C(x) → 0 as x → 2 π . It follows that C (x) → −1 and S (x) → 0 there, and because we have that S(π − x) and C(π − x) are solutions of y 00 + y = 0, we see we can extend S and C to be defined up to π.

Keep repeating this trick to get that S and C are defined for all x and are periodic with period 2π.

D. DeTurck Math 360 001 2017C: Integral/functions 29 / 32 More trig identities

Addition formulas: What is sin(a + b)? Well, the function y = sin(a + x) satisfies y 00 + y = 0 and y(0) = sin a and y 0(0) = cos a. Therefore we have

y = sin(a + x) = sin a cos x + cos a sin x

Get addition formulas for other trig functions in a similar way.

D. DeTurck Math 360 001 2017C: Integral/functions 30 / 32 Integration by substitution

We pause to do the proof of a familiar integration technique, which is the chain rule in reverse: Integration by subsitution Suppose f :[a, b] → R and g :[c, d] → R are of bounded variation, and that the derivative of g exists and is of bounded variation on (c, d). Further, assume that the image of g :(c, d) → R is contained in the interval (a, b). Then

d g(d) f (g(x))g 0(x) dx = f (x) dx. ˆc ˆg(c) Proof: Apply FTC to the function

x g(x) H(x) = f (g(t)) dt − f (t) dt. ˆc ˆg(c)

D. DeTurck Math 360 001 2017C: Integral/functions 31 / 32 What is π?

We’ve defined π/2 to be the first positive zero of cos x, but now we can give it a geometric significance. To do so, we’ll use the substitution g(x) = sin x on the following integral:

1 p π/2 p 1 − x2 dx = 1 − sin2 θ cos θ dθ ˆ0 ˆ0 π/2 π/2 1 = cos2 θ dθ = (1 + cos 2θ) dθ ˆ0 ˆ0 2   π/2 θ sin 2θ π = + = 2 4 0 4 This shows that π is the area of the unit circle. At last.

D. DeTurck Math 360 001 2017C: Integral/functions 32 / 32