Integral Formula for Logarithm

1. Jun 25, 2010

psholtz

If we define a function f(x) such that:

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

for $$x>0$$, so that:

$$f(y) = \int_{1}^y \frac{dt}{t}$$

and

$$f(xy) = \int_{1}^{xy} \frac{dt}{t}$$

is there a way, using just these "integral" definitions, to prove that:

$$f(x) + f(y) = f(xy)$$

Clearly, the function we are dealing w/ is the logarithm, but I'd like to prove this from the "definitions" given above, rather than reverting to "known properties" of the logarithm function.

2. Jun 25, 2010

Staff: Mentor

$$f(xy) = \int_{1}^{xy} \frac{dt}{t} = \int_1^x \frac{dt}{t} + \int_x^{xy} \frac{dt}{t}$$

By making a change to the dummy variable in the last integral, you can show it to be equal to f(y).

3. Jun 25, 2010

psholtz

I thought about doing a Taylor expansion about $$x=1$$, that is to say:

$$f(1+x) = f(1) + f'(1)x + \frac{1}{2!}f''(1)x^2 + \frac{1}{3!}f'''(1)x^3 + ...$$

$$f(1+x) = x - \frac{1}{2}x^2 + \frac{1}{3}x^3 - \frac{1}{4}x^4 + ...$$

but I'm not sure this leads anywhere fruitful.. First, we'd have to get an expression for $$f(x)$$, rather than $$f(1+x)$$:

$$f(x) = (x-1) - \frac{1}{2}\left(x-1\right)^2 + \frac{1}{3}\left(x-1\right)^3 - \frac{1}{4}\left(x-1\right)^4 + ...$$

but, if added to the equivalent expression for $$f(y)$$, I'm not sure this would lead to anything like an expression:

$$f(xy) = \left(xy-1\right) - \frac{1}{2}\left(xy-1\right)^2+ ...$$

4. Jun 25, 2010

psholtz

OK... yes. Thanks!

5. Jun 25, 2010

Count Iblis

Change of variables by rescaling t works. E.g. you can write:

$$f(x) = \int_{1}^{x}\frac{dt}{t} = \int_{\frac{1}{x}}^{1}\frac{du}{u} = - \int_{1}^{\frac{1}{x}}\frac{du}{u} = -f\left(\frac{1}{x}\right)$$

Last edited: Jun 25, 2010
6. Jun 25, 2010

psholtz

Which, in fact, is another useful property of the logarithm:

$$f(x) = -f\left(\frac{1}{x}\right)$$

7. Jun 25, 2010

Ben Niehoff

It was traditional in older calculus books to prove all the properties of the logarithm from the integral definition (or at least, they did in my mom's old calculus text she loaned me once). I think newer books may have decided this was "too confusing" or something.

As you can see, it's easy to show that the integral has the desired properties. You can also prove that it is the inverse function to $\exp x$.

8. Jun 25, 2010

Landau

Spivak also does it.

9. Jun 25, 2010

psholtz

Yes .. in point of fact, the question arose in my mind while I was reading one of the older calculus texts, specifically, Forsythe on Differential Equations. Forsythe mentions passingly that all the "properties" of the logarithm can be derived from its integral representation/definition, but doesn't go into any further detail.

Forgive my ignorance, but how can you derive that the exponential function is the inverse of the logarithm, given only the integral expression for logarithm?

10. Jun 25, 2010

Ben Niehoff

One way is this: We know that e^x solves the differential equation

$$\frac{dy}{dx} = y$$

Considering x as a function of y, we obtain

$$\frac{dx}{dy} = \frac{1}{y}$$

and hence the inverse function is

$$x(y) = \int_{y_0}^y \frac{dt}{t}$$

and it merely remains to fix the constant y_0.

Of course, the first differential equation is also solved by any constant multiple of e^x. With a little bit more algebra, you can take that into account.

11. Jun 30, 2010

psholtz

Interesting, thanks.

Yes, that makes sense.

Also, if we don't yet "know" what the symbol "e" is supposed to mean, I suppose one could also proceed by supposing that the solution y is in the form of an (indeterminate) polynomial, in the sense of:

$$y(x) = a_0 + a_1x + a_2x^2 + a_3x^3 + ...$$

so that from y'=y we get:

$$y'(x) = a_1 + 2a_2x + 3a_3x^2 + 4a_4x^3 + ....$$

and setting the constants equal, we get:

$$a_1 = a_0, a_2 = \frac{1}{2}a_1, a_3 = \frac{1}{3}a_2, a_4 = \frac{1}{4}a_3, ...$$

so that the solution works out to:

$$y(x) = a_0 + a_0x + \frac{1}{2!}a_0x^2 + \frac{1}{3!}a_0x^3 + ...$$

$$y(x) = a_0 \sum_{k=0}^{\infty} \frac{x^k}{k!}$$

which summation can be taken as the "definition" of $$e^x$$.

