Chapter 08 Additional Topics

8.1 The Generalized Riemann Integral

Exercise 8.1.1

Let $(P,{c_k})$ be an arbitrary tagged partition of $[a,b]$ that is $δ$-fine, and let $P' = P ∪ P_ϵ$.

(a) Explain why both the Riemann sum $R(f,P)$ and $\int_{a}^{b}f$ fall between $L(f,P)$ and $U(f,P)$.

Proof: Since $\int_{a}^{b}f$ is an upper bound of $L(f,P)$ and a lower bound of $U(f,P)$, so $\int_{a}^{b}f$ fall between $L(f,P)$ and $U(f,P)$.

Since $m_k \leq c_k \leq M_k$, $R(f,P)$ also fall between $L(f,P)$ and $U(f,P)$.

$\square$

(b) Explain why $U(f,P')−L(f,P') < ϵ/3$.

Proof:

Since $P' = P ∪ P_ϵ$, we have

$$L(f, P_ϵ) \leq L(f, P') \leq U(f, P') \leq U(f, P_ϵ)$$

$\square$

8.1.2.

Explain why $U(f,P)−U(f,P') ≥ 0$.

Proof: See previous exercise.

8.1.3.

(a) In terms of $n$, what is the largest number of terms of the form $M_k(x_k - x_{k-1})$ that could appear in one of $U(f,P)$ or $U(f,P')$ but not the other?

Solution: Note that $P \subseteq P' = P \cup P_{\epsilon }$, and since $P_{\epsilon}$ has $n$ segments, so $P'$ has at most $n$ points more than $P$.

If these $n$ points fall into $n$ different segments of $P$, then we can achive the largest number of terms of the form $M_k(x_k - x_{k-1})$ that appears in one of $U(f,P)$ or $U(f,P')$ but not the other, which is $3n$.

(b) Finish the proof in this direction by arguing that

$$U(f, P) - U(f, P') < \epsilon / 3.$$

Proof: Now we can bound

$$U(f, P) - U(f, P') < 3n \cdot M \cdot \frac{\epsilon}{9nM} = \epsilon / 3$$

Combine the previous exercises, we have

$$R(f, P), \int_{a}^{b}f \leq U(f, P) < U(f, P') + \epsilon / 3 \\ R(f, P), \int_{a}^{b}f \geq L(f, P) > L(f, P') - \epsilon / 3$$

So $\left| R(f, P) - \int_{a}^{b} f \right| < \epsilon$.

$\square$

8.1.4.

(a) Show that if $f$ is continuous, then it is possible to pick tags $\{c_k\}^n_{k=1}$ so that $R(f,P) = U(f,P)$.

Proof: Since $f$ is continuous, for each segment $[x_{k-1}, x_k]$, because of extreme value theorem, it is possible to find $c_k$ such that $f(x_k) = M_k$. Pick those $c_k$, we have $R(f,P) = U(f,P)$.

(b) If $f$ is not continuous, it may not be possible to find tags for which $R(f,P) = U(f,P)$. Show, however, that given an arbitrary $ϵ > 0$, it is possible to pick tags for $P$ so that

$$U(f,P) - R(f,P) < \epsilon.$$

The analogous statement holds for lower sums.

Proof: For each segment $[x_{k-1}, x_k]$, $M_k$ is the supremum of $f(x)$ for $x \in [x_{k-1}, x_k]$. So we can find $c_k$ such that $M_k - c_k < \epsilon / (b-a)$.

Pick those $c_k$, we have $U(f,P) - R(f,P) < \epsilon$.

$\square$

8.1.5.

Use the results of the previous exercise to finish the proof of Theorem 8.1.2.

Proof: Given $\epsilon$, we can find partition $\delta$, such that any $\delta$-fine tagged partition $|R(f, P) - A| < \epsilon /4$. $P$, such that $U(f,P) - R_1(f,P) < \epsilon / 4$. Similarly, $R_2(f,P) - L(f,P) < \epsilon / 4$.

Let $P_{\epsilon} = P_1 \cup P_2$, so

$$U(f, P) - L (f, P) = U(f, P) - R_1(f,P) + R_1(f,P) - A + A - R_2(f,P) + R_2(f,P) - L (f, P) \\ < \epsilon / 4 + \epsilon / 4 + \epsilon / 4 + \epsilon / 4 < \epsilon$$

$\square$

Gauges and δ(x)-fine Partitions

Definition 8.1.3.

A function $δ : [a,b] → \mathbf{R}$ is called a gauge on $[a,b]$ if $δ(x) > 0$ for all $x ∈ [a,b]$.

Definition 8.1.4.

Given a particular gauge $δ(x)$, a tagged partition $(P, {c_k}_{k=1}^n)$ is $\delta(x)$-fine if every subinterval $[x_{k−1},x_k]$ satisfies $x_k - x_{k−1} < \delta(c_k)$.

In other words, each subinterval $[x_{k−1},x_k]$ has width less than $\delta(c_k)$.

8.1.6.

Consider the interval [0,1].

(a) If $\delta(x) = 1/9$, find a $\delta(x)$-fine tagged partition of $[0,1]$. Does the choice of tags matter in this case?

Solution:

Let $P = \{0, 0.1, 0.2, \cdots, 0.9, 1.0\}$. The choice of tags does not matter.

$\square$

(b) Let

$$\delta (x) = \begin{cases} 1/4 &\text{if } x = 0\\ x/3 &\text{if } 0 < x \leq 1\\ \end{cases}$$

Construct a $δ(x)$-fine tagged partition of $[0,1]$.

Solution:

Let $x_0 = 0, x_1 = 1, c_1 = 0$.

$\square$

Theorem 8.1.5.

Given a gauge $δ(x)$ on an interval $[a,b]$, there exists a tagged partition $(P,\{c_k\}^n_{k=1})$ that is $δ(x)$-fine.

Proof. Let $I_0 = [a,b]$. It may be possible to find a tag such that the trivial partition $P= \{a,b\}$ works.

Specifically, if $b−a < δ(x)$ for some $x ∈ [a,b]$, then we can set $c_1$ equal to such an $x$ and notice that $(P,{c_1})$ is $δ(x)$-fine. If no such $x$ exists, then bisect $[a,b]$ into two equal halves.

Exercise 8.1.7.

Finish the proof of Theorem 8.1.5.

Proof: Assume otherwise, then this process continuous indefinitely, then we get a nested intervals. Because of NIP, we can find $c$ in all these intervals. That means $\delta (c) \geq b_n - a_n$ for all $n$. But $\lim_{n \to \infty} b_n - a_n = \lim_{n \to \infty}\frac{b-a}{2^n} = 0$ So we have a contradiction.

$\square$

Definition 8.1.6.

A function $f$ on $[a,b]$ has generalized Riemann integral $A$ if, for every $ϵ > 0$, there exists a gauge $δ(x)$ on $[a,b]$ such that for each tagged partition $(P,\{c_k\}^n_{k=1})$ that is $δ(x)$-fine, it is true that

$$|R(f,P)−A| < ϵ$$

In this case, we write $A = \int_{a}^{b}f$.

Theorem 8.1.7.

If a function has a generalized Riemann integral, then the value of the integral is unique.

Exercise 8.1.8.

Finish the argument.

Proof: Assume that a function $f$ has generalized Riemann integral $A1$ and that it also has generalized Riemann integral $A2$. We must prove $A1 = A2$.

Since $f$ is generalized Riemann integrable, given $\epsilon$, we can find $\delta_1(x)$, such that if a tagged partition $P_1$ is $\delta_1(x)$-fine, then $|R(f, P_1) - A_1| < \epsilon$.

Similarly, we can find $\delta_2(x)$, such that if a tagged partition $P_2$ is $\delta_2(x)$-fine, then $|R(f, P_2) - A_2| < \epsilon$.

Now we can define $\delta (x) = \min \{\delta_1(x), \delta_2(x)\}$. From Theorem 8.1.5, we can find a tagged partition $(P,\{c_k\}^n_{k=1})$ that is $\delta(x)$-fine.

If $(P,\{c_k\}^n_{k=1})$ is $\delta(x)$-fine, it has to be $\delta_1(x)$-fine. This is because

$$x_k - x_{k-1} < \delta (c_k) \leq \delta_1(c_k)$$

So $|R(f, P) - A_1| < \epsilon$.

For the same reason, $(P,\{c_k\}^n_{k=1})$ is $\delta(x)$-fine, it has to be $\delta_2(x)$-fine, and $|R(f, P) - A_2| < \epsilon$.

Since

$$|A_1 - A_2| < |R(f, P) - A_1| + |R(f, P) - A_2| < \epsilon /2$$

And $\epsilon$ can be arbitrary, so $A_1 = A_2$.

$\square$

Exercise 8.1.9

Explain why every function that is Riemann-integrable with $\int_{a}^{b}f = A$ must also have generalized Riemann integral $A$.

Proof: If $f$ is Riemann-integrable, then for any $\epsilon$, we can find $\delta$, any tagged partition $P$ is $\delta$-fine then $|R(f, P) - A| \leq \epsilon$.

Let this $\delta(x) = \delta$ be the gauge, we can see $f$ is generalized Riemann integrable.

$\square$

Theorem 8.1.8.

Dirichlet’s function $g(x)$ is generalized Riemann-integrable on $[0,1]$ with $\int_{0}^{1} g = 0$.

Proof: Let ϵ > 0. By Definition 8.1.6, we must construct a gauge $δ(x)$ on [0,1] such that whenever $(P,\{c_k\}^n_{k=1})$ is $\delta(x)$-fine tagged partition, it follows that

$$0 \leq \sum_{k=1}^{n} g(c_k)(x_k - x_{k-1}) < \epsilon$$

The gauge represents a restriction on the size of $Δx_k = x_k−x_{k−1}$ in the sense that $Δx_k < δ(c_k)$.

The Riemann sum consists of products of the form $g(c_k)Δx_k$. Thus, for irrational tags, there is nothing to worry about because $g(c_k) = 0$ in this case. Our task is to make sure that any time a tag $c_k$ is rational, it comes from a suitably thin subinterval.

