Analysis 2

Question 1

When does a Power Series Converge?

Answer 1

Let Σ0,∞ [anx^n] be a power series with Σant^n convergent. Then
Σ0,∞ [anx^n]
converges absolutely for all x with |x| < |t|.

Proof:
Since Σant^n converges we know that ant^n → 0 as n → ∞ and so the sequence is bounded. There is some M for which |ant^n| < M for all n. Now
Σ0,N [|anx^n|] = Σ0,N [|ant^n||x/t|^n
≤ MΣ0,N [|x/t|^n]
≤ MΣ0,∞ [|x/t|^n]
= M/1 − |x|/|t|
Hence Σ0,∞|anx^n| < ∞ and the series converges absolutely.

Question 2

What is the Theorem for the possibilities of how a power series converges?

Answer 2

Let Σ0,∞ [anx^n] be a power series. One of the
following holds.
* The series converges only if x = 0.
* The series converges for all real numbers x.
* There is a positive number R with the property that the series converges if |x| < R and diverges if |x| > R.
In the third case the number R is called the radius of convergence. In the first case we say that the radius of convergence is 0 and in the second that it is ∞.

Question 3

How does the radius of convergence for the absolute power series relate to the original series?

Answer 3

Let Σ0,∞ [anx^n] be a power series with radius of convergence R. Then Σ0,∞|an|x^n also has radius of convergence R.

Question 4

How are power series continuous?

Answer 4

Let Σ0,∞ [anx^n] with radius of convergence R.
Then the function
x → Σ0,∞ [anx^n] is continuous on the interval (−R, R).

Question 5

What is the Exponential?

Answer 5

If x ∈ R the series
1 + x + x^2/2 + x^3/6 + · · · + x^n/n! + · · ·
converges. We call the sum exp x.

Use the ratio test of successive terms to show it converges.

Question 6

What is the Characteristic Property of the Exponential?

Answer 6

If x, y ∈ R then
exp(x + y) = exp(x) exp(y).

Proof For a fixed number z consider the function
x → exp(x) exp(z − x).
We may differentiate this with respect to x using the product rule and the chain rule to get
exp(x) exp(z − x) − exp(x) exp(z − x) = 0.
By the MVT the function is constant. At x = 0 the function is exp z so we know that for all x
exp(x) exp(z − x) = exp(z).
Now if we set z = x + y we get the conclusion we want.

Question 7

What are the Inequalities for the exponential?

Answer 7

The following estimates hold for the exponential function:
1. 1 + x ≤ e^x for all real x
2. e^x ≤ 1/(1 − x) if x < 1.

Proof If x ≥ 0 then
e^x = 1 + x + x^2/2 + · · · ≥ 1 + x
and if 0 ≤ x < 1
e^x = 1 + x + x^2/2 +x^3/6 + · · · ≤ 1 + x + x^2 + x^3 + · · · = 1/1 − x.
So the inequalities are easy to establish if x ≥ 0. To obtain them for negative x we use the characteristic property of the exponential much as we did to prove positivity.
Suppose x = −u is negative. We know that e^u ≥ 1 + u and hence e^−x ≥ 1 − x. But this implies that
1/1 − x ≥ e^x
so the second inequality is now established for all x < 1.
If x ≤ −1 then 1 + x ≤ 0 whereas e^x > 0 so e^x ≥ 1 + x. It remains to prove the first inequality for −1 < x < 0. If x = −u then 0 < u < 1 and so e^u ≤ 1/(1 − u). This says
that e^−x ≤ 1/(1 + x) and this implies that
1 + x ≤ e^x.

Question 8

How does the exponential increase?

Answer 8

The exponential function is strictly increasing and its range is (0,∞).

Proof Suppose x < y. Then
e^y = e^(y−x)e^x ≥ (1 + y − x)e^x > e^x.
This shows that the exponential is strictly increasing.

Question 9

What is the Logarithm?

