Integral of 1/x: Proving Invalidity of Method

[O.J.]

I understand it now. only one more thing left, when choosing the base of the logarithm, could they have chosen any base or it strictly had to be e?
also how can u extend this definition to f'(x)/f(x)?

 

[Gib Z]

It had to be e.

e has the special unique property that its exponentials are its own derivatives, which is essential in DH's proof.

To extend it, realize that we could simply let u=f(x) and use the chain rule.

 

[HallsofIvy]

u can't just come and convince me that they JUST 'defined' the integral of 1/x to be a logarithm function whose base happens to be a number that 2.721...etc. I badly need to know how they arrived at it. how they figured it out.

No, 2.71828... etc.

 

[mathwonk]

isnt tht a song? "it had to be e, that wonderful e"?

 

[HallsofIvy]

I understand it now. only one more thing left, when choosing the base of the logarithm, could they have chosen any base or it strictly had to be e?
also how can u extend this definition to f'(x)/f(x)?

What do you mean by "the" logarithm? There are, of course, logarithms to different bases- in fact the common logarithm, base 10 came before the natural logarithm, base 2.718...

This has been pointed out before but I will repeat it:

The derivative of the general "exponential" function, y=a^x can be calculated as
$$\lim_{h\rightarrow 0}\frac{a^{x+h}-a^x}{h}= \lim_{h\rightarrow 0}\frac{a^xa^h- a^x}{h}$$
[tex]= \lim_{h\rightarrow 0}a^x\frac{a^h- a^0}{h}[/itex]
[tex]= \left(\lim_{h\rightarrow 0}\frac{a^h- 1}{h}\right)a^x[/itex]
(The hard part is showing that that limit exist- which is why I prefer defining ln(x) by the integral and then e^x as its inverse!)

Given that the limit exists, we see that the derivative of a^x is simply a constant times a^x. Also it is easy to see that if a= 2 that constant is less than 1 (taking a=2, h= 0.001 gives that fraction as about 0.6934) and that if a= 3 that constant is larger than 1 (taking a= 3, h= 0.001 gives that fraction as about 1.099). There must be a number between 2 and 3 for which that constant is exactly 1. If we call that number "e" (taking values of a between 2 and 3 and zeroing in on making that fraction equal to 1 for small h give approximately 2.718) we have that the derivative of e^x is precisely e^x itself- the world's simplest derivative!

 

[mathwonk]

i traveled around, and finally found, a number who, could make me be true,could make me be true...,

 

[Gib Z]

lol nice song. At my high school there's a song one of our teachers made to remember derivatives, it goes through all the basic ones. Its pretty funny :)

 

[duke_nemmerle]

I understand it now. only one more thing left, when choosing the base of the logarithm, could they have chosen any base or it strictly had to be e?
also how can u extend this definition to f'(x)/f(x)?

Second part first, by the chain rule. If you have a function ln(u), the derivative d ln(u)/dx will be the 1/u that we've been talking about multiplied by du/dx. Or as you've put it d ln(f(x))/dx= 1/f(x) * f'(x)=f'(x)/f(x). This is just like other compositions of functions you've differentiated before, you find the derivative as the derivative of the "outer" function evaluated at the "inner" function multiplied by the derivative of the "inner" function.


In the case of the number e, it is exactly the number that gives the area under the curve 1/t from 1 to x as 1. That is, the area under the curve 1/t on the interval [1,e] is 1.

 

[Mr.4]

Okay, I want to be least scientific here. Let's make up a story:

The fathers (or mothers) of Calculus had to invent Limits, Derivative and differentiation before they came to Integration. So one day when they were finding the derivatives of different functions for fun (:D) or whatever, they stumbled upon the derivative of log (x) (through several methods that the apparently smart people here have suggested), and they found it out to be 1/x. So when the finally invented integration they didn't have to look any further for the integral of 1/x OR was feeling too lazy to do any derivation again OR had the common sense to conclude that the integral of 1/x is indeed log (x).