Let $\{r_1,r_2,r_3,...\}$ be an enumeration of the countable set of rational numbers contained in $[0,1]$. For each $r_k$, set $δ(r_k) = ϵ/2^{k+1}$. For x irrational, set $δ(x) = 1$.

Exercise 8.1.10.

Show that if $(P,\{c_k\}^n_{k=1})$ is a $δ(x)$-fine tagged partition, then $R(g,P) < ϵ$.

Proof:

If $c_k$ is irrational, then $g(c_k)Δx_k = 0$. Otherwise, $g(c_k)Δx_k < δ(r_k) = ϵ/2^{k+1}$. So $\sum_{k=1}^{n} g(c_k)(x_k - x_{k-1}) < \epsilon$.

$\square$

Theorem 8.1.9.

Assume $F : [a,b] → \mathbf{R}$ is differentiable at each point in $[a,b]$, and set $f(x) = F'(x)$. Then, $f$ has the generalized Riemann integral

$$\int_{a}^{b} f = F(b) - F(a)$$

Proof. Let $P= {x_0,x_1,x_2,...,x_n}$ be a partition of $[a,b]$. Both this proof and the proof of Theorem 7.5.1 make use of the following fact.

Exercise 8.1.11.

Show that

$$F(b) - F(a) = \sum_{k=1}^{n} [F(x_k) - F(x_{k-1})]$$

Proof: It is obvious. $\square$

If $\{c_k\}^n_{k=1}$ is a set of tags for $P$, then we can estimate the difference between the Riemann sum $R(f,P)$ and $F(b)−F(a)$ by

$$|F(b) - F(a) - R(f, P)| = \left| \sum_{k = 1}^{n} [F(x_k) - F(x_{k-1}) - f(c_k)(x_k - x_{k-1})] \right| \\ \leq \sum_{k = 1}^{n} |F(x_k) - F(x_{k-1}) - f(c_k)(x_k - x_{k-1})|$$

Let $ϵ > 0$. To prove the theorem, we must construct a gauge $δ(c)$ such that

$$|F(b) - F(a) - R(f, P)| < ϵ$$

for all $(P,{c_k})$ that are $δ(c)$-fine. (Using the variable $c$ in the gauge function is more convenient than $x$ in this case.)

Exercise 8.1.12.

For each $c ∈ [a,b]$, explain why there exists a $δ(c) > 0$ (a $δ > 0$ depending on $c$) such that

$$\left| \frac{F(x) - F(c)}{x - c} - f(c) \right| < \epsilon$$

for all $0 < |x-c| < \delta (c)$.

Proof:

Since $F(x)$ is differential at $c$, that means

$$\lim_{x \to c} \frac{F(x) - F(c)}{x - c} = f(c)$$

i.e. given $\epsilon$, for $c$, we can find a $\delta (c)$, as long as $0 < |x-c| < \delta (c)$,

$$\left| \frac{F(x) - F(c)}{x - c} - f(c) \right| < \epsilon$$

$\square$

Exercise 8.1.13.

(a) For a particular $c_k ∈ [x_{k−1},x_k]$ of $P$, show that

$$|F(x_k)−F(c_k)−f(c_k)(x_k−c_k)| < ϵ(x_k−c_k)$$

and

$$|F(c_k)−F(x_{k−1})−f(c_k)(c_k−x_{k−1})| < ϵ(c_k−x_{k−1}).$$

Proof: Since $0 < x_k - c_k < \delta (c_k)$, then

$$\left| \frac{F(x_k) - F(c_k)}{x_k - c_k} - f(c_k) \right| < \epsilon$$

i.e.

$$|F(x_k)−F(c_k)−f(c_k)(x_k−c_k)| < ϵ(x_k−c_k)$$

The other part is the same.

$\square$

(b) Now, argue that

$$|F(x_k)−F(x_{k−1})−f(c_k)(x_k−x_{k−1})| < ϵ(x_k−x_{k−1})$$

and use this fact to complete the proof of the theorem.

Proof:

$$|F(x_k)−F(x_{k−1})−f(c_k)(x_k−x_{k−1})| =\\ \left| F(x_k)−F(c_k)−f(c_k)(x_k−c_k) + F(c_k)−F(x_{k−1})−f(c_k)(c_k−x_{k−1}) \right| \\ \leq |F(x_k)−F(c_k)−f(c_k)(x_k−c_k)| + |F(x_k)−F(c_k)−f(c_k)(x_k−c_k)| \\ \leq ϵ(x_k−c_k) + ϵ(c_k−x_{k−1})\\ = ϵ(x_k−x_{k−1})$$

Then we have

$$|F(b) - F(a) - R(f, P)| \\ \leq \sum_{k = 1}^{n} |F(x_k) - F(x_{k-1}) - f(c_k)(x_k - x_{k-1})| \\ \leq \sum_{k = 1}^{n} ϵ(x_k−x_{k−1}) \\ = \epsilon$$

Then we proved Theorem 8.1.9.

$\square$

If we consider the function

$$F(x) = \begin{cases} x^{3/2} \sin (1/x) &\text{if } x \not = 0\\ 0 &\text{if } x = 0\\ \end{cases}$$

then it is not too difficult to show that $F$ is differentiable everywhere, including $x = 0$, with

$$F'(x) = \begin{cases} (3/2) \sqrt[]{x} \sin (1/x) - (1/\sqrt[]{x}) \cos (1/x) &\text{if } x \not = 0\\ 0 &\text{if } x = 0\\ \end{cases}$$

What is notable here is that the derivative is unbounded near the origin. The theory of the ordinary Riemann integral begins with the assumption that we only consider bounded functions on closed intervals, but there is no such restriction for the generalized Riemann integral. Theorem 8.1.9 proves that $F'$ has a generalized integral.

Now, improper Riemann integrals have been created to extend Riemann integration to some unbounded functions, but it is another interesting fact about the generalized Riemann integral that any function having an improper integral must already be integrable in the sense described in Definition 8.1.6.

Theorem 8.1.10 (Change-of-variable Formula).

Let $g : [a,b] → \mathbf{R}$ be differentiable at each point of $[a,b]$, and assume $F$ is differentiable on the set $g([a,b])$. If $f(x) = F'(x)$ for all $x ∈ g([a,b])$, then

$$\int_{a}^{b} (f◦g)·g' = \int_{g(a)}^{g(b)} f.$$

Proof. The hypothesis of the theorem guarantees that the function $(F ◦g)(x)$ is differentiable for all $x ∈ [a,b]$.

Exercise 8.1.14.

(a) Why are we sure that $f$ and $(F◦g)'$ have generalized Riemann integrals?

Proof: This is because $F$ is differentiable and $f(x) = F'(x)$. Since $g$ is also differentialble, then $F◦g$ is differentialble. And $(F◦g)'$ is its derivative.

$\square$

(b) Use Theorem 8.1.9 to finish the proof.

Proof:

On the left side

$$(F◦g)' = (f◦g) \cdot g'$$

So

$$\int_{a}^{b} (f◦g)·g' = F(g(b)) - F(g(a))$$

On the right side

$$\int_{g(a)}^{g(b)} f = F(g(b)) - F(g(a))$$

$\square$

8.3 Euler’s Sum

In Section 6.1 we saw Euler’s first and most famous derivation of the formula

$$1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \frac{1}{25} + \cdots = \frac{\pi ^2}{6}$$

At the crux of this argument are two representations for the function $sin(x)$. The first is the standard Taylor series representation

$$\tag{1} \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots ,$$

and the second is an infinite product representation

$$\tag{2} \sin x = x \left( 1 - \frac{x}{\pi } \right) \left( 1 + \frac{x}{\pi } \right) \left( 1 - \frac{x}{2 \pi } \right) \left( 1 + \frac{x}{2 \pi } \right) \cdots .$$

Wallis’s Product

Exercise 8.3.1.

Supply the details to show that when $x = π/2$ the product formula in (2) is equivalent to

$$\tag{3} \frac{\pi }{2} =\lim_{n \to \infty} \left( \frac{2 \cdot 2}{1 \cdot 3} \right) \left( \frac{4 \cdot 4}{3 \cdot 5} \right) \left( \frac{6 \cdot 6}{5 \cdot 7} \right) \cdots \left( \frac{2n \cdot 2n}{(2n - 1) \cdot (2n + 1)} \right)$$

Proof:

We plug the $\frac{\pi}{2}$ into equation (2).

$\square$

Exercise 8.3.2.

Assume $h(x)$ and $k(x)$ have continuous derivatives on $[a,b]$ and derive the familiar integration-by-parts formula

$$\int_{a}^{b} h(t) k'(t) dt = h(b) k(b) - h(a) k (a) - \int_{a}^{b} h'(t) k(t) dt .$$

Proof: See exercise 7.5.6.

$\square$

Exercise 8.3.3.