Answer 9

There is a continuous strictly increasing function x → log x
defined on (0,∞) satisfying
e^log x = x
for all positive x and
log(e^y) = y
for all real y. We have that for all positive u and v,
log(uv) = log u + log v.

Proof We just need to check the last assertion. But
e^log u+log v = e^log u * e^log v = uv.

Question 10

What are Powers?

Answer 10

If x > 0 and p ∈ R we define
x^p = exp(p log x).
We have the usual rules
1. If n is a positive integer then x^n as defined here is indeed the product x.x. . . . .x of n copies of x.
2. x^p+q = x^px^q for all x > 0 and p, q ∈ R.
3. log(x^) = p log x for x > 0 and p ∈ R.
4. x^pq = (x^p)^q for all x > 0 and p, q ∈ R.
5. exp(p) = e^p for all p ∈ R.

Question 11

What is the Inequality for the logarithm?

Answer 11

If x > 0 then log x ≤ x − 1.

Question 12

What is the Limit of a function?

Answer 12

Let I be an open interval, c ∈ I and f a real valued function defined on I except possibly at c. We say that
lim(x→c) f(x) = L
if for every ε > 0 there is a number δ > 0 so that if 0 < |x − c| < δ then |f(x) − L| < ε.
Thus we can guarantee that f(x) is close to L by insisting that x is close to c, but we exclude the possibility that x = c: we don’t care what f does at c itself nor even whether f is defined at c.

Question 13

How do limits relate to continuity?

Answer 13

If f : I → R is defined on the open interval I and c ∈ I then f is continuous at c if and only if
lim(x→c) f(x) = f(c).

Question 14

How do limits relate to sequences?

Answer 14

If f : I{c} → R is defined on the interval I except at c ∈ I then
lim(x→c) f(x) = L
if and only if for every sequence (xn) in I/{c} with xn → c we have
f(xn) → L.

Question 15

What are the Algebra of Limits?

Answer 15

If f, g : I{c} → R are defined on the interval I except at c ∈ I and limx→c f(x) and limx→c g(x) exist then
1. limx→c(f(x) + g(x)) = limx→c f(x) + limx→c g(x)
2. limx→c f(x)g(x) = limx→c f(x) limx→c g(x)
3. if limx→c g(x) /= 0 then
limx→c f(x)/g(x) = (limx→c f(x))/(limx→c g(x)).

Question 16

What is the Derivative?

Answer 16

Suppose f : I → R is defined on the open interval I and c ∈ I. We say that f is differentiable at c if
lim(h→0) f(c + h) − f(c)/h
exists. If so we call the limit f’(c).

We can rewrite the derivative
f’(c) = limx→c f(x) − f(c)/x − c
since if we put x = c + h and h → 0 we have x → c

Question 17

How does differentiability imply continuity?

Answer 17

If I is an open interval, f : I → R is differentiable at c ∈ I then f is continuous at c.

Proof We know that
f(x) − f(c)/x − c → f’(c)
as x → c. Hence
f(x) − f(c) = f(x) − f(c)/x − c * (x − c) → f’(c)*0 = 0
as x → c which implies that f(x) → f(c) as required.

Question 18

What are the sum and product rules of the derivative?

Answer 18

Suppose f, g : I → R are defined on the open interval I and are differentiable at c ∈ I. Then f + g and fg are differentiable at c and (f + g)’(c) = f’(c) + g’(c) and (fg)’(c) = f’(c)g(c) + f(c)g’(c).

Proof of product rule
f(x)g(x) − f(c)g(c)/x − c
=(f(x) − f(c))g(x) + f(c)(g(x) − g(c))/x − c
=f(x) − f(c)/x − c * g(x) + f(c)* g(x) − g(c)/x − c.
The algebra of limits tells us that as x → c this expression approaches
f’(c)g(c) + f(c)g’(c).

Question 19

What are the derivatives of the monomials?

Answer 19

If n is a positive integer then the derivative of x → x^n is x → nx^(n−1).