See that's a good explanation! Maybe I shd write my own Math book one day :D

 

[Schrodinger's Dog]

An interesting note is that if

$$\frac{dy}{dx}\frac{f(x)}{g(x)}$$

Where f(x) is the derivative of g(x)

then

$$\int h(x)=log(g(x))+C$$

ie

$$\int\frac{2x}{x^2}= log(x^2)+C$$

The trick is to spot when derivative of bottom is the top.

if it is of the form

$$\int \frac{f(x)}{(g(x)){^2}}$$

then a version of the chain rule is used on the denominator.

$$\int \frac {2x}{(x^2+1)^{2}}=?$$

 

[alice22]

No they don't and no they don't. Actually the lower limit in the integral is 1, therefore

$$\ln x=\int_{1}^{x} \frac{1}{t}{}dt$$.

Firstly sorry to drag up an old thread.

Actually I was wondering about that last night, I was thinking the limit was from 0,
and then I though that at 0, 1/x = 1/0 = infinite!


And hence all integrals of 1/x= infinity - infinity!

So why start at 1? You could avoud the 1/0 by starting at 1/2 or 0.1 or 00000001 etc...


But I still have problems with it
http://www.animations.physics.unsw.edu.au/jw/graphics/ln(x).gif

Here you can see the graphs.

Now you can see for x>0 you can see the area is always positive, yet for x<1 (from 0-1 anyway) the log is negative, that means negative area, which is wrong I think.

So that needs explaining.
Clearly the area under 1/x at 1 is not zero.

Also what about values of 1/x where are x<1?

How are they defined?

Something seems wrong here. (probably me lol).
I can't follow some of the thread (without more time, also a key link is dead).

 

[losiu99]

So why start at 1? You could avoud the 1/0 by starting at 1/2 or 0.1 or 00000001 etc...

For natural log of 1 is zero, and no other number has that property. Defining log as such integral starting from other number would lead to different, messier properties, and, more importantly, it wouldn't be inverse of exponential function.

for x<1 (from 0-1 anyway) the log is negative, that means negative area, which is wrong I think.

Values of log are negative, but it's still an increasing function. Remember the fundamental theorem of calculus - to determine the area, we substract log(b)-log(a). Since for a < b there is log(a) < log(b), the area is positive.

Clearly the area under 1/x at 1 is not zero.

What do you mean by area at the point?

Also what about values of 1/x where are x<1?

How are they defined?

Just like everywhere else.

 

[arildno]

alice:
Letting k>1, we can see that:
$$\ln(\frac{1}{k})=\int_{1}^{\frac{1}{k}}\frac{dt}{t}$$
Set
$$u=kt\to\int_{1}^{\frac{1}{k}}\frac{dt}{t}=\int_{k}^{1}\frac{\frac{du}{k}}{\frac{u}{k}}=\int_{k}^{1}\frac{du}{u}=-\int_{1}^{k}\frac{du}{u}=-\ln(k)$$

That is to say, we have proven, for all (in fact) numbers k, that:
$$\ln(\frac{1}{k})=-\ln(k)$$

 

[alice22]

For natural log of 1 is zero, and no other number has that property. Defining log as such integral starting from other number would lead to different, messier properties, and, more importantly, it wouldn't be inverse of exponential function.


Values of log are negative, but it's still an increasing function. Remember the fundamental theorem of calculus - to determine the area, we substract log(b)-log(a). Since for a < b there is log(a) < log(b), the area is positive.

What do you mean by area at the point?

Just like everywhere else.

"Just like everywhere else."

If they are the same as everywhere else then why not start from less than zero.



For example what is the integral of 1/x from 0.4 to 0.8, if you have to start from 1 you cannot do this?

"What do you mean by area at the point? "

Well the area from 0-1, although this area would in fact be infinite I believe.

"For natural log of 1 is zero, and no other number has that property. Defining log as such integral starting from other number would lead to different, messier properties, and, more importantly, it wouldn't be inverse of exponential function."


