Why does $\log{x}$ represent the area below the graph of $f(x)=\frac{1}{x}$ from $1$ to $x$? What's so special about $1$ in this case?

Of course I understand that $\log{1}=0$ and I also understand that you cannot start at $x=0$ because $f(0)$ is not defined.

Still I can't get my head around of why it has to be $1$.

Also, this further implies that part of the antiderivative of $f(x)=\frac{1}{x}$ has to be negative (as part of the function of $\log{x}$ is negative). But why is this necessary?

The background of this question is that my calculus-book (Calculus, a complete course) starts with noticing that $f(x)=\frac{1}{x}$ is not an antiderivative of a polynomial function and then attempts to define a antiderivative which ends up being $\log{x}$. It does this all before even addressing the fundamental theorem of calculus or techniques of integration. They then simply define $log(1)$ to be $0$ without even knowing what that function really is yet. So I am stuck with kind of a circular reasoning where log(1)=0 because we defined it that way, but I don't understand why we define it that way.

When calculating areas under graphs by taking the limit of a sum (instead of by integration), you would start at $x=0$ right?

So in short: Why does the antiderivative of $\frac{1}{x}$ have to be $0$ at precisely $x=1$? (when $C = 0$). Why do we define it that way?

I am looking for some kind of deeper understanding; something still didn't "click". So some real background on what's going on here would be much appreciated.



What is "the" anti-derivative of $x\mapsto x+3$? Well, there isn't really a "the" here. For any constant $C$, the map $x\mapsto\frac12x^2+3x+C$ ia an anti-derivative. It looks like $x\mapsto \frac12x^2+3x$, obtained by picking $C=0$ plays a special role among the anti-derivatives, but it doesn't. Somebody else might rightfully describe the family of anti-derivatives of $x\mapsto x+3$ as $x\mapsto \frac12(x+3)^2+C$ (and in fact this can happen in a systematic way by using the substitution rule when integrating), again with $C$ an arbitrary constant, but unrelated to the $C$ used above. Of course, if we set $C=0$ in this description, we arrive at $x\mapsto (x+3)^2$, not the same function as $x\mapsto \frac12x^2+3x$. So is one of them "better" than the other? This may be the case depending on context (e.g., the second form tells us immediately where the apex of the parabola is), but there is no general notion of better.

So what is an anti-derivative of $f\colon(0,\infty)\to \Bbb R$, $x\mapsto \frac1x$? It is any function $F\colon(0,\infty)\to \Bbb R$ with $F'(x)=f(x)$ for all $x>0$. If we do not have a suitable function with this property already in our utility belt, we might simply define a specific one this way and give it a name. To get rid of the $C$, we rely on the definite integral, i.e., we fix some $a$ and define $F(x)=\int_a^xf(t)\,\mathrm dt$. This merely shifts the burden: What $a$ do we pick? $a=0$ would be nice, but $0$ is not in the domain of $f$; and not even $\lim_{a\to0}\int_a^xf(t)\,\mathrm dt$ exists. We might even try $a=+\infty$, but alas, $\lim_{a\to+\infty }\int_a^xf(t)\,\mathrm dt$ doesn't exist either (but for other $f$ this might still be a viable choice; for example, in physics, this is the default way to pick the integration constant when defining potentials). So we better pick $a$ somewhere between $0$ and $\infty$. But as $f(x)>0$ for all $x$, this automatically implies $F(x)<F(a)=0$ for $0<x<a$ and $0=F(a)<F(x)$ for all $x>a$, i.e., we cannot achieve that $F$ is non-negative (or non-positive) throughout.

But does that mean $a=1$ is the best choice? Maybe $a=\pi$ woul dmake things particularly elegant? Well, as above, this may depend on the context, but there are some overwhelming contexts: By substituting $x\leftarrow \frac1cx$ (with $c>0$), we see that $$F(b)=\int_{a}^bf(x)\,\mathrm dx=\frac1c\int_{ac}^{bc}f({\tfrac xc})\,\mathrm dx=\frac1c\int_{ac}^{bc}\frac1c{x}\,\mathrm dx=\int_{ac}^{bc}\frac1{x}\,\mathrm dx=F(bc)-F(ac)$$ or: $F(bc)=F(b)+F(ac)$. This formula becomes most beautiful (namely: symmetric) if we agree on $a=1$: $F(bc)=F(b)+F(c)$. This is one compelling reason to define our "go-to" anti-derivative (which we decide to call $\ln$) as $\ln x:=\int_1^x\frac 1t\,\mathrm dt$.


Why does the antiderivative of $\frac{1}{x}$ have to be $0$ at precisely $x=1$?

Are you familiar with the fundamental theorem of calculus? In your case it can be stated as

$$\int_y^x \frac{1}{t}dt=\ln(x)-\ln(y)$$

Because $\ln(t)$ is an antiderivative of $\frac{1}{t}$. Now for $x=1$ we have $\ln(x)=0$ and thus

$$\int_1^x \frac{1}{t}dt=\ln(x)$$

So the right side is very simple. That's why the choice of $1$ is so convenient.

So the question should not be "why antiderivative is $0$ at $1$" but rather "why we chose $\ln(x)$ to be the very special antiderivative of $\frac{1}{x}$". And the reason is stated above. Simplicity.

