Exercise 7.5.8

Question

[Natural Logarithm and Euler's Constant]
    Let
$$
L(x)=\int_{1}^{x} \frac{1}{t} d t
$$
where we consider only $x>0$.
\begin{enumerate}[label=(\alph*)] 
\item What is $L(1)$? Explain why $L$ is differentiable and find $L^{\prime}(x)$.
\item Show that $L(x y)=L(x)+L(y)$. (Think of $y$ as a constant and differentiate $g(x)=L(x y)$)
\item Show $L(x / y)=L(x)-L(y)$.
\item Let
$$
\gamma_{n}=\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right)-L(n)
$$
Prove that $\left(\gamma_{n}\right)$ converges. The constant $\gamma=\lim \gamma_{n}$ is called Euler's constant.
\item Show how consideration of the sequence $\gamma_{2 n}-\gamma_{n}$ leads to the interesting identity
$$
L(2)=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\cdots .
$$
 \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
\item $L(1) = 0$. Since $1/t$ is continuous, by Theorem 7.5.1 (ii) $L'(x) = 1/t$.
\item Differenting $L(xy)$ with respect to $x$ leaves us with $1/x$, so by Theorem 7.5.1 (i) we have
$$\int^x_1 \frac{1}{t}dt = L(x) - L(1) = L(x) = L(xy) - L(y) \implies L(xy) = L(x) + L(y)$$
\item Differenting $L(1/x)$ with respect to $x$ leaves us with $-1/x$, so
$$\int^x_1 \frac{-1}{t}dt = L(1/x) = -L(x) = -\int^x_1 \frac{1}{t} dt $$
Then $L(x/y) = L(x) + L(1/y) = L(x) - L(y)$ as desired.
\item For conciseness let $h_n = \sum^n_{i=1} \frac{1}{n}$ and $f(t) = \frac{1}{t}$. Consider
$\int^n_1 f$.
By definition, this is $L(n)$. Consider the partition $P = \{x : 1 \leq x \leq n,\ x \in \mathbf{N} \}$. Since $1/t$ is decreasing, we have
$$U(f,P) = \sum^{n-1}_{i=1} f = h_n - \frac{1}{n} \text{ and } L(f,P) = \sum^n_{i=2} f = h_n - 1$$
We therefore have $h_n - L(n) \geq h_n - U(f, P) = \frac{1}{n}$ and $h_n - L(n) \leq h_n - L(f,P) = 1$. This indicates $(\gamma_n)$ is bounded.  Now note that

$$\gamma_{n+1} - \gamma_n = \frac{1}{n+1} - \int^{n+1}_n \frac{1}{x}dx \leq 0$$
where $\frac{1}{n+1} = L(f, \{n,n+1\})$. This indicates that the sequence $(\gamma_n)$ is also monotone, so by the Monotone Convergence Theorem $(\gamma_n)$ converges.

\item $$\gamma_{2n} - \gamma_n = h_{2n} - h_n - L(2n) + L(n) = h_{2n} - h_n - L(2)$$
If we pair every element of $h_n$ with every other element of $h_{2n}$ we get
$$\gamma_{2n} - \gamma_n + L(2) = \sum^{2n}_{i=1} \frac{(-1)^{i+1}}{i} $$
Since $(\gamma_n)$ converges, the left side converges to $L(2)$, giving us the desired identity.
 \end{enumerate}

Exercise 7.5.8

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Exercise 7.5.8

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