I accept the point that whilst the logs (ln) below 1 may be negative but as you are doing a subtraction you will get a positive area, I guess I was careless to overlook that.

I don't know what messier properties are so I will have to pass on that, unless someone can explain that in more basic language.


The inverse of the exponential function has values below 1 so when you say it would not be the inverse of the exponential function I can "well it's not the inverse of the exponential function anyway because that is defined below 1" ?

 

[Mark44]

"Just like everywhere else."

If they are the same as everywhere else then why not start from less than zero.



For example what is the integral of 1/x from 0.4 to 0.8, if you have to start from 1 you cannot do this?

Sure you can.
$$\int_{.4}^{.8} \frac{dt}{t} + \int_{.8}^{1} \frac{dt}{t} = \int_{.4}^{1} \frac{dt}{t}$$
$$\Rightarrow \int_{.4}^{.8} \frac{dt}{t} = \int_{.4}^{1} \frac{dt}{t} - \int_{.8}^{1} \frac{dt}{t}$$
$$= -\int_{1}^{.4} \frac{dt}{t} + \int_{1}^{.8} \frac{dt}{t}$$
= - ln(.4) + ln(.8) = ln(.8/.4) = ln 2 ~.693

"What do you mean by area at the point? "

Well the area from 0-1, although this area would in fact be infinite I believe.

"For natural log of 1 is zero, and no other number has that property. Defining log as such integral starting from other number would lead to different, messier properties, and, more importantly, it wouldn't be inverse of exponential function."

You're quoting someone in this thread, but what is said in the quote is incorrect. For any base b, with b > 0 and b != 1, log_b 1 = 0. This has nothing to do with whatever the base happens to be.

I accept the point that whilst the logs (ln) below 1 may be negative but as you are doing a subtraction you will get a positive area, I guess I was careless to overlook that.

I don't know what messier properties are so I will have to pass on that, unless someone can explain that in more basic language.


The inverse of the exponential function has values below 1 so when you say it would not be the inverse of the exponential function I can "well it's not the inverse of the exponential function anyway because that is defined below 1" ?

I don't understand what you're asking here. The natural exponential function e^x has all real numbers as its domain and the positive reals as its range. The inverse of this function (the natural log function) has a domain of the positive reals and its range is all real numbers.

 

[alice22]

Sure you can.
$$\int_{.4}^{.8} \frac{dt}{t} + \int_{.8}^{1} \frac{dt}{t} = \int_{.4}^{1} \frac{dt}{t}$$
$$\Rightarrow \int_{.4}^{.8} \frac{dt}{t} = \int_{.4}^{1} \frac{dt}{t} - \int_{.8}^{1} \frac{dt}{t}$$
$$= -\int_{1}^{.4} \frac{dt}{t} + \int_{1}^{.8} \frac{dt}{t}$$
= - ln(.4) + ln(.8) = ln(.8/.4) = ln 2 ~.693
You're quoting someone in this thread, but what is said in the quote is incorrect. For any base b, with b > 0 and b != 1, log_b 1 = 0. This has nothing to do with whatever the base happens to be.


I don't understand what you're asking here. The natural exponential function e^x has all real numbers as its domain and the positive reals as its range. The inverse of this function (the natural log function) has a domain of the positive reals and its range is all real numbers.

What I am saying is the exponential function does not stop at 1, however the integral of 1/x
does not seem to be defined below 1.
You seem to have ignored that bit and it was that bit I was specifically concerned with.

 

[losiu99]

For any base b, with b > 0 and b != 1, log_b 1 = 0. This has nothing to do with whatever the base happens to be.

What I ment was that 1 is an unique zero of natural log. Of course, any other log as well. My point is that antiderivative of 1/x, namely natural log, defined as the inverse of the exponential, must be zero at the lower limit of this integration, for otherwise the integral does not represent natural log.

Area under 1/x between 0 and 1 is indeed infinite.

