Logarithm defined using the definite integral without Fundamental Theorem

If one wants to define the natural logarithm using (Riemann) integrals, he could do as follows:

$$\log(x) := \int_{1}^{x} \frac{1}{t} dt$$

Let's assume we hadn't defined the $\exp$-function yet. How can one prove some basic logarithmic identities WITHOUT using the Fundamental Theorem of Calculus.

1. $\log (xy) = \log (x) + \log (y)$
2. $\log '(1) = 1$

Edit: This exercise came up as an homework for a Calculus I class at my university. They explicitly stated not to use the exp-function or the FTC. Since I didn't know how they were supposed to solve this with these restrictions, I thought I post this problem here. (Note: This homework was due a few weeks ago. So posting the solution here shouldn't be a problem.)

• The integral cannot start from $0$. Feb 11 '18 at 21:51
• @VonNeumann Indeed, thx. Feb 11 '18 at 21:52
• Why would anyone doing this Riemann integral's exercises not use the FTC...? Feb 11 '18 at 22:10
• @DonAntonio This exercise came up as an homework for a Calculus I class at my university. They explicitly stated not to use the $exp$-function or FTC. Since I didn't know how they were supposed to solve this with these restrictions, I thought I post this problem here. (Note: This homework was long due a few weeks ago. So posting the solution here should't be a problem.) Feb 11 '18 at 22:14
• Well, in my answer I use FTC (and I acknowledge this explicitly). I've no pale idea how to do $\;f'(1)=1\;$ without using FTC, either explicitly or implicitly...Perhaps someone else can help here. Feb 11 '18 at 22:16

3 Answers

\begin{align} \log(xy) =& \int_1^{xy}\dfrac{1}{t}dt \\ =& \int_1^{x}\dfrac{1}{t}dt+\int_x^{xy}\dfrac{1}{t}dt \\ =& \int_1^{x}\dfrac{1}{t}dt+\int_x^{xy}\dfrac{1}{xt}d(xt) \\ =& \int_1^{x}\dfrac{1}{t}dt+\int_1^{y}\dfrac{1}{t}dt \\ =& \log x+\log y \end{align} For the second \begin{align} \log'(1) &= \lim_{h\to0}\dfrac{\log(1+h)-\log(1)}{h} \\ &= \lim_{h\to0}\dfrac{1}{h}\int_1^{1+h}\dfrac{1}{t}dt \\ &= \lim_{h\to0}\dfrac{1}{\xi}~~~~~~\text{MVT for integrals}~,~1\leq\xi\leq1+h \\ &= 1 \end{align}

• The pass between the third line and the fourth line is exactly substitution...but without giving it that name: the variable changed, the limits changed. I just point out because someone thought (see comment below my answer) that substitution requires FTC, which I think it doesn't. For the second part: I'd love (honest, no sarcasm here) to see a proof of the integral MVT that doesn't use the FTC. Feb 11 '18 at 22:55
• @DonAntonio I am not sure about the substitution rule, but as for MTV, I know what it doesn't require the FTC. In fact, I think we used MTV to prove FTC. Using the definition of step functions you can easily get that $m \leq f(x) \leq M$ implies $m(b-a) \leq \int_{a}^{b} f(x) dx \leq M(b-a)$. Using that, you can prove MVT quite easily as well. Feb 11 '18 at 23:20
• Sry, Regulated functions, not step functions. Feb 11 '18 at 23:29
• @ClassicEndingMusic: linear substitution follows directly from the definition of an integral as limit of Riemann sums. No need of FTC there. The second part can be established by integrating inequalities (see end of my answer). Use of MVT or inequalities is the basis of a proof of FTC so the process is nothing but just spilling out the details of FTC. Therefore I have replaced integrals altogether in my answer. Feb 12 '18 at 7:48
• @ClassicEndingMusic Perhaps. Do you have any source where the integral MVT is proved without the FTC? Feb 12 '18 at 7:56

Here is another technique which goes by the basics and does not use FTC (in fact it does not use anything related to derivatives or integrals apart from the bare definition of an integral). Since the integral in question exists we can think of it as a limit of Riemann sum and thus if $x>1$ we can take a partition of $[1,x]$ given by points $x_k=q^k$ so that $q^n=x, q=\sqrt[n] {x}$ and then $$\log x=\lim_{n\to\infty} \sum_{k=1}^{n}\frac{x_{k}-x_{k-1}}{x_{k-1}}=\lim_{n\to\infty}\sum_{k=1}^{n}(q-1)=\lim_{n\to\infty}n(x^{1/n}-1)$$ One can show that the same limit formula holds for $0<x\leq 1$ also. And thus we get an alternative expression for $\log x$ in terms of a limit.

