${\log}_{a}{x}\neq {\int}^{x}_{1}{\frac{1}{t}}dt$

Question

In most calculus textbooks, $\ln{x}$ is defined to be ${\int}^{x}_{1}{\frac{1}{t}}dt$. Some textbooks validate this definition by demonstrating that this function $\int^{x}_{1}{\frac{1}{t}}dt$ has all the properties of a logarithmic function (I've included pictures of this). I'm skeptical to this particular approach, since we could also define ${\log}_{a}{x}={\int}^{x}_{1}{\frac{1}{t}}dt$. We can still show that the laws of logarithms are properties of this integral, it's also obvious how the algebra will work out. And that means we've justified our claim?

Hell no! The derivative of ${\log}_{a}{x}$ is $\frac{1}{x}{\log}_{a}{e}$. Isn't this approach erroneous then? How then, could we show that this integral does not equal ${\log}_{a}{x}$? We could try showing that some of the properties of the logarithmic functions ($a\neq{e}$) do not hold for ${\int}^{x}_{1}{\frac{1}{t}}dt$. But how do we go about it?

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Answer 1

Your critique is completely correct: the properties shown in those proofs are not enough to distinguish between $\ln x$ and $\log_a x$ for any $a>1$. Indeed, all of the proofs would go through for the function $\int_1^x \frac Ct\,dt$ for any positive constant $C$ as well. (This is secretly the same ambiguity in disguise....)

So yes, you're right that this is not a proof that $\int_1^x \frac 1t\,dt$ must equal $\ln x$ instead of some other $\log_a x = \frac1{\ln a}\ln x$. To be fair, the textbook didn't claim it was such a proof—only that the integral does have logarithm-like properties.

A proof that we really do get $\ln x$ itself, instead of some multiple of it, would need to use some property of the number $e$—which itself depends on what definition of $e$ you choose. One common definition is of all the exponential functions $a^x$ with $a>1$, the number $e$ is the only base $a$ with the property that $\frac d{dx}(b^x)\big|_{x=0} = 1$.

From this one can derive (using the relationship between the derivatives of a function and its inverse function) that $e$ is the only base $a$ for which the inverse function $\log_a x$ satisfies $\frac d{dx} \log_a x\big|_{x=1}=1$. And this additional property is enough to show that $\int_1^x \frac1t\,dt$ is equal to $\log_e x=\ln x$, since the derivative of $\int_1^x \frac1t\,dt$ equals $\frac 1x$ by the fundamental theorem of calculus.

Answer 2

The function $t \mapsto \to \frac 1t$ is continuous at the interval $ (0,+\infty)$, it has then many antiderivative at $(0,+\infty)$. The antiderivative which takes $ 0 $ at $ x=1 $ and $ 1 $ at $ x=e $ is Nepierian logarithm $ (\ln) $.

The one which takes $ 1$ at $x=a$ is $\log_a$.

$$\ln(x)=\log_e(x)=\int_1^x \frac{dt}{t}$$

$${\log_a(x)=\frac{\displaystyle\int_1^x\frac{dt}{t}}{\displaystyle\int_1^a\frac{dt}{t}}}$$

Answer 3

Note that $\log_a a = 1$. If we want $\log_a x = \int_1^x\frac{dt}{t}$ we will need $\int_1^a\frac{dt}{t} = 1$. There is a unique solution $a$ for that. If we like, we could take that as the definition of the number $e$.

Answer 4

Define $L(x) := \int_1^x \frac1t dt$. Then clearly by the fundamental theorem of calculus $L$ is differentiable with $$\frac{dL}{dx}(x)=\frac1x \quad \text{ for all $x>0$}$$ In particular, since $\frac{dL}{dx}>0$, $L$ is strictly increasing and has a strictly increasing inverse function $E=L^{-1}$. By the inverse function theorem we can find the derivative of $L^{-1}$ as $$\frac{dL^{-1}}{dx}(x) = \frac{1}{L'(L^{-1}(x))}=L^{-1}(x)$$ thus by defining $E(x)=L^{-1}(x)$ we see that $E$ satisfies the differential equation $E'=E$ with $E(0)=1$, which means that $E(x)=e^x$ (in fact we may define $e^x$ as the solution to this exact differential equation).

Since $E$ and $L$ are inverses we can conclude, that $L(x)$ is the inverse of $e^x$, but that is by definition the natural logarithm, hence $L(x)=\ln(x)$.

Answer 5

Let us put it this way. Any continuous function $f:(0, \infty) \to\mathbb {R} $ with the property $$f(xy) =f(x) +f(y)$$ is said to be a logarithmic function.

If we start in this manner then there is no unique logarithmuc function. If $f$ satisfies the above property then so does any multiple of it. And there is a trivial function $f(x) =0$ which also works fine.

The key part is the following

Theorem: Let a function $f:(0, \infty) \to\mathbb {R} $ be continuous on $(0,\infty) $ and satisfy $f(xy) =f(x) +f(y) $. Then either $f(x) =0$ for all $x>0$ or there is a unique positive number $a\neq 1$ such that $f(a) =1$. The function $f$ is determined uniquely by the value of $a$. Let's use symbol $f_a$ for the function determined by $a$. Then $f_a$ is differentiable and further there is unique number $e>1$ with $f_e'(1)=1$ and more generally $$f_a(x) =\frac{f_e(x)} {f_e(a)}, \, f_a'(x) =\frac{1}{f_e(a)x}$$

The function $f_a(x) $ is called the logarithm of $x$ to the base $a$ and conventionally denoted by $\log_a x$. The function $\log_e x$ is simply denoted by $\log x$ or $\ln x$.

A proof of some parts of the theorem above is presented here. The introduction of the number $e$ and the fact that $f_a'(1)=1/f_e(a)$ can be proved with a little more effort and thus the formula in the theorem above can be established.

Also observe that $e$ can be introduced only when we consider the derivative of the logarithmiic function $f_a$. As long as we are working with algebraic properties we don't get to $e$. A similar treatment can be given for exponential functions using the property $f(x+y) =f(x) f(y) $.