(a) Using the simple identity $\sin^n(x) = \sin^{n−1}(x)\sin(x)$ and the previous exercise, derive the recurrence relation

$$b_n = \frac{n-1}{n} b_{n-2} \text{ for all } n \geq 2$$

Proof:

$$\int_{0}^{\frac{\pi}{2}} \sin^n(x) = \int_{0}^{\frac{\pi}{2}} \sin^{n-1} (x) \sin (x)\\ = \sin^{n-1} (\frac{\pi}{2}) (-\cos (\frac{\pi}{2})) - \sin^{n-1} (0) (-\cos (0)) - \int_{0}^{\frac{\pi}{2}} (n - 1) \sin ^{n-2} (x) \cos (x) (-\cos (x))\\ = (n - 1)\int_{0}^{\frac{\pi}{2}} \sin ^{n-2} (x) \cos ^2 (x) \\ = (n - 1) \int_{0}^{\frac{\pi}{2}} \sin ^{n-2} (x) - \sin ^{n} (x)$$

So we have

$$\int_{0}^{\frac{\pi}{2}} \sin^n(x) + (n - 1) \int_{0}^{\frac{\pi}{2}} \sin^n(x) = (n - 1) \int_{0}^{\frac{\pi}{2}} \sin^{n-2}(x)$$

So,

$$b_n = \frac{n-1}{n} b_{n-2} \text{ for all } n \geq 2$$

$\square$

(b) Use this relation to generate the first three even terms and the first three odd terms of the sequence $(b_n)$.

$$b_2 = \frac{1}{2} b_0 = \frac{\pi }{4}\\ b_4 = \frac{3}{4} b_2 = \frac{3 \pi }{16}\\ b_6 = \frac{5}{6} b_4 = \frac{15 \pi }{96}\\ b_3 = \frac{2}{3} b_1 = \frac{2}{3}\\ b_5 = \frac{4}{5} b_3 = \frac{8}{15}\\ b_7 = \frac{6}{7} b_4 = \frac{48}{105}\\ $$

$\square$

(c) Write a general expression for $b_{2n}$ and $b_{2n+1}$

$$b_{2n} = \frac{\pi}{2} \cdot \frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdots \cdot \frac{2n-1}{2n}$$

and

$$b_{2n+1} = 1 \cdot \frac{2}{3} \cdot \frac{4}{5} \cdot \frac{6}{7} \cdots \cdot \frac{2n}{2n+1}$$

Exercise 8.3.4.

Show

$$\lim_{n \to \infty} \frac{b_{2n}}{b_{2n+1}} = 1$$

and use this fact to finish the proof of Wallis’s product formula in (3).

Proof:

Because $b_{2n} \geq b_{2n+1} \geq b_{2n+2}$, so

$$1 \leq \frac{b_{2n}}{b_{2n+1}} \leq \frac{b_{2n}}{b_{2n+2}} = \frac{2n+2}{2n+1}$$

So

$$\lim_{n \to \infty} \frac{b_{2n}}{b_{2n+1}} \leq \lim_{n \to \infty} \frac{2n+2}{2n+1} = 1 \\ \lim_{n \to \infty} \frac{b_{2n}}{b_{2n+1}} \geq \lim_{n \to \infty} 1 = 1$$

Then it's easy to see product formula in (3) holds.

$\square$

Exercise 8.3.5.

Derive the following alternative form of Wallis’s product formula:

$$\sqrt[]{\pi} = \lim_{n \to \infty} \frac{ 2^{2n} (n!)^2 }{ (2n)! \sqrt[]{n} }$$

Proof:

We know

$$\frac{\pi }{2} =\lim_{n \to \infty} \left( \frac{2 \cdot 2}{1 \cdot 3} \right) \left( \frac{4 \cdot 4}{3 \cdot 5} \right) \left( \frac{6 \cdot 6}{5 \cdot 7} \right) \cdots \left( \frac{2n \cdot 2n}{(2n - 1) \cdot (2n + 1)} \right) \\ = \frac{ (2^{2n} (n!)^2)^2 }{ (2n!)^2 (2n+1) }$$

So

$$\pi = \lim_{n \to \infty} \frac{ (2^{2n} (n!)^2)^2 2 }{ (2n!)^2 (2n+1) }$$

So

$$\sqrt[]{\pi} = \lim_{n \to \infty} \frac{ 2^{2n} (n!)^2 }{ (2n)! \sqrt[]{n} }$$

$\square$

Taylor Series

Exercise 8.3.6.

Show that $1/\sqrt[]{1−x}$ has Taylor expansion $\sum_{n=0}^{∞} c_nx_n$, where $c_0 = 1$ and

$$c_n = \frac{ (2n)! }{2^{2n}(n!)^2} = \frac{ 1 \cdot 3 \cdot 5 \cdots (2n - 1) }{ 2 \cdot 4 \cdot 6 \cdots 2n }$$

for $n \geq 1$.

Solution:

$$f'(0) = \frac{1}{2} (1 - x)^{-3/2} = \frac{1}{2} \\ f''(0) = \frac{3}{4} (1-x)^{-5/2} = \frac{3}{4} \\ \cdots \\ f^{(n)}(0) = \frac{1 \cdot 3 \cdots(2n-1)}{2^n}$$

Then

$$c_n = \frac{f^{(n)}(0)}{n!} = \frac{ 1 \cdot 3 \cdot 5 \cdots (2n - 1) }{ 2 \cdot 4 \cdot 6 \cdots 2n }$$

$\square$

Exercise 8.3.7.

Show that $\lim c_n = 0$ but $\sum_{n=0}^{∞} c_n$ diverges.

Proof:

$$c_n = \frac{ 1 \cdot 3 \cdot 5 \cdots (2n - 1) }{ 2 \cdot 4 \cdot 6 \cdots 2n }$$

$c_n$ is a decreasing and bigger than $0$, so $c_n$ converges. if $(c_n) \rightarrow c > 0$, then

$$\lim_{n \to \infty} \frac{1}{c_n \sqrt[]{n}} \\ = \frac{1}{c} \times 0 = 0 \not = \sqrt[]{\pi}$$

Now note

$$c_n > \frac{ 1 \cdot 2 \cdot 4 \cdots (2n - 2) }{ 2 \cdot 4 \cdot 6 \cdots 2n } = \frac{1}{2n}$$

So we have

$$\sum_{k=1}^{n} c_n > \frac{1}{2} (1 + 1/2 + 1/4 + \cdots + 1/n)$$

So $\sum_{n=0}^{∞} c_n$ diverges.

$\square$

The divergence of $\sum_{n=0}^{∞} c_n$ makes sense when we consider the Taylor series for $1/$. We want to determine the values of $x$ for which

$$\tag{4} \frac{1}{\sqrt[]{1−x}} = \sum_{n = 1}^{\infty} c_nx^n,$$

Exercise 8.3.8.

Using the expression for $E_N(x)$ from Lagrange’s Remainder Theorem, show that equation (4) is valid for all $|x| < 1/2$. What goes wrong when we try to use this method to prove (4) for $x ∈ (1/2,1)$?

Solution:

The Lagrange's Remainder is

$$E_N(x) = \frac{f^{(N+1)}(c)}{(N+1)!}x^{N+1} = \frac{1 \cdot 3 \cdots (2N+1) }{2^{N+1} (N+1)!} \frac{x^{N+1}}{(1-c)^{N+1} \sqrt[]{1-c}}$$

Fix $x < 1/2$, then $c < x < 1/2$. So $1 - c > x$. $1 - c > 1/2$. So $\frac{1}{\sqrt[]{1-c}} < \frac{1}{\sqrt[]{2}}$.

$$\frac{x}{1-c} < 1$$

So $E_N(x) \rightarrow 0$.

When $x ∈ (1/2,1)$, we cannot guarantee

$$\frac{x}{1-c} < 1.$$

$\square$

Theorem 8.3.1 (Integral Remainder Theorem).

Let $f$ be differentiable $N + 1$ times on $(−R,R)$ and assume $f^{(N+1)}$ is continuous. Define $a_n = f^{(n)}(0)/n!$ for $n = 0,1,...,N$, and let

$$S_N(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \cdots + a_N x^N$$

For all $x ∈ (−R,R)$, the error function $E_N(x) = f(x)−S_N(x)$ satisfies

$$E_N(x) = \frac{1}{N!} \int_{0}^{x} f^{(N+1)}(t) (x - t)^N dt .$$

Proof. The case $x = 0$ is easy to check, so let’s take $x \not = 0$ in $(−R,R)$ and keep in mind that $x$ is a fixed constant in what follows. To avoid a few technical distractions, let’s just consider the case $x > 0$.

Exercise 8.3.9.

(a) Show

$$f(x) = f(0) + \int_{0}^{x} f'(t) dt$$

Proof: $f'(t)$ is differentiable, so it's continuous. So it's integrable. From theorem 7.5.1, then

$$\int_{0}^{x} f'(t) dt = f(x) - f(0)$$

$\square$

(b) Now use a previous result from this section to show

$$f(x) = f(0)+f'(0)x + \int_{0}^{x} f''(t) (x-t) dt .$$

Proof:

We use integration-by-parts, and let $h(x) = f'(x), k(x) = (x-t)$

$$-\int_{0}^{x} f'(t) dt = \int_{0}^{x} h(t) k'(t) dt \\ =h(x)k(x) - h(0)k(0) - \int_{0}^{x} h'(t) k(t) dt \\ = -f'(0) x- \int_{0}^{x} f''(t) (x-t) dt \\ $$

So

$$\int_{0}^{x} f'(t) dt = f'(0)x + \int_{0}^{x} f''(t) (x-t) dt$$

$\square$

(c) Continue in this fashion to complete the proof of the theorem.

Proof:

Let $h(t) = f^{(N)}(t), k(t) = -\frac{(x-t)^{N}}{N}$.

Consider

$$\int_{0}^{x} f^{(N)}(t) (x - t)^{N-1} dt = \int_{0}^{x} h(t) k'(t) dt = h(x)k(x) - h(0)k(0) - \int_{0}^{x} h''(t) k(t) dt \\ = \frac{1}{N} f^{(N)}(0) x^N + \frac{1}{N} \int_{0}^{x} f^{(N+1)}(t) (x-t)^N dt$$

So we have

$$E_N(x) = \frac{1}{N!} \int_{0}^{x} f^{(N+1)}(t) (x - t)^N dt .$$

$\square$

Exercise 8.3.10.

(a) Make a rough sketch of $1/\sqrt[]{1-x}$ and $S_2(x)$ over the interval $(−1,1)$, and compute $E_2(x)$ for $x = 1/2,3/4$, and $8/9$.

Solution:

$$a_0 = 1\\ a_1 = \frac{f'(0)}{1!} = \frac{1}{2} \\ a_2 = \frac{f''(0)}{2!} = \frac{1 \cdot 3}{2 \cdot 4} \\ $$

So $S_2(x) = 1 + \frac{1}{2}x + \frac{3}{8} x^2$.