By messier properties I mean things like this:
Suppose we define
$$f(x)=\int_{0.5}^{x} \frac{dx}{x}$$
Then instead of nice addition formula for log, we get
$$f(x)+f(y)=\log2 +f(xy)$$

So defined function also is not inverse of the exponential. Not because of domain, but because e^0=1, and f(1)=log 2.

Which does not of course mean you cannot integrate 1/x over intervals containing numbers less than one. You cannot start integration from zero, for answer will be undefined, but any other intervals are perfectly ok.

 

[Mark44]

What I am saying is the exponential function does not stop at 1, however the integral of 1/x
does not seem to be defined below 1.
You seem to have ignored that bit and it was that bit I was specifically concerned with.

No, I didn't ignore that, but you must have ignored what I wrote. You asked about
$$\int_{.4}^{.8} \frac{dt}{t}$$

and I showed how that could be evaluated using the integral definition of the natural log function. In the example, both .4 and .8 are less than 1.

The natural log function ln(x) is defined for any x > 0.

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

The only restriction on x in the integral above is that it must be positive.

One other thing, don't confuse "area" and "value of a definite integral." Area is always nonnegative, but a definite integral can have a negative value. In this integral, if x is between 0 and 1,

$$ln(x) = \int_{1}^{x} \frac{dt}{t} < 0$$

 

[HallsofIvy]

No they don't and no they don't. Actually the lower limit in the integral is 1, therefore

$$\ln x=\int_{1}^{x} \frac{1}{t}{}dt$$.

Ouch! Thanks!

 

[alice22]

No, I didn't ignore that, but you must have ignored what I wrote. You asked about
$$\int_{.4}^{.8} \frac{dt}{t}$$

and I showed how that could be evaluated using the integral definition of the natural log function. In the example, both .4 and .8 are less than 1.

The natural log function ln(x) is defined for any x > 0.

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

The only restriction on x in the integral above is that it must be positive.

One other thing, don't confuse "area" and "value of a definite integral." Area is always nonnegative, but a definite integral can have a negative value. In this integral, if x is between 0 and 1,

$$ln(x) = \int_{1}^{x} \frac{dt}{t} < 0$$

Well it just seems a bit of a 'cheat' to me.
I mean if you defined the integral to be from 0 to x, you would come up with the
same answers wouldn't you so why bother saying it is from 1 to x?

 

[losiu99]

If we define log to be this integral from zero to x, it will be divergent for every real number. That's not the same answer, I'm afraid
The fact is: if we want to represent natural log understood as an inverse of the exponential function by integral of 1/t with x as the upper limit, lower limit must be 1. For no other number the equality holds.

 

[Bohrok]

Well it just seems a bit of a 'cheat' to me.
I mean if you defined the integral to be from 0 to x, you would come up with the
same answers wouldn't you so why bother saying it is from 1 to x?

You know, that's pretty much exactly what I was wondering in post #19 here
https://www.physicsforums.com/showthread.php?t=412403&page=2

If we define log to be this integral from zero to x, it will be divergent for every real number. That's not the same answer, I'm afraid
The fact is: if we want to represent natural log understood as an inverse of the exponential function by integral of 1/t with x as the upper limit, lower limit must be 1. For no other number the equality holds.

But how did we know to choose 1 in the first place? Just luck, or kind of working backwards from what we know the answer to be?

 

[losiu99]

Well, from the definition of natural log as inverse of the exponential, it is easy to show that it's derivative is 1/x. So, by fundamental theorem of calculus,
$$\int_{a}^x \frac{dt}{t}=\log x - \log a$$
Since log 1 = 0 (directly from previously mentioned definition), it is obvious that
$$\log x = \int_{1}^x \frac{dt}{t}$$
Natural log is increasing (once again, definition), so no number other than one in place of $$a$$ fits.

Edit: of course, if we didn't want log to be inverse of the exponential, it doesn't really matter where do we start integrating from.