EDIT: So now I understand that actually

$$\ln(x):=\int_1^x \frac{1}{t}dt$$

is the definition of logarithm in your book. So now the question "why we pick $1$?" makes more sense. The reason is not obvious even though it's simple: because it works. In the sense that this definition of $\ln(x)$ has some good properties, in particular it's a homomorphism


Also it's the inverse of the classical exponential map:


These properties won't hold if you use any other number then $1$ in the integral.

I understand the confusion now. And IMO it is not good (from educational point of view) to define the logarithm like that.

Side note: starting from $0$ is not good since $\int_{0}^x\frac{1}{t}dt=\infty$ for any $x>0$. This corresponds to the fact that the tangent line to $\ln(x)$ at $0$ is vertical.

  • $\begingroup$ But that is because we defined log(1) to be 0 in the first place. The point is that the definition of log(x) is the surface between 1 and x. Why define it that way? $\endgroup$ – GambitSquared May 25 '17 at 11:19
  • $\begingroup$ @ImreVégh You usually define $\ln(x)$ as the inverse of $e^{x}$. It's a corollary that $\ln(x)$ calculates area under $\frac{1}{x}$. We don't define $\ln(1)$ to be $0$. It's a consequence. $\endgroup$ – freakish May 25 '17 at 11:21
  • $\begingroup$ In my book they do it the other way around. Start with log(x) (as a function that represents the area under 1/x) and then define e^x from log(x) $\endgroup$ – GambitSquared May 25 '17 at 11:23
  • $\begingroup$ @ImreVégh Fair enough. So if you define $\ln(x)$ as an integral (just like in my answer) then obviously $\ln(1)=0$ because $\int_1^1 f(t)dt = 0$. $\endgroup$ – freakish May 25 '17 at 11:25
  • $\begingroup$ Can you explain why? Isn't the integral from 2 to 2 also 0? $\endgroup$ – GambitSquared May 25 '17 at 11:26

From your comments it seems like the part of your question not yet answered is "why is $\log(1)$ defined to be $0$? The answer here lies in the purpose of the logarithm.

Logarithms convert multiplication into addition: $\log(xy) = \log(x)+\log(y)$, and they do that because it's easier for humans to add things up than it is to multiply them together. In order for this to work the logarithm has to map the multiplicative identity to the additive identity: i.e. $log(1) = 0$. If it didn't then we'd get $\log(x) = \log(1\cdot x) = \log(1) + \log(x) \not= \log(x)$ which would be a problem.

Logarithms are often described (see wikipedia for example) as inverting the exponentiation operator, which maps addition into multiplication.


If you have a look at this page of epistemology (sorry it is in French butI'll try to summarize the main points)


The article says that at the beginning $\log(1)$ was not zero, but instead is was $\log(10^7)=0$

Around $1615-1616$ Briggs and Neper had some discussion about the pros and cons of their own choices about the new tool (the logarithms) and the point was that multiplying quantities by various powers of $10$ to fit such or such pre-calculated table was not pleasant, and it seems they agreed on setting $\log(1)=0$.

An additional advantage was the fundamental relation $\mathbf{\log(a\times b)=\log(a)+\log(b)}$

(for which $\log(1)=2\log(1)$ forces $\log(1)=0$)

This break through allowing to manipulate products likes sums via tables of anti-logarithms.$^*$

$(*)$ exponential was developed much later in history, with Euler and the value of $e$, standing for $\bar euler$ I presume...


The antiderivate is not unique. For $x>0$, the general antiderivate of $\frac{1}{x}$ is $\ln(x)+C$

If we require that the antiderivate contains a point (in this case $(1/0)$) , the antiderivate gets unique. Inserting the point in $y=\ln(x)+C$ gives $0=\ln(1)+C=C$, and vice versa, if we take $C=0$, we have $\ln(x)$ which contains the point $(1/0)$


As other answers mention, the choice is in some sense arbitrary but note that $\log x+C=\log Ax,$ where $A=e^C.$ When deciding which of these options we should arbitrarily use, the only real consideration is that we should use whichever one makes algebra (and integrating more complex expressions) easier. The choice $C=0,$ i.e. $A=1$ minimises the complexity of the expression, making things easier to deal with when finding the antiderivative of more complex expressions.


In short: It doesn't! 1 was chosen arbitrarily for convenience, and any other number would work. The antiderivative of $1/x$ is $\log x + C$ (in $x>0$) and it's simply the choice of $C=0$ that forces the antiderivative to be zero at $x=1$. Choosing a different $C$, which is entirely sensible, would give a different point where the antiderivative is zero.

  • $\begingroup$ I understand, but if we take C=0, then why do we have to start at x=1? $\endgroup$ – GambitSquared May 25 '17 at 11:16
  • $\begingroup$ Simply because the antiderivative is then $log x$ exactly, and $log 1 =0$. $\endgroup$ – B. Mehta May 25 '17 at 11:18
  • $\begingroup$ Yes, but that's because we defined logx that way! My question is: why did we define it that way? $\endgroup$ – GambitSquared May 25 '17 at 11:18
  • $\begingroup$ Typically just for convenience, and to give an inverse to the exponential function. Notice exp(0)=1 so for log to be an inverse, log(1)=0 $\endgroup$ – B. Mehta May 25 '17 at 11:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.