Here are the python code to plot and compute $E_2(x)$.

import numpy as np
import matplotlib.pyplot as plt

# Define the range of x values, avoiding the point where the function is undefined
x = np.linspace(-1, 0.99, 400)  # 400 points between -1 and 0.99

# Compute the function values
y = 1 / np.sqrt(1 - x)
z = 1 + (1/2) * x + (3/8) * (x**2)
# Plot the function
plt.plot(x, y)
plt.plot(x, z)
plt.title('Plot of $1/\sqrt{1-x}$ and $1 + (1/2)x + (3/8)x^2$')
plt.xlabel('x')
plt.ylabel('1/sqrt(1-x)')
plt.grid(True)
plt.show()

x = np.array([1/2, 3/4, 8/9])
y = 1 / np.sqrt(1 - x)
z = 1 + (1/2) * x + (3/8) * (x**2)
e = y - z
print(e)

The $E_2(x)$ for $x = 1/2,3/4$, and $8/9$ are $[0.07046356, 0.4140625, 1.25925926]$

$\square$

(b) For a general $x$ satisfying $−1 < x < 1$, show

$$E_2(x) = \frac{15}{16} \int_{0}^{x} \left( \frac{x-t}{1-t} \right) ^ 2 \frac{1}{ (1-t)^{3/2} } dt$$

Proof:

$$E_2(x) = \frac{1}{2!} \int_{0}^{x} f^{(2+1)}(t) (x - t)^2 dt$$

$$f^{(1)}(t) = \frac{1}{2} (1-x)^{-\frac{3}{2}} \\ f^{(2)}(t) = \frac{1 \cdot 3}{2 \cdot 2} (1-x)^{-\frac{5}{2}} \\ f^{(3)}(t) = \frac{1 \cdot 3 \cdot 5}{2 \cdot 2 \cdot 2} (1-x)^{-\frac{7}{2}} \\ $$

So we proved.

$\square$

(c) Explain why the inequality

$$\left| \frac{x-t}{1-t} \right| \leq |x|$$

is valid, and use this to find an overestimate for $|E_2(x)|$ that no longer involves an integral. Note that this estimate will necessarily depend on $x$. Confirm that things are going well by checking that this overestimate is in fact larger than $|E_2(x)|$ at the three computed values from part (a).

Proof:

Assume $0 \leq t \leq x < 1$. Then $xt \leq t$, $x - xt \geq x - t$, so $\frac{x-t}{1-t} \leq x$

If $-1 < x \leq t \leq 0$, then $xt \geq t$ so $x - xt \leq x - t$, so $x \leq \frac{x-t}{1-t}$, i.e. $-x \geq -\frac{x-t}{1-t}$

In both cases, we proved it.

$$|E_2(x)| = \frac{15}{16} \left| \int_{0}^{x} \left( \frac{x-t}{1-t} \right) ^ 2 \frac{1}{ (1-t)^{3/2} } dt \right| \\ \leq \frac{15}{16} x^3 \frac{1}{(1-|x|)^{3/2}}$$

For $x = 1/2,3/4$, and $8/9$ are, the errors are

$$[0.3314563, 3.1640625, 17.77777778]$$

$\square$

(d) Finally, show $E_N(x) → 0$ as $N → ∞$ for an arbitrary $x ∈ (−1,1)$.

Proof:

We have

$$f^{(N+1)}(t) = \frac{1 \cdot 3 \cdots (2N+1) }{2^{N+1}} \cdot \frac{1}{(1-t)^{N}} \cdot \frac{1}{(1-t)^{3/2}}$$

So

$$E_N(x) = \frac{1}{N!} \int_{0}^{x} \frac{1 \cdot 3 \cdots (2N+1) }{2^{N+1}} \cdot \frac{1}{(1-t)^{N}} \cdot \frac{1}{(1-t)^{3/2}} (x-t)^N dt \\ \leq C_N \cdot (2N+1)|x|^N \frac{|x|}{(1-|x|)^{3/2}}$$

Since $C_N \rightarrow 0$ and $(2N+1)|x|^N \rightarrow 0$, then $E_N(x)$.

$\square$

The next step is to take the term-by-term anti-derivative of this series. Any time we start manipulating infinite series as though they were finite in nature we need to pause and make sure we are on solid footing.

Exercise 8.3.11.

Assuming that the derivative of $\arcsin (x)$ is indeed $1 / \sqrt[]{1−x^2}$ supply the justification that allows us to conclude

$$\tag{5} \arcsin (x) = \sum_{n = 0}^{\infty} \frac{c_n}{2n+1} x^{2n+1} \text{ for all } |x| < 1.$$

Proof: Since $\sum_{n = 0}^{\infty}c_nx^{2n}$ converges for $|x| < 1$, and $\frac{c_n}{2n+1} x^{2n+1} < c_nx^{2n}$, then $\sum_{n = 0}^{\infty} \frac{c_n}{2n+1} x^{2n+1}$ converges for $|x| < 1$.

Let

$$f(x) = \sum_{n = 0}^{\infty} \frac{c_n}{2n+1} x^{2n+1}$$

From theorem 6.5.7, $f(x)$ is differentiable and $f'(x) = \arcsin' (x)$. So $f(x) = \arcsin (x) + C$. Since $f(0) = \arcsin (0) = 0$, then $f(x) = \arcsin (x)$, for $|x| < 1$.

$\square$

Exercise 8.3.12.

Our work thus far shows that the Taylor series in (5) is valid for all $|x| < 1$, but note that $\arcsin(x)$ is continuous for all $|x| ≤ 1$. Carefully explain why the series in (5) converges uniformly to $\arcsin(x)$ on the closed interval $[−1,1]$.

Proof: Consider

$$f(1) = \sum_{n = 0}^{\infty} \frac{c_n}{2n+1}$$

Since

$$\lim_{n \to \infty} \frac{1}{c_n \sqrt[]{n}} = \sqrt[]{\pi }$$

Then

$$\lim_{n \to \infty} c_n \sqrt[]{n} = 1/\sqrt[]{\pi }$$

when $n$ is large, $c_n < \frac{1 + 1/\sqrt[]{\pi}}{\sqrt[]{n}} < \frac{2}{\sqrt[]{n}}$.

So

$$\frac{c_n}{2n+1} < \frac{2}{(2n+1)\sqrt[]{n}} \lt n^{-3/2}$$

So

$$\sum_{n = 0}^{\infty} \frac{c_n}{2n+1}$$

converges. Since it's all positive, it converges absolutely. Then (5) also converges in $-1$.

$\square$

Summing $\sum_{n = 1}^{\infty} 1/n^2$

Every proof of Euler’s sum contains a moment of genuine ingenuity at some point, and this is where our proof takes an unanticipated turn.

Let’s make the substitution $x = \sin(θ)$ in (5) where we restrict our attention to $−π/2 ≤ θ ≤ π/2$. The result is

$$\theta = \arcsin (\sin (\theta) ) = \sum_{n = 1}^{\infty} \frac{c_n}{2n+1} \sin ^{2n+1} (\theta )$$

which converges uniformly on $[−π/2,π/2]$.

Exercise 8.3.13.

(a) Show

$$\int_{0}^{\pi /2} \theta d \theta = \sum_{n = 1}^{\infty} \frac{c_n}{2n+1} b_{2n+1}$$

being careful to justify each step in the argument. The term $b_{2n+1}$ refers back to our earlier work on Wallis’s product.

Proof: let

$$f_n(\theta ) = \frac{c_n}{2n+1} \sin ^{2n+1} (\theta ) \\ f(\theta ) = \theta$$

So $\sum_{n = 1}^{\infty}f_n(\theta ) \rightarrow f(\theta )$ uniformly, then

$$\int_{0}^{\pi /2} \theta d \theta = \sum_{n = 1}^{\infty} \int_{0}^{\pi /2} f_n(\theta) = \sum_{n = 1}^{\infty} \frac{c_n}{2n+1} \int_{0}^{\pi /2} \sin ^{2n+1} (\theta ) \\ = \sum_{n = 1}^{\infty} \frac{c_n}{2n+1} b_{2n+1}$$

$\square$

(b) Deduce

$$\frac{\pi ^2}{8} = \sum_{n = 0}^{\infty} \frac{1}{(2n+1)^2},$$

and use this to finish the proof that $\frac{\pi ^2}{6} = \sum_{n = 1}^{\infty} 1/n^2$.

Proof:

$$c_n = \frac{ 1 \cdot 3 \cdot 5 \cdots (2n - 1) }{ 2 \cdot 4 \cdot 6 \cdots 2n }$$

$$b_{2n+1} = 1 \cdot \frac{2}{3} \cdot \frac{4}{5} \cdot \frac{6}{7} \cdots \cdot \frac{2n}{2n+1}$$

So

$$\frac{c_n}{2n+1} b_{2n+1} = \frac{1}{(2n+1)^2}$$

On the left side, $\int_{0}^{\pi /2} \theta d \theta = \frac{1}{2}(\frac{\pi}{2})^2 - 0 = \frac{\pi ^2}{8}$.

So we proved the first part.

Let

$$A = \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \cdots$$

Then

$$A = (\frac{1}{1^2} + \frac{1}{3^2} + \cdots) + (\frac{1}{2^2} + \frac{1}{4^2} + \cdots) \\ = \frac{\pi^2}{8} + \frac{1}{4} A$$

So $A = \frac{\pi ^2}{6}$.

$\square$

8.4 Inventing the Factorial Function

Exercise 8.4.2.

Verify that the series converges absolutely for all $x ∈ R$, that $E(x)$ is differentiable on $R$, and $E'(x) = E(x)$.

Proof: Fix any $x \in \mathbf{R}$, we can find $N > 2x$. Let $a_n = \frac{x^n}{n!}$ and

$$A = \left| \frac{x^N}{N!} \right|$$

For $n > N$, $|a_n| < \frac{A}{2^{n-N}}$. So $\sum_{k = 1}^{\infty}|a_n|$ converges absolutely. Then we can apply Theorem 6.5.7 to know that $E(x)$ is differentiable on $R$, and $E'(x) = E(x)$.

