Logarithms

[!math|{“type”:“definition”,“number”:“auto”,“setAsNoteMathLink”:false,“title”:“Natural Logarithm”,“label”:“natural-logarithm”,“_index”:0}] Definition 1 (Natural Logarithm). For any $x>0$, we define its natural logarithm by

$$\log x=\ln x={\int }_1^x\frac{dt}{t}$$

Properties §

[!math|{“type”:“proposition”,“number”:“auto”,“setAsNoteMathLink”:false,“_index”:1}] Proposition 2. $\log 1=0$

[!math|{“type”:“proposition”,“number”:“auto”,“setAsNoteMathLink”:false,“_index”:2}] Proposition 3. $\log x$ is differentiable on $(0,\infty )$ with derivative $\frac{1}{x}$

\begin{proof}For any $x$, the function $\frac{1}{t}$ is continuous on $[1,x]$, so by the first fundamental theorem of calculus, $\log x$ is differentiable with derivative $\frac{1}{x}$.\end{proof}

[!math|{“type”:“proposition”,“number”:“auto”,“setAsNoteMathLink”:false,“_index”:3}] Proposition 4. For all $x,y>0$, $\log xy=\log x+\log y$.

\begin{proof} Fix $y>0$ and consider the function given by

$$l(x)=\log (xy)$$

for $x>0$.

Since $l(x)$ is the composition of the differentiable functions $x\mapsto xy$ and $u\mapsto \log u$, it is also differentiable. By chain rule,

$$l^{\prime }(x)=y\left(\frac{1}{yx}\right)=\frac{1}{x}.$$

Thus, $l$ and $\log x$ have the same derivative, so

$$l(x)=\log x+c.$$

Taking $x=1$, we get

$$\log y=l(1)=0+c⟹\log y=c,$$

so that

$$\log xy=l(x)=\log x+c=\log x+\log y.$$

\end{proof}

[!math|{“type”:“proposition”,“number”:“auto”,“setAsNoteMathLink”:false,“_index”:4}] Proposition 5. $\log x$ is strictly increasing on $(0,\infty )$.

\begin{proof} Since $\frac{d}{dx}\log x=\frac{1}{x}>0$ for all $x>0$, the function is strictly increasing. \end{proof}

[!math|{“type”:“proposition”,“number”:“auto”,“setAsNoteMathLink”:false,“_index”:5}] Proposition 6. $\log x\to \infty$ as $x\to \infty$ and $\log x\to -\infty$ as $x\to 0$.

\begin{proof} $\log x$ is strictly increasing and $\log 1=0$, so $\log 2>0$. Then,

$$\log (2^n)=n\log 2\to \infty$$

as $n\to \infty$.

In the other direction, notice that $\log \frac{1}{2}<0$, so

$$\log \left(\frac{1}{2^n}\right)=n\log \frac{1}{2}\to -\infty$$

as $n\to \infty$. \end{proof}

[!math|{“type”:“proposition”,“number”:“auto”,“setAsNoteMathLink”:false,“_index”:6}] Proposition 7. $\log x$ is integrable on any finite subinterval of $(0,\infty )$ and

$$\int \log xdx=x\log x-x+C$$

\begin{proof} Since $\log x$ is continuous, it is integrable on any finite subinterval of $0,\infty$. By integration by parts,

$$\begin{array}{rl}\int \log xdx& =x\log x-\int x\frac{d}{dx}(\log x)dx\\ & =x\log x-\int 1dx\\ & =x\log x-x+c\end{array}$$

\end{proof}

[!math|{“type”:“proposition”,“number”:“auto”,“setAsNoteMathLink”:false,“_index”:7}] Proposition 8.

$$\underset{x\to 0}{\lim }x\log x=0\text{and}\underset{x\to 0}{\lim }\frac{\log x}{x}=0$$

\begin{proof} Let $L={\lim }_{x\to 0}x\log x$. If $u=\frac{1}{x}$, $f(u)=-\log (\frac{1}{u})$ and $g(u)=u$, then

$$L=\underset{u\to \infty }{\lim }\frac{1}{u}\log \left(\frac{1}{u}\right)=-\underset{u\to \infty }{\lim }\frac{f(u)}{g(u)}$$

Using L’Hopital’s Rule,

$$\begin{array}{rl}-\underset{u\to \infty }{\lim }\frac{f(u)}{g(u)}& =-\underset{u\to \infty }{\lim }\frac{f^{\prime }(u)}{g^{\prime }(u)}\\ & =-\underset{u\to \infty }{\lim }\frac{1}{u}\\ & =0\end{array}$$

We can similarly prove

$$\underset{x\to 0}{\lim }\frac{\log x}{x}.$$

\end{proof}

[!math|{“type”:“proposition”,“number”:“auto”,“setAsNoteMathLink”:false,“_index”:8}] Proposition 9. Improper integral of $\log x$ over $(0,b]$ exists for any $b>0$ and

$${\int }_0^b\log xdx=b\log b-b.$$

\begin{proof}

$$\begin{array}{rl}L& =\underset{a\to 0}{\lim }{\int }_a^b\log tdt\\ & =\underset{a\to 0}{\lim }[t\log t-t{]}_a^b\\ & =\underset{a\to 0}{\lim }b\log b-b-(a\log a-a)\\ & =b\log b-b-\underset{a\to 0}{\lim }(a\log a-a)\\ & =b\log b-b.\end{array}$$

\end{proof}