12. Jun 30, 2010

The Chaz

Could you please elaborate (or just do it outright!) ?
Thanks

13. Jun 30, 2010

arildno

Set u=t/x

Then, we have:
$$t=xu, \frac{dt}{du}=x\to{dt}=xdu$$
We therefore get:
$$\int_{x}^{xy}\frac{dt}{t}=\int_{1}^{y}\frac{xdu}{xu}=\int_{1}^{y}\frac{du}{u}=f(y)$$

14. Jul 1, 2010

HallsofIvy

Oddly enough, I was thinking of this as how newer calculus books did it! I may just be showing my age.

15. Jul 1, 2010

arildno

I thought so.

Back in your days, you had to make to do with tiny counting stones, right?

16. Jul 1, 2010

HallsofIvy

That is an insult! We used the abacus! (I knew the man who invented them.)

17. Jul 1, 2010

HallsofIvy

By the way, here is an important property of the logarithm not often shown in texts that define the logarithm in terms of the integral.

Since ln(x) is defined as an integral it is clearly differentiable for all positive numbers and, in particular, is differentiable on [1, 2] so we can apply the mean value theorem to that interval: there exist c in [1, 2] such that
$$\frac{ln(2)- ln(1)}{2- 1}= (ln)'(c)$$

Since
$$ln(x)= \int_1^t \frac{1}{x}dt$$
by the "fundamental theorem of caluculus" $(ln(x))'= 1/x$ for x positive.

Further
$$ln(1)= \int_1^1 \frac{1}{t}dt= 0[/itex] because $\int_a^a f(t)dt= 0$ for any integrable function f. So that gives [tex]\frac{ln(2)- ln(1)}{2- 1}= \frac{ln(2)- 0}{1}= ln(2)= \frac{1}{c}$$
for some c between 1 and 2. But if $c\le 2$, then $1/c\ge 1/2$ so that $ln(2)\ge 1/2$!

Why is that important? Because if X is any positive number, then $ln(2^{2X})= 2X ln(2)\ge X$. That is, given any positive number there exist x so that ln(x) is larger than that: ln(x) does not have an upper bound. Since the derivative of ln(x) is 1/x> 0, ln(x) is an increasing function. Since there is no upper bound, we must have $\lim_{x\to\infty}ln(x)= \infty$. Since ln(1/x)= -ln(x), we also have $\lim_{x\to -\infty} ln(x)= -\infty$.

That is, ln(x) maps the set of positive real numbers one-to-one and onto the set of all real numbers. Because that is true, it has an inverse function, which we can call "Exp(x)".

Now, suppose y= Exp(x). Then x= ln(y). If $x\ne 0$, $1= (1/x)ln(y)= ln(y^{1/x})$. Now going back to the exponential form, $Exp(1)= y^{1/x}$ so that $y= (Exp(1))^x$. Of course, if x= 0, y= Exp(0)= 1 and $Exp(1)^0= 1$ so this is still true.

That is, from the integral definition of ln(x) we can recover the fact that it is the inverse function to some number to the x power.

Last edited by a moderator: Jul 3, 2010
18. Jul 1, 2010

arildno

19. Jul 1, 2010

Bohrok

How would you know to make the lower limit 1 when first defining ln x
$$\ln x = \int_1^x \frac{1}{t} dt$$?
Why not 2, e, or 10? Or is it like the same kind of magic sometimes used in limit proofs for picking delta? :tongue:

20. Jul 1, 2010

arildno

If you pick 2, you will have a different function, that's all.

21. Jul 1, 2010

The Chaz

@Halls...
Double-check those limits. I think you stated in English something different than you wrote in symbols.

22. Jul 2, 2010

Landau

To prevent confusing, this should of course read
$$\ln x = \int_1^x \frac{1}{t}dt$$

23. Jul 2, 2010

Gib Z

We can also establish this by comparison to the Harmonic Series.

24. Jul 3, 2010

Bohrok

Ok, so can you still derive ex easily from that function using 2 as the lower limit? I'd try it myself but I don't really know how to start it...

25. Jul 3, 2010

HallsofIvy

If you use "2" as a lower limit you get
[tex]f(x)= \int_2^x dt/t= \int_1^x dt/t- \int_1^2 dt/t= ln(x)- ln(2)[/itex]

Of course, that's now different function and has a different inverse. If y= f(x)= ln(x)- ln(2) then ln(x)= y+ ln(2) and so $x= e^{y+ ln(2)}= e^y e^{ln(2)}= 2e^x$ so $f^{-1}(x)= 2e^x$.