$\square$

Exercise 8.4.3.

(a) Use the results of Exercise 2.8.7 and the binomial formula to show that $E(x+y) = E(x)E(y)$ for all $x,y ∈ \mathbf{R}$.

Proof: Fix $x$ and $y$. Let

$$a_n = \frac{x^n}{n!} \\ b_n = \frac{y^n}{n!} \\ $$

Then note

$$d_k = a_{0}b_{k} + a_{1}b_{k-1} + \cdots + a_{k-1}b_{1} + a_k b_0 \\ = \frac{y^k}{0!k!} + \frac{xy^{k-1}}{1!(k-1)!} + \cdots + \frac{x^{k-1}y}{(k-1)!1!} + \frac{x^k}{k!0!} \\ = \frac{(x+y)^k}{k!}$$

Use the results of Exercise 2.8.7, $\sum_{k = 1}^{\infty}d_k$ converges and it converges $E(x)E(y)$.

On the other hand, $\sum_{k = 1}^{\infty}d_k = E(x+y)$.

So $E(x+y) = E(x)E(y)$.

$\square$

(b) Show that $E(0) = 1, E(−x) = 1/E(x)$, and $E(x) > 0$ for all $x ∈ \mathbf{R}$.

Proof:

$$E(0) = 1 + \frac{0}{1!} + \frac{0^2}{2!} + \cdots = 1$$

Also

$$1 = E(0) = E(x + (-x)) = E(x)E(-x)$$

So $E(-x)=1/E(x)$.

Also since if $x > 0$, then $a_n = \frac{x^n}{n!} > 0$, then $E(x) > 0$.

$\square$

Exercise 8.4.4.

Define $e = E(1)$. Show $E(n) = e^n$ and $E(m/n) = (\sqrt[n]{e})^m$ for all $m,n ∈ Z$.

Proof:

$$E(n) = E(1+1+\cdots +1) = E(1)\cdot E(1) \cdots E(1)\\ = e^n$$

We also have

$$e = E(1) = E(1/n + \cdots +1/n) = E(1/n)^n$$

So $E(1/n) = \sqrt[n]{e}$. So $E(m/n) = (\sqrt[n]{e})^n$.

$\square$

Definition 8.4.1.

Given $f : [a,∞] → R$, we say that $\lim_{x \to \infty} f(x) = L$ if for all $ϵ > 0$, there exists $M > a$ such that whenever $x ≥ M$ it follows that

$$|f(x) - L| < \epsilon$$

Exercise 8.4.5.

Show $\lim_{x \to \infty} x^ne^{−x} = 0$ for all $n = 0,1,2,...$.

Proof:

$$\frac{x^n}{e^x} = \frac{x^n}{1 + \frac{x}{1} + \frac{x^2}{2!}+ \cdots + \frac{x^{n+1}}{(n+1)!} + \cdots} \leq \frac{(n+1)!}{x} \rightarrow 0$$

$\square$

Exercise 8.4.6.

(a) Explain why we know $e^x$ has an inverse function—let’s call it $\log x$—defined on the strictly positive real numbers and satisfying

(i) $\log (e^y) = y$ for all $y \in \mathbf{R}$ and

(ii) $e^{\log x} = x$, for all $x > 0$.

(b) Prove $(\log x)' = 1/x$. (See Exercise 5.2.12.)

Proof:

Let $y = e^x$ $$\log'y = \frac{1}{(e^x)'} = \frac{1}{e^x} = 1/y$$

$\square$

(c) Fix $y > 0$ and differentiate $\log(xy)$ with respect to $x$. Conclude that

$$\log_{} (xy) = \log x + \log y \text{ for all } x, y > 0.$$

Proof: Also see exercise Exercise 7.5.8 (Natural Logarithm and Euler’s Constant).

$$\log_{}' (xy) = \frac{1}{xy} \cdot y = \frac{1}{x}$$

Since $\log_{}' x = \frac{1}{x}$, so

$$\log xy = \log x + C$$

Since $\log 1 = 0$, we can plug $x = 1$ and get $\log y = C$.

$\square$

(d) For $t > 0$ and $n ∈ N$, $t_n$ has the usual interpretation as $t·t···t$ ($n$ times). Show that

$$\tag{2} t^n = e^{n \log_{} t} \text{ for all } n \in \mathbf{N}$$

Proof:

From (c), we know

$$n \log_{} t = \log_{} t^n$$

And from (a), we know

$$e^ {\log_{} t^n } = t^n$$

So (2) holds.

$\square$

Definition 8.4.2.

Given $t > 0$, define the exponential function $t^x$ to be

$$t^x = e^{x \log_{} t} \text{ for all } x \in \mathbf{R}$$

Exercise 8.4.7.

(a) Show $t^{m/n} = (\sqrt[n]{t})^m$ for all $m, n \in \mathbf{N}$.

Proof:

$$(\sqrt[n]{t})^m = e^{m \log_{} \sqrt[n]{t}}$$

Let $p = \log_{} \sqrt[n]{t}$, then $e^p = \sqrt[n]{t}$, $e^{pn} = t$, i.e. $n \log_{} e^p = \log_{} t$, i.e. $\log_{} e^p = \frac{1}{n} \log_{} t$, i.e. $\log_{} \sqrt[n]{t} = \frac{1}{n} \log_{} t$.

So

$$e^{m \log_{} \sqrt[n]{t}} = e^{\frac{m}{n} \log_{} t}\\ = t^{\frac{m}{n}}$$

$\square$

(b) Show $\log(t^x) = x\log t$, for all $t > 0$ and $x ∈ \mathbf{R}$.

Proof: Note

$$e^{x\log t} = t^x \\ e^{\log(t^x)} = t^x$$

So $\log(t^x) = x\log t$.

$\square$

(c) Show $t^x$ is differentiable on $\mathbf{R}$ and find the derivative.

Proof: Because $t^x = e^{x \log_{} t}$, let

$$g(x) = x \log_{} t$$

which is differentiable, and let $f(x) = e^x$ which is also differentiable, then their composite function $f \circ g$ is also differentiable.

$$(e^{x \log_{} t})' = e^{x \log_{} t} \log_{} t = t^x \log_{} t$$

$\square$

Exercise 8.4.8.

Inspired by the fact that $0! = 1$ and $1! = 1$, let $h(x)$ satisfy

(i) $h(x) = 1$ for all $0 ≤ x ≤ 1$, and

(ii) $h(x) = xh(x−1)$ for all $x ∈ \mathbf{R}$.

(a) Find a formula for $h(x)$ on $[1,2]$, $[2,3]$, and $[n,n + 1]$ for arbitrary $n \in \mathbf{N}$.

$$h(x) = x h(x - 1) = x, x \in [1,2] \\ h(x) = x h(x - 1) = x(x-1), x \in [2,3] \\ h(x) = x h(x - 1) = x(x-1) \cdots 1, x \in [n, n+1] \\ $$

$\square$

(b) (c) skipped

Improper Riemann Integrals

Definition 8.4.3.

Assume $f$ is defined on $[a,∞)$ and integrable on every interval of the form $[a,b]$. Then define $\int_{a}^{\infty }f$ to be

$$\lim_{b \to \infty} \int_{a}^{b}f,$$

provided the limit exists. In this case we say the improper integral $\int_{a}^{\infty }f$ converges.

Exercise 8.4.9.

(a) Show that the improper integral $\int_{a}^{\infty }f$ converges if and only if, for all $ϵ > 0$ there exists $M > a$ such that whenever $d > c \geq M$ it follows that

$$\left| \int_{c}^{d} f \right| < \epsilon$$

(In one direction it will be useful to consider the sequence $a_n = \int_{a}^{a+n} f$.

Proof:

$\Rightarrow$ Assume $\int_{a}^{\infty }f$ converges, and assume $\lim_{b \to \infty} \int_{a}^{b}f = B$, then we can find $M$, if $b \geq M$, then

$$\left| \int_{a}^{b} f - B \right| < \epsilon / 2$$

Consider $d > c \geq M$,

$$\left| \int_{c}^{d} f \right| = \left| \int_{a}^{d} f - \int_{a}^{c} f \right| \leq \left| \int_{a}^{d} f - B \right| + \left| B - \int_{a}^{c} \right| < \epsilon / 2 + \epsilon / 2 = \epsilon$$

$\Leftarrow$ Consider $a_n = \int_{a}^{a+n} f$ then $a_n$ is Cauchy sequence and assume $\lim_{n \to \infty} a_n = B$.

For $\epsilon$, we can find $M_1$, such that $d > c \geq M_1$,

$$\left| \int_{c}^{d} f \right| < \epsilon / 2$$

We can also find $M_2$, if $a+n > M_2$, then $|a_n - B| < \epsilon / 2$. Then let $M = \max \{M_1, M_2\}$.

If $b > M$, and find $n$ such that $a+n > M$ then

$$\left| \int_{a}^{b} f - B \right| = \left| \int_{a}^{b} f - \int_{a}^{a+n} f + \int_{a}^{a+n} f - B \right| < \\ \left| \int_{a}^{b} f - \int_{a}^{a+n} f \right| + \left| \int_{a}^{a+n} f - B \right| < \\ \epsilon / 2 + \epsilon / 2 = \epsilon$$

So

$$\lim_{b \to \infty} \int_{a}^{b} f = B$$

$\square$

(b) Show that if $0 ≤ f ≤ g$ and $\int_{a}^{b}g$ converges then $\int_{a}^{b}f$ converges.

Proof: Since $\int_{a}^{b}g$ converges, then we can find $M$, if $d > c \geq M$ then $|\int_{c}^{d} g | = \int_{c}^{d} g < \epsilon$. Since $0 ≤ f ≤ g$, then $|\int_{c}^{d} f | = \int_{c}^{d} f \leq \int_{c}^{d} g < \epsilon$

$\square$

(c) Part (a) is a Cauchy criterion, and part (b) is a comparison test. State and prove an absolute convergence test for improper integrals.