 

[HallsofIvy]

Well it just seems a bit of a 'cheat' to me.
I mean if you defined the integral to be from 0 to x, you would come up with the
same answers wouldn't you so why bother saying it is from 1 to x?

I can't help but wonder what kind of Calculus course you took!

$\int_a^b f(x)dx$ does NOT mean that b must be larger than a.

Surely one of the first things you learned about the integral is that
[tex]\int_a^b f(x)dx= -\int_b^a f(x) dx[/itex].

The only thing starting the integration at x= 1 causes is that ln(1)= 0.
Since, by the fundamental theorem of calculus, the derivative of $\int_1^x (1/T)dt$ is 1/x> 0 for x> 0 so ln(x) is an increasing function and so if x< 1, ln(x)< 0 and if x> 1, ln(x)> 0.

The only thing that restricts what x can be is that we cannot integrate across a point where f(x) becomes unbounded- which is why ln(x) is not defined for x< 0.
That is also why you cannot start the integral at 0- 1/t is ubounded in any neighborhood of t= 0.

"If you defined the integral to be from 0 to x, you would come up with the same answers wouldn't you?" No, you wouldn't! Especially because 1/t is not bounded in any neighborhood of 0 and so that integral is impossible.

You know, that's pretty much exactly what I was wondering in post #19 here
https://www.physicsforums.com/showthread.php?t=412403&page=2



But how did we know to choose 1 in the first place? Just luck, or kind of working backwards from what we know the answer to be?

Not "luck" but just because 1 is an easy number (and 0 won't work as I showed above).

If we were to define
$$L_a(x)= \int_a^x \frac{1}{t} dt$$
then we have
$$L_a(x)= \int_1^x \frac{1}{t}dt- \int_1^a \frac{1}{t}dt$$
so that $L_a(x)= ln(x)- ln(a)= ln(x/a)$.

If $y= L_a(x)$ then we would have $y= e^{x/a}$ which is the same as $(e^{1/a})^x$. That is, choosing "a" as the lower limit of the integral would just lead to a different base, $e^{1/a}$- and so, perhaps, to a different choice of numerical value for "e".

Historically, of course, it worked the other way. The function $f(x)= a^x$ can be defined, without using logarithms, for any positive number, a. The derivative of $a^x$ would be given by
$$\lim_{h\to 0}\frac{a^{x+h}- a^x}{h}= \lim_{h\to a}\frac{a^xa^h- a^x}{h}$$
$$= \lim_{h\to a}\left(a^x\right)\frac{a^h- 1}{h}= \left(a^x\right)\lim_{h\to 0}\frac{a^h- a}{h}$$
that is, the derivative of $a^x$, for any positive a, is some constant (which depends on a), $C_a$, times $a^x$ itself.

Now, it is easy to see that if a= 2,
$$C_2= \lim_{h\to 0}\frac{2^h- 2}{h}$$
is close to .69 (take h= .001) and that if a= 3,
$$C_3= \lim_{h\to 0}\frac{3^h- 3}{h}$$
is close to 1.099.

$C_2$ is less than 1 and $C_3$ is larger than 1. It doesn't take much (essentially showing that this limit process is a continuous function of a) to see that there is some number between 2 and 3 such that the limit is 1. We call that number "e" and have the result that $(e^x)'= e^x$ (instead of some weird constant times $a^x$).

But either way, our choice of numerical value for "e" and the definition of ln(x) are based on the fact that "1" is a convenient number.

If you were able to choose a value for a positive constant that would keep popping up in your work, what value would you choose?

(By the way, I just noticed that alice22 resurrected this from a thread that had been closed three years ago.)

 

[Unit]

Surely one of the first things you learned about the integral is that
$$\int_a^b f(x)dx= \int_b^a f(x) dx$$.

I think you meant

$$\int_a^b f(x) dx = - \int_b^a f(x) dx$$

with the minus sign :) Other than that, I really like your explanation!