Proof We already saw this for n = 1. Assume inductively that we have the result for f(x) = x^n. Then
x^n+1 = xf(x) so by the product rule its derivative is
1.f(x) + xf’(x) = x^n + xnx^(n−1) = (n + 1)x^n
completing the inductive step.

Question 20

What is the Chain Rule?

Answer 20

Suppose I and J are open intervals, f : I → R and g : J → I,
that g is differentiable at c and f is differentiable at g(c). Then the composition f ◦ g is differentiable at c and
(f ◦ g)’(c) = f’(g(c))* g’(c).

Proof.
f is differentiable at g(c) so for all y
f(y) − f(g(c)) = f’(g(c))(y − g(c)) + ε(y)(y − g(c)).
where ε(g(c)) = 0 and ε is continuous at g(c). Hence
f(g(x)) − f(g(c)) = f’(g(c))(g(x) − g(c)) + ε(g(x))(g(x) − g(c)).
Consequently if x /= c
f(g(x)) − f(g(c))/x − c = f’(g(c)) * g(x) − g(c)/x − c + ε(g(x)) * g(x) − g(c)/x − c.
As x → c,
g(x) − g(c)/x − c → g’(c)
while ε ◦ g is continuous at c so ε(g(x)) → ε(g(c)) = 0. Hence
f(g(x)) − f(g(c))/x − c → f’(g(c)).g’(c) as required.

Question 21

What is the Mean Value Theorem?

Answer 21

Suppose f : [a, b] → R is continuous on the closed interval [a, b] and differentiable on the open interval (a, b). Then there is a point c ∈ (a, b) where
f’(c) = f(b) − f(a)/b − a.

Question 22

What is Rolle’s Theorem?

Answer 22

Suppose f : [a, b] → R is continuous on the closed interval [a, b] and differentiable on the open interval (a, b) and that f(a) = f(b). Then there is a point c in the open interval where
f’(c) = 0.

Proof.
If f is constant on the interval then its derivative is zero everywhere. If not it takes values different from f(a) = f(b). Assume it is somewhere larger than f(a).
Since f is continuous on the closed interval it attains its maximum value at some point c and this cannot be a or b: so c lies in (a, b). If x > c then f(x)−f(c) ≤ 0 while x−c > 0
so the ratio
f(x) − f(c)/x − c ≤ 0.
So f’(c) is a limit of non-positive values and so is not positive.
On the other hand if x < c then f(x) − f(c) ≤ 0 while x − c < 0 so the ratio
f(x) − f(c)/x − c ≥ 0.
So f’(c) is a limit of non-negative values and so is not negative. Therefore f’(c) = 0.

Question 23

What is the proof of the MVT?

Answer 23

Consider the function given by
g(x) = f(x) − x[f(b) − f(a)/b − a].
Then
g(b) − g(a) = f(b) − f(a) − (b − a)[f(b) − f(a)/b − a] = 0.
So by Rolle’s Theorem there is a point c where g’(c) = 0. But this implies
f’(c) = f(b) − f(a)/b − a.

Question 24

What is the derivative of the maximum in an interval?

Answer 24

Suppose f : [a, b] → R is continuous and that it is differentiable on the open interval. Then f attains its maximum and minimum either at points in (a, b) where f’ = 0 or at one of the ends a or b.

Question 25

What is the derivative of an inverse?

Answer 25

Let f : (a, b) → R be differentiable with positive derivative. Then g = f^−1 is differentiable and
g’(x) = 1/f’(g(x)).

Question 26

How are power series differentiable?

Answer 26

Let f(x) = Σanx^n be a power series with radius of convergence R. Then f is differentiable on (−R, R) and
f’(x) = Σn=1,∞ [nan x^n−1].

Question 27

What are the Trig Functions?