Proof: The absolute convergence test for improper integrals is: if $f$ is integrable on any interval $[a,b]$ and $\lim_{b \to \infty} \int_{a}^{b} |f|$ converges, then $\lim_{b \to \infty} \int_{a}^{b} f$ converges.

Since $f$ is integrable on any interval $[a,b]$, then from Theorem 7.4.2. (v), $|f|$ is also integrable on any interval $[a,b]$. Since $\lim_{b \to \infty} \int_{a}^{b} f$ converges, we can find a $M$ such that $d > c \geq M$, then

$$\int_{c}^{d} |f| < \epsilon$$

Again from Theorem 7.4.2. (v)

$$\left| \int_{c}^{d} f \right| \leq \int_{c}^{d} |f| < \epsilon$$

Then from part (a) $\lim_{b \to \infty} \int_{a}^{b} f$ converges.

$\square$

Exercise 8.4.10.

(a) Use the properties of $e^t$ previously discussed to show

$$\int_{0}^{\infty} e^{-t} dt = 1$$

Proof: $e^t$ is continuous on $[0,b]$, so it's integrable. Then we can use the first part of theorem 7.5.1, let

$$F(t) = - e^{-t}$$

and $F'(t) = e^{-t}$, so

$$\int_{0}^{\infty} e^{-t} dt = \lim_{b \to \infty} F(b) - F(0) = 1 - e^{-b} = 1$$

$\square$

(b) Show

$$\tag{3} \frac{1}{\alpha } = \int_{0}^{\infty}e^{-\alpha t} dt, \text{ for all } \alpha > 0.$$

Proof:

$e^{-\alpha t}$ is continuous on $[0,b]$, so it's integrable.

Let

$$F(t) = - \frac{1}{\alpha} e^{-\alpha t}$$

and $F'(t) = f(t)$.

$$\int_{0}^{\infty} e^{-\alpha t} dt = \lim_{b \to \infty} F(b) - F(0) = \frac{1}{\alpha}$$

$\square$

Exercise 8.4.11.

(a) Evaluate $\int_{0}^{b} t e^{-\alpha t}$ using the integration-by-parts formula from Exercise 7.5.6. The result will be an expression in α and b.

Solution:

Let

$$h(t) = - \frac{1}{\alpha } e^{-\alpha t} \\ k(t) = t$$

Then

$$ \int_{0}^{b} t e^{-\alpha t} = \int_{0}^{b} k(t) h'(t) = h(b) k(b) - h(0) k(0) - \int_{0}^{b} k'(t) h(t) = \ - \frac{1}{\alpha } e^{-\alpha b} \cdot b -

\int_{0}^{b} -\frac{1}{\alpha } e^{-\alpha t} dt \

\frac{1}{\alpha } \int_{0}^{b} e^{-\alpha t} dt - \frac{b}{\alpha } e^{-\alpha b} \ = \frac{1}{\alpha } (\frac{1}{\alpha} - \frac{1}{\alpha } e^{-\alpha b}) - \frac{b}{\alpha } e^{-\alpha b} $$

$\square$

(b) Now compute

$$\int_{0}^{\infty} t e^{-\alpha t}$$

and verify equation (4).

$$\tag{4} \frac{1}{\alpha ^2} = \int_{0}^{\infty} t e^{-\alpha t} dt$$

Solution:

$$\lim_{b \to \infty} \frac{1}{\alpha } (\frac{1}{\alpha} - \frac{1}{\alpha } e^{-\alpha b}) - \frac{b}{\alpha } e^{-\alpha b} = \frac{1}{\alpha ^2}$$

$\square$

Differentiating Under the Integral

• Our sense of distance between points $(x_0,t_0)$ and $(x,t)$ with the familiar Euclidean distance formula

$$\| (x,t) - (x_0,t_0) \| = \sqrt[]{(x - x_0^2) + (t - t_0)^2}$$

Definition 8.4.4.

A function $f : D → \mathbf{R}$ is continuous at $(x_0,t_0)$ if for all $ϵ > 0$, there exists $δ > 0$ such that whenever $\| (x,t) - (x_0,t_0) \| < \delta$, it follows that

$$\left| f(x, t) - f(x_0, t_0) \right| < \epsilon.$$

Exercise 8.4.12.

Assume the function $f(x,t)$ is continuous on the rectangle $D = \{(x,t) : a \leq x \leq b, c \leq t \leq d\}$. Explain why the function

$$F(x) = \int_{c}^{d} f(x, t)dt$$

is properly defined for all $x \in [a,b]$.

Proof: Fix $x$, $f(x, t)$ is a continuous function defined on $[c,d]$. So it is integrable. So

$$\int_{c}^{d} f(x, t)dt$$

exists.

$\square$

It should not be too surprising that Theorem 4.4.7 has an analogue in the $\mathbf{R}^2$ setting. The set $D$ is compact in $\mathbf{R}^2$, and a continuous function on $D$ is uniformly continuous in the sense that the $δ$ in Definition 8.4.4 can be chosen independently of the point $(x_0,t_0)$.

Theorem 8.4.5.

If $f(x,t)$ is continuous on $D$, then $F(x) = \int_{c}^{d} f(x,t) dt$ is uniformly continuous on $[a,b]$.

Exercise 8.4.13.

Prove Theorem 8.4.5.

Proof: We will prove $F(x)$ is continuous first. Given $x_0$ in $[a, b]$ and $\epsilon > 0$, since $f(x,t)$ is uniformly continuous on $D$, we can find $\delta$, as long as $x \in U_{\delta}(x_0)$, $\left| f(x, t) - f(x_0, t) \right| < \frac{\epsilon}{d-c}$ for all $t \in [c,d]$.

So

$$|F(x) - F(x_0)| = \left| \int_{c}^{d} f(x, t) - \int_{c}^{d} f(x_0, t) \right| \leq \int_{c}^{d} \left| f(x, t) - f(x_0, t) \right| \\ \leq \frac{\epsilon }{d-c} (d-c) = \epsilon$$

Since $F(x)$ is continuous then $F(x)$ is uniformly continuous.

$\square$

Taking inspiration from equations (3) and (4), let’s add the assumption that for each fixed value of $t$ in $[c,d]$, the function $f(x,t)$ is a differentiable function of $x$; that is,

$$f_x(x, t) = \lim_{z \to x} \frac{ f(z,t) - f(x,t) }{z-x}$$

exists for all $(x,t) ∈ D$. In addition, let’s assume that the derivative function $f_x(x,t)$ is continuous.

Theorem 8.4.6.

If $f(x,t)$ and $f_x(x,t)$ are continuous on $D$, then the function $F(x) = \int_{c}^{d} f(x,t) dt$ is differentiable and

$$F'(x) = \int_{c}^{d} f_x(x,t) dt.$$

Proof. Fix $x$ in $[a,b]$ and let $ϵ > 0$ be arbitrary. Our task is to find a $δ > 0$ such that

$$\tag{5} \left| \frac{F(z) - F(x)}{z-x} - \int_{c}^{d} f_x(x,t)dt \right| < \epsilon$$

whenever $0 < |z-x| < \delta$.

Exercise 8.4.14.

Finish the proof of Theorem 8.4.6

Proof:

Because

$$\left| \frac{F(z) - F(x)}{z-x} - \int_{c}^{d} f_x(x,t)dt \right| \\ = \left| \frac{\int_{c}^{d} f(z,t) - \int_{c}^{d} f(x,t)}{z-x} - f_x(x,t) \right| \\ = \left| \int_{c}^{d} \frac{f(z,t) - f(x,t)}{z-x} - f_x(x,t) \right| \\ \leq \int_{c}^{d} \left| \frac{f(z,t) - f(x,t)}{z-x} - f_x(x,t) \right| < \frac{\epsilon}{d-c}(d-c) = \epsilon$$

$\square$

Improper Integrals, Revisited

Theorem 8.4.6 is a formal justification for differentiating under the integral sign, but we need to extend this result to the case where the integral is improper. Looking back one more time to our motivating example in equation (3), we see that what we have is a function $f(x,t)$ where the domain of the variable $t$ is the unbounded interval $c ≤ t < ∞$.

Let’s fix $x$ from some set $A ⊆ R$. For such an $x$, we define

$$\tag{6} F(x) = \int_{c}^{\infty} f(x,t)dt = \lim_{d \to \infty} \int_{c}^{d} f(x,t)dt,$$

provided the limit exists.

As we have seen on numerous occasions, the elixir required to ensure that good behavior in the finite setting extends to the infinite setting is uniformity.

Definition 8.4.7.

Given $f(x,t)$ defined on $D= \{(x,t) : x ∈ A,c ≤ t\}$, assume $F(x) = \int_{c}^{\infty}f(x,t)dt$ exists for all $x ∈ A$. We say the improper integral converges uniformly to $F(x)$ on $A$ if for all $ϵ > 0$, there exists $M > c$ such that

$$\left| F(x) - \int_{c}^{d} f(x,t) dt \right| < \epsilon$$

for all $d ≥ M$ and all $x ∈ A$.

Exercise 8.4.15.

(a) Show that the improper integral $\int_{0}^{\infty} e^{-xt} dt$ converges uniformly to $1/x$ on the set $[1/2,∞)$.

Proof:

$$\int_{0}^{b} e^{-xt} dt = 1 - \frac{1}{x} e^{-xb} \geq 1 - 2 e ^{-b/2}$$

It does not depend on $x$.

$\square$

(b) Is the convergence uniform on $(0,∞)$?

Solution: No. For any fixed $x$, $\int_{0}^{\infty} e^{-xt} dt = 1$, but give any $b$, let $x = 1/b$

$$\int_{0}^{b} e^{-xt} dt = 1 - b/e$$

$\square$

Exercise 8.4.16.

Prove the following analogue of the Weierstrass M-Test for improper integrals: If $f(x,t)$ satisfies $|f(x,t)| ≤ g(t)$ for all $x ∈ A$ and $\int_{a}^{\infty} g(t)dt$ converges, then $\int_{a}^{\infty} f(x,t)dt$ converges uniformly on $A$.