 

[alice22]

(By the way, I just noticed that alice22 resurrected this from a thread that had been closed three years ago.)

Yes I said that it the first line of my post.
The thread was not closed though, it was still open for posting.

 

[HallsofIvy]

I think you meant

$$\int_a^b f(x) dx = - \int_b^a f(x) dx$$

with the minus sign :) Other than that, I really like your explanation!

I think that is the second time in a few days I have made that mistake.

Oh, well, I'll go back and edit and then no one will know!

Thanks for the heads-up.

 

[alice22]

OK I get lost sometimes in the notation of maths, it makes it hard to follow, but what I am saying is this, what is so special about 1?
OK, ln(x) passed though y=0 at this point, but so what?
Lot's of functions pass through zero, that's no big deal.

I will concede one thing however, maybe I am answering my own question, and that is
that all the powers of x, x^2, x^3, x^7847 all pass through (0,0), however a lot of other functions do not, for example cos x. We do not do that integral from pi/4 do we?
Answers containing the fewest mathematical symbols would be the easiest to follow I think! Sometimes it is difficult to what point someone is making when it is merely mathematical shorthand!

Maybe I am making a mountain out of a molehill?

 

[Mark44]

OK I get lost sometimes in the notation of maths, it makes it hard to follow, but what I am saying is this, what is so special about 1?
OK, ln(x) passed though y=0 at this point, but so what?
Lot's of functions pass through zero, that's no big deal.

And lots of them don't cross the x-axis anywhere, one of them being y = 1/x, the function that plays a role in this definition of ln(x). What's so special about 1 is that ln(1) = log(1) = log_b(1) = 0, so 1 seems as good a starting point for the lower limit of integration as any.

I will concede one thing however, maybe I am answering my own question, and that is
that all the powers of x, x^2, x^3, x^7847 all pass through (0,0), however a lot of other functions do not, for example cos x. We do not do that integral from pi/4 do we?

But we don't ordinarily define cos(x) as an integral.

Answers containing the fewest mathematical symbols would be the easiest to follow I think! Sometimes it is difficult to what point someone is making when it is merely mathematical shorthand!

If you ask a very specific question about a definite integral (as you have done), we will naturally assume that you have some knowledge of the notation that is used in calculus. Although it would be possible to explain everything in English without resorting to symbols, the explanations would be much more verbose and likely not as precise.

That's not to say that the best explanation is one that is purely equations and symbols.

Maybe I am making a mountain out of a molehill?

Maybe. The equation
$$ln(x) = \int_1^x \frac{dt}{t}$$
is one definition for the natural log function.

 

[Mute]

OK I get lost sometimes in the notation of maths, it makes it hard to follow, but what I am saying is this, what is so special about 1?
OK, ln(x) passed though y=0 at this point, but so what?
Lot's of functions pass through zero, that's no big deal.

Choosing 1 as the lower limit makes this definition of the logarithm correspond exactly with our other definitions.

Even if the integral definition had been the first definition of the logarithm known, I would guess that 1 still would have been chosen as the lower limit initially. Why? Well, zero obviously wouldn't work, so might as well just pick the next natural number, which is one. People like natural numbers much more than fractions, and skipping 1 to go to 2 seems even more arbitrary than picking 1 in this scenario. Even if some other number had been picked, once the function's relation to the exponential function had been realized (for example), the definition probably would have been changed to have the lower limit of 1 so that it corresponded exactly to the inverse of the exponential function.

 

[HallsofIvy]

OK I get lost sometimes in the notation of maths, it makes it hard to follow, but what I am saying is this, what is so special about 1?
OK, ln(x) passed though y=0 at this point, but so what?
Lot's of functions pass through zero, that's no big deal.

I will concede one thing however, maybe I am answering my own question, and that is
that all the powers of x, x^2, x^3, x^7847 all pass through (0,0), however a lot of other functions do not, for example cos x. We do not do that integral from pi/4 do we?
Answers containing the fewest mathematical symbols would be the easiest to follow I think! Sometimes it is difficult to what point someone is making when it is merely mathematical shorthand!