Proving the first and fundamental property of $\log x$ is easy and we can just note that $$\log xy=\lim_{n\to\infty} n((xy) ^{1/n}-1)=\lim_{n\to\infty} y^{1/n}n(x^{1/n}-1)+n(y^{1/n}-1)=1\cdot \log x+\log y$$ and we are done. Putting $y=1/x$ and noting that $\log 1=0$ gives us $\log(1/x)=-\log x$ (this will be used later).

The second property is non-trivial and requires some more work. One can prove with some effort that $n(x^{1/n}-1)$ is decreasing as $n$ increases and hence it is less than or equal to $(x-1)$. Therefore its limit $\log x$ is less than or equal to $(x-1)$. Thus we have $$\log x\leq x-1,x>0$$ Next note that $$\log x=-\log(1/x)\geq - ((1/x)-1)=\frac{x-1}{x}$$ and thus we have $$\frac{x-1}{x}\leq \log x\leq x-1$$ and therefore if $x\neq 1$ the ratio $(\log x) /(x-1)$ lies between $1/x$ and $1$. By Squeeze Theorem we can see that $$\lim_{x\to 1}\frac{\log x} {x-1}=1$$ and the second property is also established.

The fact that $s_n=n(x^{1/n}-1)$ is decreasing as $n$ increases is not difficult to prove and one just needs to start with the number $y=x^{1/n(n+1)}$ so that $$s_n=n(y^{n+1}-1),s_{n+1}=(n+1)(y^n-1)$$ and we have $s_n\geq s_{n+1}$ if $$ny^{n+1}-ny^n-y^n+1\geq 0$$ ie $$ny^n(y-1)-(y^n-1)\geq 0$$ ie $$(y-1)\left(\sum_{k=1}^{n}(y^{n}-y^{k-1})\right)\geq 0$$ and it is easy to see that both factors are of same sign so that their product is non-negative.

The second part can also be done by just noting that if $x\geq 1$ then $1/x\leq 1/t\leq 1$ for all $t\in[1,x]$ and integrating this inequality with respect to $t$ on interval $[1,x]$ we get $$\frac{x-1}{x}\leq \log x\leq x-1,x\geq 1$$ If $0<x<1$ then we can use $1\leq 1/t\leq 1/x$ for all $t\in[x, 1]$ and integrating this on interval $[x, 1]$ we get $$1-x\leq - \log x\leq \frac{1-x}{x}$$ which is same as previous inequality and the proof proceeds as given earlier.

We assume, for simplicity, that $\;x>1\;$ (You do the other case...):$$\log xy:=\int_1^{xy}\frac1t\,dt\;\;(**)$$

Substitute n

$$u:=\frac tx\implies dt=xdu\implies (**)=\int_{1/x}^y\frac{x\,du}{xu}=\int_{1/x}^1\frac1u\,du+\int_1^y\frac1u\,du=$$

$$-\int_1^{1/x}\frac1u\,du+\log y=-\log\frac1x+\log y\;\; (***)$$

and now, substituting $\;s:=\frac1u\implies du=-\frac1{s^2}ds\;$:

$$\int_1^{1/x}\frac1u\,du=\int_1^x s\cdot\frac{-ds}{s^2}=-\int_1^x\frac{ds}s=-\log x$$

so we actually got in

$$(***)=-(-\log x)+\log y=\log x+\log y$$

The following uses the FTC. I know no other way to solve it:

Now use the FTC: if $\;F\;$ is a primitive function of $\;\cfrac1t\;$ in $\;[1,x]\;$ , then

$$\log x:=\int_1^x\frac{dt}t=F(x)-F(1)\implies (\log x)'=F'(x)=\frac1x \implies (\log 1)'=1$$

• Doesn't the proof of the substitution rule rely on FTC as well? Feb 11 '18 at 22:15
• @ClassicEndingMusic Not at all. Substitution is also done in indefinite integration = antidifferentiation ,where there is no primitives and stuff. Of course, the actual difference between "primitive" and "antiderivative" is, in some sense, rather cosmetic...but we don't use "primitive function" in indefinite integrals. Feb 11 '18 at 22:18