Proof:

Given $\epsilon$, since $\int_{a}^{\infty} g(t)dt$ converges, we can find $M$, if $d > c \geq M$, we have $|\int_{c}^{d} g(t) dt | = \int_{c}^{d} |g(t)| dt < \epsilon / 2$.

Then

$$\left| \int_{c}^{d} f(x,t) dt \right| \leq \int_{c}^{d} |f(x,t)| dt \leq \int_{c}^{d} g(t) dt < \epsilon / 2$$

Then given any $x_0$, we can find $d_{x_0}$, such that

$$\left| \int_{a}^{d_{x_0}} f(x_0, t) dt - F(x_0) \right| < \epsilon / 2$$

Then

$$\left| \int_{a}^{c} f(x_0, t) dt - F(x_0) \right| \leq \left| \int_{a}^{c} f(x_0, t) dt - \int_{a}^{d_{x_0}} f(x_0, t) dt \right| + \left| \int_{a}^{d_{x_0}} f(x_0, t) dt - F(x_0) \right| \\ \leq \epsilon / 2 + \epsilon / 2 = \epsilon$$

$\square$

An immediate consequence of Definition 8.4.7 is that if the improper integral converges uniformly then the sequence of functions defined by

$$F_n(x) = \int_{c}^{c+n} f(x,t) dt$$

converges uniformly to $F(x)$ on $[a,b]$. This observation gives us access to the host of useful results we developed in Chapter 6.

Theorem 8.4.8.

If $f(x,t)$ is continuous on $D= \{(x,t) : a ≤ x ≤ b,c ≤ t\}$, then

$$F(x) = \int_{c}^{\infty} f(x,t)dt$$

is uniformly continuous on $[a,b]$, provided the integral converges uniformly.

Proof:

Let

$$F_n(x) = \int_{c}^{c+n} f(x,t) dt$$

From theorem 8.4.5., then $F_n(x)$ is continuous. Since the integral converges uniformly, then $F_n(x)$ converges to $F(x)$ uniformly. Then according to theorem 6.2.6, $F(x)$ is continuous.

$\square$

Theorem 8.4.9.

Assume the function $f(x,t)$ is continuous on $D= \{(x,t) : a ≤ x ≤ b,c ≤ t\}$ and $F(x) = \int_{c}^{\infty} f(x,t)dt$ exists for each $x ∈ [a,b]$. If the derivative function $f_x(x,t)$ exists and is continuous, then

$$\tag{7} F'(x) = \int_{c}^{\infty} f_x(x,t) dt,$$

provided the integral in (7) converges uniformly.

Exercise 8.4.18.

Prove Theorem 8.4.9.

Proof:

Let $G(x) = \int_{c}^{\infty} f_x(x,t) dt$.

Again, consider

$$F_n(x) = \int_{c}^{c+n} f(x,t) dt \text{ and } G_n(x) = \int_{c}^{c+n} f_x(x,t) dt$$

And let $D_n = \{(x,t) : a ≤ x ≤ b,c ≤ t ≤ c+n\}$.

Since $f(x,t)$ and $f_x(x,t)$ are continuous on $D_n$, then from theorem 8.4.6, $F_n'(x) = G_n(x)$.

Since $F_n(x)$ converges to $F(x)$ and $G_n(x)$ converges uniformly to $G(x)$, then from theorem 6.3.1, $F(x)$ is differentiable and $F'(x) = G(x)$.

$\square$

The Factorial Function

It’s time to return our attention to equation (3) from earlier in this section:

$$\frac{1}{\alpha } = \int_{0}^{\infty}e^{-\alpha t} dt, \text{ for all } \alpha > 0.$$

Exercise 8.4.19.

(a) Although we verified it directly, show how to use the theorems in this section to give a second justification for the formula

$$\frac{1}{\alpha ^2} = \int_{0}^{\infty} t e^{-\alpha t} dt, \text{ for all } \alpha > 0.$$

Proof: Consider $f(x,t) = e^{-xt}$, and $D = \{(x,t) : 1/2 ≤ x ≤ 3/2, 0 ≤ t\}$.

$$f_x(x,t) = -t e^{-xt}$$

is also continuous on $D$.

$$ \int_{0}^{b} t e^{-x t} = \int_{0}^{b} k(t) h'(t) = h(b) k(b) - h(0) k(0) - \int_{0}^{b} k'(t) h(t) = \ - \frac{1}{x } e^{-x b} \cdot b -

\int_{0}^{b} -\frac{1}{x } e^{-x t} dt \

\frac{1}{x } \int_{0}^{b} e^{-x t} dt - \frac{b}{x } e^{-x b} \ = \frac{1}{x } (\frac{1}{x} - \frac{1}{x } e^{-x b}) - \frac{b}{x } e^{-x b} \ = \frac{1}{x^2} - \frac{1}{x^2} e^{-x b} - \frac{b}{x } e^{-x b} $$

So

$$\int_{0}^{\infty} f_x(x, t) dt$$

converges uniformly. We can apply theorem 8.4.9:

$$-\frac{1}{x^2} = F'(x) = \int_{0}^{\infty} f_x(x,t) dt = \int_{0}^{\infty} -t e^{-xt} dt$$

So,

$$\frac{1}{x^2} = \int_{0}^{\infty} t e^{-xt} dt$$

Then we can plug $\alpha$ back.

$\square$

(b) Now derive the formula

$$\tag{8} \frac{n!}{\alpha ^ {n+1}} = \int_{0}^{\infty}t^n e^{-\alpha t} dt \text{ for all } \alpha > 0.$$

Proof:

First, we prove $\int_{0}^{\infty}t^n e^{- x t} dt$ for any $0 < a \leq x \leq b$.

$$\left| t^n e^{- x t} \right| = t^n e^{- x t} \leq t^n e^{- a t}$$

Since

$$e^{at} = 1 + (at) + \frac{(at)^2}{2!} + \cdots + \frac{(at)^{n+2}}{(n+2)!} + \cdots > \frac{(at)^{n+2}}{(n+2)!}$$

So

$$t^n e^{- a t} \leq g(t) = \begin{cases} 1 &\text{if } 0 \leq t \leq 1\\ \frac{(n+2)!}{a^{n+2} t^2} &\text{if } t > 1\\ \end{cases}$$

The improper integral $\lim_{t \to \infty} g(t)$ converges. So from exercise 8.4.16, $\int_{0}^{\infty}t^n e^{- x t} dt$ converges uniformly.

We can use induction, assume the improper integral holds

$$\frac{(n-1)!}{x^{n}} = \int_{0}^{\infty} t^{n-1} e^{-xt} dt$$

Then $f(x,t)$ is continuous, $f_x(x,t) = -t^n e^{-xt}$ continuous and $\int_{0}^{\infty}f_x(x,t) dt$ converges uniformly.

So we have

$$\frac{n!}{x^{n+1}} = \int_{0}^{\infty} t^n e^{-xt} dt$$

$\square$

Definition 8.4.10.

For $x ≥ 0$, define the factorial function

$$x! = \int_{0}^{\infty} t^x e^{-t} dt .$$

Exercise 8.4.20.

(a) Show that $x!$ is an infinitely differentiable function on $(0,∞)$ and produce a formula for the nth derivative. In particular show that $(x!)'' > 0$.

Proof: Let $f(x,t) = t^x e^{-t}$ . Given $x$, since $t^x$ and $e^{-t}$ are continuous functions on $[0, c]$, so they are integrable.

Also, if $0 < a \leq x \leq b$, we can find $k$, such that $0 < 1/k < a \leq x \leq b < k$. then for $t > 1$,

$$t^x e^{-t} = \frac{ t^x }{1 + \frac{t}{1!} + \frac{t^2}{2!} + \cdots + \frac{t^{k+2}}{(k+2)!} + \cdots} \\ < \frac{(k+2)!}{t^2}$$

define $g(x)$ as

$$g(t) = \begin{cases} 1 &\text{if } 0 \leq t \leq 1\\ \frac{(k+2)!}{t^2} &\text{if } t > 1\\ \end{cases}$$

So

$$|t^x e^{-t}| = t^x e^{-t} \leq g(x)$$

Since the improper integral $\lim_{t \to \infty} g(t)$ converges, then $f(x,t)$ converges uniformly.

Now, let's consider the nth derivative of $f(x,t)$ w.r.t $x$:

$$f_x^{(n)}(x, t) = t^x (\log_{} (t))^n e^{-t}$$

Note that $f_x^{(n)}(x, t)$ is not defined at $t = 0$. But we can use L’Hospital’s Rule: ∞/∞ case:

$$\lim_{t \to 0} \frac{(\log_{} (t))^n}{t^{-x}} = \lim_{t \to 0} \frac{n (\log_{} (t))^{n-1} (1/t)}{(-x)t^{-1-x}} \\ = \lim_{t \to 0} \frac{n}{(-x)} \frac{\log_{} (t))^{n-1}}{t^{-x}} \\ \lim_{t \to 0} = \frac{n!}{(-x)^n} \frac{1}{t^{-x}} = 0$$

Thus we can define $f_x^{(n)}(x, 0) = 0$.

When $x \in [a,b]$, and $t > 1$

$$\left| t^x (\log_{} (t))^n e^{-t} \right| = t^x (\log_{} (t))^n e^{-t} \\ \leq t^b t^n e^{-t} \leq \frac{C}{t^2}$$

where $C$ is a constant independent of $x$. So the improper integral $\int_{0}^{\infty} t^x (\log_{} (t))^n e^{-t}$ converges uniformly.

So the nth derivative is

$$\int_{0}^{\infty} t^x (\log_{} (t))^n e^{-t} dt$$

And the 2nd derivative is

$$\int_{0}^{\infty} t^x (\log_{} (t))^2 e^{-t} dt$$

For any $x > 0$, $t^x (\log_{} (t))^2 e^{-t} >0$, so $(x!)'' > 0$.