Maybe I am making a mountain out of a molehill?

You have now been told repeatedly that the choice of "1" as the lower limit for the integral defining ln(x) is simply because it is convenient. Choosing that lower limit to be any positive number other than 1 would just change the numerical value of "e".

If you could choose a positive constant that will keep popping up in your calculations, don't you think "1" would be the most reasonable choice?

Do you see why "0" cannot be the lower bound? (Because 1/t is not bounded in any neighborhood of t= 0.)

Finally, Mark44 refers to the fact that ln(1)= 0 but, of course, that is due to the definition of ln(x) as the inverse of $e^x$ and I am referring to it as defined by the integral (and $e^x$ then defined as its inverse).

 

[alice22]

You have now been told repeatedly that the choice of "1" as the lower limit for the integral defining ln(x) is simply because it is convenient. Choosing that lower limit to be any positive number other than 1 would just change the numerical value of "e".

If you could choose a positive constant that will keep popping up in your calculations, don't you think "1" would be the most reasonable choice?

Do you see why "0" cannot be the lower bound? (Because 1/t is not bounded in any neighborhood of t= 0.)

Finally, Mark44 refers to the fact that ln(1)= 0 but, of course, that is due to the definition of ln(x) as the inverse of [itex]e^x[/math] and I am referring to it as defined by the integral (and [itex]e^x[/math] then defined as its inverse).

I am looking for a viable explanation not to be 'told' this is the answer.
I don't think convenience is a credible answer.

I also do not see how the value of "e" can be changed no more than the value of pi can be changed.

It might be convenient to say pi=3 however, convenience is not what is required, the correct
answer is the only acceptable answer.

So no I do not see convenience as an acceptable answer.

 

[losiu99]

Okay then, answer to what problem is this 1 here? If the question is "Why is 1 the lower limit of this integral representing natural log (defined as the inverse of the exponential)?", then I believe I answered it in my previous posts (In fact, this integral with other lower limit doesn't define log of any base). If the question is "why log is defined as this integral with 1 as lower limit, and not some other real number", the answer is: someone wanted log to be inverse of exponential. If the question is "Why did he?", I think "He thought it would be nice to find some integral representation for inverse of such a fundamental function as exp". It is perfectly ok to define some function as the same integral with other lower limit, it simply will be different function, satisfying different properties etc. Could you formulate your question once again? I feel a little lost, I no longer understand what are you asking for.

 

[Char. Limit]

Actually, convenience is an acceptable answer. Much of what mathematicians do is based off of convenience. Just think about the table of integrals in any calculus book. A mathematician could solve any of those functions by himself every time he needed them, but why would he when he could just solve the general case and use it?

It's the same general idea of convenience, picking the number that makes things easier.

Also, just out of curiousity, is it possible to define the sine function as an integral of the cosine function? Just a little side question.

 

[alice22]

Okay then, answer to what problem is this 1 here? If the question is "Why is 1 the lower limit of this integral representing natural log (defined as the inverse of the exponential)?", then I believe I answered it in my previous posts (In fact, this integral with other lower limit doesn't define log of any base). If the question is "why log is defined as this integral with 1 as lower limit, and not some other real number", the answer is: someone wanted log to be inverse of exponential. If the question is "Why did he?", I think "He thought it would be nice to find some integral representation for inverse of such a fundamental function as exp". It is perfectly ok to define some function as the same integral with other lower limit, it simply will be different function, satisfying different properties etc. Could you formulate your question once again? I feel a little lost, I no longer understand what are you asking for.

No I have problems with that.
For starters show me the mathematics equation for "someone wanted something"?
I do not recall covering the that in my math course, maybe I missed the maths of "someone wanted".

example x^2 = 3 (because someone wanted it to be).

The inverse of the exponential function does not stop at one so that again is a wholely invalid reason.