Answer 27

For x ∈ R we define
cos x = 1 − x^2/2 + x^4/24 − · · · + (−1)^k x^2k/(2k)! + · · ·
sin x = x − x^3/6 + x^5/120 − · · · + (−1)k x^2k+1/(2k + 1)! + · · ·

Question 28

What are the addition formulae?

Answer 28

For all real x and y
cos(x + y) = cos x cos y − sin x sin y
sin(x + y) = sin x cos y + cos x sin y

Proof We shall check the first: the second is similar. For a fixed z let
f(x) = cos x cos(z − x) − sin x sin(z − x).
Using the derivatives above it is easy to check that f’(x) = 0 for all x and so f is constant.
When x = 0 we have
f(0) = cos 0 cos z − sin 0 sin z = cos z.
Hence f(x) = cos z for all x. Now if we set z = x + y we get
cos x cos y − sin x sin y = f(x) = cos z = cos(x + y).

Question 29

What is the Circular Property of the Trig Functions?

Answer 29

For all real x
cos^2 x + sin^2 x = 1.

Proof Set y = −x in the addition formula for cos. We know that cos(−x) = cos x and sin(−x) = − sin x and so we get
1 = cos 0 = cos(x − x) = cos x cos(−x) − sin x sin(−x) = cos2 x + sin2 x.

Question 30

What is Cauchy’s MVT?

Answer 30

If f, g : [a, b] → R are continuous, are differentiable on (a, b) and g’(t) /= 0 for t between a and b then there is a point t where
f’(t)/g’(t) = f(b) − f(a)/g(b) − g(a).

Proof Consider the function
x → h(x) = f(x)(g(b) − g(a)) − (f(b) − f(a))g(x).
At x = a the value of the function is
h(a) = f(a)g(b) − f(a)g(a) − f(b)g(a) + f(a)g(a) = f(a)g(b) − f(b)g(a).
Similarly
h(b) = f(b)g(b) − f(b)g(a) − f(b)g(b) + f(a)g(b) = f(a)g(b) − f(b)g(a).
So h(b) = h(a) and by Rolle’s Theorem there is a point t between a and b where h’(t) = 0.
Thus we have h(x) = f(x)(g(b) − g(a)) − (f(b) − f(a))g(x)
and there is a point t where h’(t) = 0. But this means that
f’(t)(g(b) − g(a)) = (f(b) − f(a))g’(t).
Since g’is non-zero on (a, b) Rolle’s Theorem applied to g shows that g(b) − g(a) /= 0
as well and we can rearrange to get the conclusion of the theorem.

Question 31

What is L’Hopital’s Rule?

Answer 31

If f, g : I → R are differentiable on the open interval I containing c and f(c) = g(c) = 0 then
limx→c f(x)/g(x) = limx→c f’(x)/g’(x)
provided the second limit exists.

Proof
Suppose that
limx→c f’(x)/g’(x)
does indeed exist. Then it cannot be that g’(x) = 0 at a sequence of points converging to c. So there is some interval around c on which g’ is non-zero (except perhaps at c itself).
So g’ is non-zero on an interval each side of c. This enables us to apply Cauchy’s MVT.
Because f(c) = g(c) = 0
limx→c f(x)/g(x) = limx→c f(x) − f(c)/g(x) − g(c).
As long as x is in the region around c where g’ /= 0 Cauchy’s MVT ensures that there is a point t (depending upon x) between c and x where
f(x) − f(c)/g(x) − g(c) = f’(t)/g’(t).
As x → c the corresponding t is forced to approach c as well and so
f(x)/g(x) = f(x) − f(c)/g(x) − g(c) → limt→c f’(t)/g’(t).

Question 32

What is L’Hopital’s Rule at infinity?

Answer 32

If f, g : R → R are differentiable and
limx→∞ f(x) = 0 and limx→∞ g(x) = 0
or
limx→∞ f(x) = ∞ and limx→∞ g(x) = ∞
then
limx→∞ f(x)/g(x) = limx→∞ f’(x)/g’(x)
provided the second limit exists.