$\square$

(b) Use the integration-by-parts formula employed earlier to show that $x!$ satisfies the functional equation.

$$(x+1)! = (x+1)x!$$

Proof:

$$(x+1)! = \int_{0}^{\infty} t^{x+1} e^{-t} dt$$

So let $h(t) = \frac{t^{x+1}}{(x+1)}, k(t) = e^{-t}$ then

$$\int_{0}^{b} t^{x} e^{-t} = \int_{0}^{b} h'(t) k(t) \\ = h(b) k(b) - h(0) k(0) - \int_{0}^{b} h(t) k'(t) \\ = \frac{b^{x+1} e^{-b}}{x+1} - 0 + \int_{0}^{b} \frac{t^{x+1}}{x+1} e^{-t} dt$$

When $b \rightarrow \infty$, we have

$$x! = \frac{(x+1)!}{x+1}$$

$\square$

Theorem 8.4.11 (Bohr–Mollerup Theorem).

There is a unique positive function $f$ defined on $x ≥ 0$ satisfying

(i) $f(0) = 1$

(ii) $f(x+1) = (x+1)f(x)$, and

(iii) $\log(f(x))$ is convex.

Because $x!$ satisfies properties (i), (ii), and (iii), it follows that $f(x) = x!$.

Proof: We need one more geometrically plausible fact about convex functions. If $[a,b]$ and $[a',b']$ are two intervals in the domain of a convex function $φ$, and $a ≤ a'$ and $b ≤ b'$, then the slopes of the chords over these intervals satisfy

$$\frac{\phi(b)-\phi(a)}{b-a} \leq \frac{\phi(b')-\phi(a')}{b'-a'}$$

Because $f$ satisfies properties (i) and (ii) we know $f(n) = n!$ for all $n ∈ N$. Now fix $n ∈ N$ and $x ∈ (0,1]$.

Exercise 8.4.21.

(a) Use the convexity of $\log(f(x))$ and the three intervals $[n−1,n]$, $[n,n+x]$, and $[n,n+1]$ to show

$$x \log_{} (n) \leq \log_{} (f(n+x)) - \log_{} (n!) \leq x \log_{} (n+1)$$

Proof:

$$\frac{\log(f(n)) - \log_{} f((n-1))}{1} \leq \frac{\log(f(n+x)) - \log_{} f((n))}{x} \leq \frac{\log(f(n+1)) - \log_{} f((n))}{1}$$

This is the same as it asks.

$\square$

(b) Show $\log(f(n+x)) = \log(f(x))+\log((x+1)(x+2)···(x+n))$

Proof:

Since

$$f(x+n) = (x+n)f(x+n-1) = (x+n)(x+n-1)f(x+n-2) \cdots = (x+n)\cdots (x+1) f(x)$$

Then with exercise 8.4.6 (c), so we proved.

$\square$

(c) Now establish that

$$0 \leq \log_{} (f(x)) - \log_{} \left( \frac{n^xn!}{(x+1)(x+2)\cdots (x+n)} \right) \leq x \log_{} (1+\frac{1}{n})$$

Proof:

We use (a) - $x \log_{} n$ and get

$$\log_{} (f(n+x)) - \log_{} (n!) - x\log_{} n =\\ \log_{} f(x) + \log_{} ((x+1)(x+2)\cdots (x+n)) - \log_{} (n!) - x\log_{} n \\ = \log_{} f(x) - \log_{} \left( \frac{n^xn!}{(x+1)(x+2)\cdots (x+n)} \right)$$

$\square$

(d) Conclude that

$$f(x) = \lim_{n \to \infty} \frac{n^xn!}{(x+1)(x+2)\cdots (x+n)}$$

Proof:

This is because

$$\lim_{n \to \infty} x \log_{} (1 + \frac{1}{n}) = 0$$

$\square$

(e) Finally, show that the conclusion in (d) holds for all $x ≥ 0$.

Proof:

Assume $x \in (1, 2]$ then $x = 1 + x_0$, $x_0 \in (0,1]$.

$$f(x) = f(x_0+1) = (x_0+1) f(x_0) \\ = (x_0+1) \lim_{n \to \infty} \frac{n^{x_0}n!}{(x_0+1)(x_0+2)\cdots (x_0+n)} \\ = \lim_{n \to \infty} \frac{n^{x_0+1}(n-1)!}{(x_0+2)(x_0+3)\cdots (x_0+n)} \\ = \lim_{n \to \infty} \frac{n^{x}(n-1)!}{(x+1)(x+2)\cdots (x+(n-1))} \\ = \lim_{n \to \infty} \frac{n^{x}n!}{(x+1)(x+2)\cdots (x+n)} \\ $$

Then we can use induction for all $x > 0$.

$\square$

Because we have arrived at an explicit formula for $f(x)$, the function $f(x)$ must be unique. By virtue of the fact that $x!$ satisfies conditions (i), (ii), and(iii) of the theorem, we can conclude that $x!$ is this unique function; i.e., $f(x) = x!$. Thus, not only have we proved the theorem, but we have also discovered analternate representation for the factorial function called the Gauss product formula:

$$\tag{9} x! = \int_{0}^{\infty} t^x e^{-t} dt = \lim_{n \to \infty} \frac{n^{x}n!}{(x+1)(x+2)\cdots (x+n)}$$

For all $x \geq 0$.

Recall that when $x!$ is extended to all of $\mathbf{R}$ via the functional equation $x! = x(x− 1)!$ we get asymptotes at every negative integer. Thus, there is a compelling reason to consider the reciprocal function $1/x!$ which we can take to be zero for $x =−1,−2,−3,...$.

Exercise 8.4.22.

(a) Where does $g(x) = \frac{x}{x!(−x)!}$ equal zero? What other familiar function has the same set of roots?

Solution: $g(x) = 0$ for every integer. $\sin \pi x$ also has the same set of roots.

(b) The function $e^{−x^2}$ provides the raw material for the all-important Gaussian bell curve from probability, where it is known that

$$\int_{-\infty}^{\infty} e^{−x^2} dx = \sqrt[]{\pi}$$

Use this fact (and some standard integration techniques) to evaluate $(1/2)!$.

Solution:

$$(1/2)! = \int_{0}^{\infty} t^{1/2} e^{-t} dt$$

Consider the integral

$$\int_{0}^{b^2} t^{1/2} e^{-t} dt$$

We apply 7.5.10 Change-of-variable Formula and let $g(x) = x^2$. So

$$\int_{0}^{b} f(g(x)) g'(x) dx = \int_{0}^{b} (x^2)^{1/2} e^{-x^2} 2x dx \\ = \int_{0}^{b} 2x^2e^{-x^2} dx \\ = \int_{-b}^{b} x^2e^{-x^2} dx$$

Let $h(x) = - \frac{1}{2} e^{-x^2}, k(x) = x$, then

$$\int_{-b}^{b} x^2e^{-x^2} = \int_{-b}^{b} h'(x) k(x) \\ = h(b)k(b) - h(-b)k(-b) - \int_{-b}^{b} (- \frac{1}{2} e^{-x^2}) dx \\ = (-e^{-b^2} \cdot (b)) - (-e^{-b^2} \cdot (-b)) + \int_{-b}^{b} \frac{1}{2} e^{-x^2} dx$$

So

$$\lim_{b \to \infty} \int_{0}^{b^2} t^{1/2} e^{-t} dt \\ = \lim_{b \to \infty} \int_{-b}^{b} e^{-x^2} dx \\ = \frac{1}{2} \sqrt[]{\pi}$$

$\square$

(c) Now use (a) and (b) to conjecture a striking relationship between the factorial function and a well-known function from trigonometry.

Solution:

$$(1/2)! = 1/2 \cdot (-1/2)! \\ (-1/2)! = 2 \cdot (1/2)!$$

So

$$g(1/2) = \frac{1/2}{(1/2)! \cdot 2 \cdot (1/2)!} \\ = \frac{1}{\pi}$$

So my conjecture is

$$\frac{x}{x!(−x)!} = \frac{1}{\pi} \sin (\pi x)$$

$\square$

Exercise 8.4.23.

As a parting shot, use the value for $(1/2)!$ and the Gauss product formula in equation (9) to derive the famous product formula for $π$ discovered by John Wallis in the 1650s

$$\frac{\pi }{2} =\lim_{n \to \infty} \left( \frac{2 \cdot 2}{1 \cdot 3} \right) \left( \frac{4 \cdot 4}{3 \cdot 5} \right) \left( \frac{6 \cdot 6}{5 \cdot 7} \right) \cdots \left( \frac{2n \cdot 2n}{(2n - 1) \cdot (2n + 1)} \right)$$

Proof:

$$x! = \lim_{n \to \infty} \frac{n^{x}n!}{(x+1)(x+2)\cdots (x+n)}$$

We plug $x = 1/2$, and take square on both side. Then we have

$$\frac{\pi}{2} = \lim_{n \to \infty} \left( \frac{2 \cdot 2}{1 \cdot 3} \right) \left( \frac{4 \cdot 4}{3 \cdot 5} \right) \left( \frac{6 \cdot 6}{5 \cdot 7} \right) \cdots \left( \frac{2n \cdot 2n}{(2n - 1) \cdot (2n + 1)} \right) \cdot \frac{2n}{2n+1} \\ = \lim_{n \to \infty} \left( \frac{2 \cdot 2}{1 \cdot 3} \right) \left( \frac{4 \cdot 4}{3 \cdot 5} \right) \left( \frac{6 \cdot 6}{5 \cdot 7} \right) \cdots \lim_{n \to \infty} \frac{2n}{2n+1} \\ = \lim_{n \to \infty} \left( \frac{2 \cdot 2}{1 \cdot 3} \right) \left( \frac{4 \cdot 4}{3 \cdot 5} \right) \left( \frac{6 \cdot 6}{5 \cdot 7} \right) \cdots \left( \frac{2n \cdot 2n}{(2n - 1) \cdot (2n + 1)} \right)$$

$\square$