Question 33

What is Taylor’s Theorem with Remainder?

Answer 33

If f : I → R is n times differentiable on the open interval I containing a and b then
f(b) = f(a) + f’(a)(b − a) + f’‘(a)^2(b − a)/2 + · · · + f^(n−1)(a)/(n − 1)! * (b − a)^n−1 + f^(n)(t)/n! * (b − a)^n
for some point t between a and b.

Question 34

What are the Upper and Lower Sums?

Answer 34

Let f : [a, b] → R be bounded and P = {x0, x1, . . . , xn} be a partition of [a, b]. The upper and lower Riemann sums of the function f with respect to P are
U(f, P) = Σ1,n Mi (xi − xi−1) and L(f, P) = Σ1,n mi (xi − xi−1)
respectively, where for each i
mi = inf{f(x) : xi−1 ≤ x ≤ xi}
and
Mi = sup{f(x) : xi−1 ≤ x ≤ xi}.

Question 35

What are the Upper and Lower Integrals?

Answer 35

Let f : [a, b] → R be bounded. The upper and lower Riemann integrals of the function f are
∫~f = infP U(f, P)
and
∫_f = supP L(f, P)
where the sup and inf are taken over all partitions of the interval [a, b].

Ie, the upper integral is equal to the lowest upper sum over all partitions, similarly for the lower integral.

Question 36

What is the Riemann Integral?

Answer 36

Let f : [a, b] → R be bounded. Then f is said to be Riemann integrable if
∫~f = ∫_f
and in this case we write
∫a,b f(x) dx
for the common value.

Question 37

What are Refinements to partitions and how are they better?

Answer 37

If P and Q are partitions of an interval [a, b] then Q is said
to be a refinement of P if every point of P belongs to Q.

Suppose f : [a, b] → R is bounded, P and Q are partitions of [a, b] and Q refines P. Then
L(f, P) ≤ L(f, Q) ≤ U(f, Q) ≤ U(f, P).

Proof We will just check the lower sums: the upper ones are similar. Suppose I is an interval in P and that Q breaks it into intervals J1, J2, . . . , Jm. Then the infimum of f on I will be at most the infimum on each Jj . So the sums in L(f, Q) based on the Jj have a total at least as big as the sum in L(f, P).

If P and Q are partitions of an interval [a, b] then we can find a common refinement of P and Q by including all points of P and Q.

Question 38

Are upper sums bigger than lower sums?

Answer 38

Suppose f : [a, b] → R is bounded and P and Q are partitions of [a, b]. Then
L(f, P) ≤ U(f, Q).

Proof Choose R to be a refinement of both P and Q. Then
L(f, P) ≤ L(f, R) ≤ U(f, R) ≤ U(f, Q).

Question 39

When is a function integrable?

Answer 39

Suppose f : [a, b] → R is bounded. Then f is Riemann integrable if (and only if ) for every ε > 0 we can find a partition P of [a, b] with
U(f, P) − L(f, P) < ε.

We check the easy direction first. Suppose that the condition holds. Then for any partition P
L(f, P) ≤ ∫_ f
and
∫~f ≤ U(f, P).
Therefore
∫~f − ∫_f ≤ U(f, P) − L(f, P).
Since the right side can be made smaller than any positive ε the left side must be 0.
For the other direction suppose f is integrable. Then
∫f = ∫~f = infQ U(f, Q)
so we can choose a partition Q1 with
U(f, Q1) < ∫f + ε/2.
Similarly we can choose Q2 so that
L(f, Q2) > ∫f − ε/2.
Now choose P to be a common refinement of Q1 and Q2. It will satisfy both inequalities and hence
U(f, P) − L(f, P) < ε.

Question 40

What is Uniform Continuity?

Answer 40

If f : [a, b] → R is continuous then for any ε > 0 we
can find δ so that if |x − y| < δ then
|f(x) − f(y)| < ε.

Question 41

How are continuous function integrable?

Answer 41

If f : [a, b] → R is continuous then it is integrable.

Proof The function is certainly bounded so we just need to show that the integrability condition is satisfied. Given ε > 0 choose δ so that if x, y ∈ [a, b] satisfy
|x − y| < δ
then
|f(x) − f(y)| < ε/b − a.
Now let P be a partition of the interval with each gap xi − xi−1 less than δ. Then for each i
Mi − mi = sup{f(x) : xi−1 ≤ x ≤ xi} − inf{f(x) : xi−1 ≤ x ≤ xi}
≤ ε/b − a.
Therefore
U(f, P) − L(f, P) = Σ1,n (Mi − mi) (xi − xi−1) ≤ ε/b − a * Σ1,n (xi − xi−1) = ε.

Question 42

How are monotone functions integrable?

Answer 42

If f : [a, b] → R is bounded and either increasing or decreasing, then it is integrable.

Proof Suppose f is increasing. Then for any partition, on each interval Mi = f(xi) and mi = f(xi−1). Therefore
U(f, P) − L(f, P) = Σ1,n (f(xi) − f(xi−1)) (xi − xi−1).
Now suppose we take a partition into n equal intervals.
U(f, P) − L(f, P) = Σ1,n (f(xi) − f(xi−1)) (xi − xi−1)
= (b − a)/n Σ1,n (f(xi) − f(xi−1))
= (b − a)/n (f(xn) − f(x0))
= (b − a)/n (f(b) − f(a)).
This approaches 0 as n → ∞.

Question 43

What is the Linearity of the integral?

Answer 43

If f, g : [a, b] → R are integrable and λ ∈ R then λf and f + g are integrable and
∫a,b λf(x) dx = λ * ∫a,b f(x) dx
and
∫a,b (f(x) + g(x)) dx =∫a,b f(x) dx + ∫a,b g(x) dx.

Question 44

What is Monotonicity of the integral?

Answer 44

If f, g : [a, b] → R are integrable and for all x ∈ [a, b] we have f(x) ≤ g(x) then
∫a,b f(x) dx ≤ ∫a,b g(x) dx.

Question 45

What is the Theorem for Continuous Function of an Integrable Function?

Answer 45

If f : [a, b] → R is integrable and φ : R → R is continuous then φ ◦ f is integrable on [a, b]

Question 46

How are products integrable?

Answer 46

If f, g : [a, b] → R are integrable then f.g is integrable on [a, b].

Question 47

What is the Triangle Inequality for integrals?

Answer 47

If f : [a, b] → R is integrable then |f| is integrable and
|∫a,b f(x) dx| ≤ ∫a,b |f(x)| dx.

Question 48

What is the Addition of Ranges?

Answer 48

Let f : [a, c] → R be bounded and a < b < c. Then f is integrable on [a, c] if and only if it is integrable on [a, b] and [b, c] and if so
∫a,c f(x) dx = ∫a,b f(x) dx +∫b,c f(x) dx

Question 49

What is the Fundamental Theorem of Calculus?

Answer 49

Let f : [a, b] → R be integrable. Then the function
F : x → ∫a,x f(t) dt
is continuous. If f (the integrand) is continuous at a point u ∈ (a, b) then F is differentiable at that point and
F’(u) = f(u).

Proof If x and x + h are in [a, b] with h > 0 then
F(x + h) − F(x) = ∫x,x+h f(t) dt.
Let M be a bound for f: so |f(x)| ≤ M for all x ∈ [a, b]. Then
|F(x + h) − F(x)| = |∫x,x+h f(t) dt| ≤ ∫x,x+h |f(t)|dt ≤ Mh.
Similarly for |F(x − h) − F(x)|. So F is continuous.
Now if f is continuous at u then given ε > 0 choose δ > 0 so that if |t − u| < δ we have |f(t) − f(u)| < ε. Then if 0 < h < δ
|F(u + h) − F(u)/h − f(u)|
= |1/h ∫u,u+h f(t) dt − 1/h ∫u,u+h f(u) dt|
= |1/h ∫u,u+h f(t) - f(u) dt|
≤ 1/h ∫u,u+h |f(t) - f(u)| dt
≤ 1/h ∫u,u+h ε dt = ε.
Similarly for h < 0. Hence F’(u) exists and equals f(u).

Question 50

What is the FTC 2?

Answer 50

Let F : [a, b] → R be continuous on [a, b] and differentiable on (a, b) with F’ = f. Then if f is Riemann integrable we have
∫a,b f(t) dt = F(b) − F(a).

Proof It suffices to show that for each partition P
L(f, P) ≤ F(b) − F(a) ≤ U(f, P)
because we then get
∫_f ≤ F(b) − F(a) ≤ ∫~f
and if f is integrable the upper and lower integrals are equal. This is where we use the integrability of f.
Let P = {x0, x1, . . . , xn} be a partition of [a, b]. For each i, the function F is continuous on [xi−1, xi] and differentiable on (xi−1, xi) so by the MVT there is a point ci ∈ (xi−1, xi) with
F(xi) − F(xi−1) = F’(ci)(xi − xi−1) = f(ci)(xi − xi−1).
Hence for each i
mi(xi − xi−1) ≤ F(xi) − F(xi−1) ≤ Mi(xi − xi−1).
Summing over all i gives
L(f, P) ≤ F(b) − F(a) ≤ U(f, P).

Question 51

What is Integration by Parts?

Answer 51

Suppose f, g : [a, b] → R are differentiable on an open interval including [a, b] and that f’ and g’ are integrable on [a, b]. Then
∫a,b f(x)g’(x) dx = f(b)g(b) − f(a)g(a) − ∫a,b f’(x)g(x) dx.

Proof By the product rule (fg)’ = f’g + fg’ and each term is integrable. So
∫a,b f(x)g’(x) dx + ∫a,b f’(x)g(x) dx = ∫a,b (fg)’(x) dx = f(b)g(b) − f(a)g(a).

Question 52

What is Integration by Substitution?

Answer 52

Suppose u : [a, b] → R is differentiable on an open interval including [a, b] and that u’ is integrable on [a, b]. Suppose that f is a continuous function on the bounded set u([a, b]). Then
∫a,b f(u(x))u’(x) dx = ∫u(a),u(b) f(t) dt.

For each x in u([a, b]) define
F(x) = ∫u(a),x f(t) dt.
By FTC I we know that F is differentiable and F’(x) = f(x) for each x. By the chain rule we have
d/dx F(u(x)) = F’(u(x))u’(x) = f(u(x))u’(x).
The function f ◦ u is continuous and u’ is integrable so this derivative is integrable and by FTC II we have
∫a,b f(u(x))u’(x) dx = F(u(b)) − F(u(a))
= ∫u(a),u(b) f(t) dt − ∫u(a),u(a) f(t) dt
= ∫u(a),u(b) f(t) dt.

Question 53

What is the Comparison Test for Improper Integrals?

Answer 53

Suppose f, g : [a,∞) → R are integrable on each interval [a, b], that |f(x)| ≤ g(x) for all x ≥ a and that
∫a,∞ g(x) dx converges. Then
∫a,∞ f(x) dx converges.

Proof The functions |f| and f + |f| are integrable on each interval [a, b]. For each b
∫a,b |f(x)| dx ≤ ∫a,b g(x) dx ≤ ∫a,∞ g(x) dx.
and
∫a,b (f(x) + |f(x)|) dx ≤ 2∫a,b g(x) dx ≤ 2∫a,∞ g(x) dx.
Both functions |f| and f + |f| are non-negative so the functions
b → ∫a,b |f(x)| dx
and
b → ∫a,b (f(x) + |f(x)|) dx
are bounded increasing functions of b. So both have limits as b → ∞ and hence so does their difference
b → ∫a,b f(x) dx.