Integral of 1/x: Proving Invalidity of Method

[O.J.]

ur not gettin my point... let me put it this way. how did the founders of calculus explain their approach to calculating the int of 1/x. they must've provided an explanation along with it.

 

[D H]

Define a function f(x) such that f(0)=1 and \frac{d\,f(x)}{dx} = f(x). This function has Taylor expansion
f(x) = \sum_{n=0}^\infty \frac{x^n}{n!}

This is the exponential function \exp(x). The natural logarithm function, \log(x) is its inverse. In other words, if x = \exp(t), then

\log(x) = \log(\exp(t)) \equiv tWhat is the derivative of \log(x)? Differentiating with respect to t as defined above,

\frac{d\,\log(x)}{dt} = 1

Expanding the left-hand side,

\frac{d\,\log(x)}{dx}{\frac{dx}{dt} = \frac{d\,\log(x)}{dx}\;{\frac{d\;\exp(t)}{dt} = \frac{d\,\log(x)}{dx}\,\exp(t) = \frac{d\,\log(x)}{dx}\;x[/itex]<br /> <br /> Thus<br /> <br /> \frac{d\, \log(x)}{dx}\; x = 1[/itex]&lt;br /&gt; &lt;br /&gt; or&lt;br /&gt; &lt;br /&gt; \frac{d\,\log(x)}{dx} = \frac 1 x[/itex]&amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; By the fundamental theory of calculus,&amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; \int \frac 1 x dx = \log(x) + C[/itex]

 

[mathwonk]

they certainly did not define the log as that integral.

rather they knew the exponential function E(x) satisfied the laws

E(x+y) = E(x) + E(y), an d it follows easily from this that if diferentiable, the derivatiove of E(x) is a constant times E(x) itself.

thus its inverse function log, using the inverse function rule of derivatives, must be some function L such that L(xy) = L(x) + L(y) and L'(x) = c/x for some constant c.

but since it is so hard to define A^x for irrational powers x, it then dawned on someone to do it all backwards and simply define L as the area function of 1/x. then after proving abstractly that it satisfies the law of changng products into sums, one knows it must really be our old friend the log function.

i myself prefer to call this a theorem, and simp,y say the usuallog functon can be defiend as an area function.

of cousre if you do not know the difference between an integral as a limit of riemann sums, and the "integral" as an antiderivative, all this is incomprehensible to you.

 

[mathwonk]

maybe if you read sections 5.2 and 5.5 of the calulus bible at this address it will help.

http://www.math.byu.edu/Math/CalculusBible/ how quaint, the site is at byu, a "religious" school.

 

[Schrodinger's Dog]

ur not gettin my point... let me put it this way. how did the founders of calculus explain their approach to calculating the int of 1/x. they must've provided an explanation along with it.

I am all constants are easily proven and 1/x is simillar in that it is based on a constant. Proving constants e^x=? prove it. How easy is that and by extension how easy is it to prove that 1/e^x or ln(e^x)= is ?, It's not just provable, it's self evident. it's the same as saying prove that 1/pi is ?

http://www.karlscalculus.org/explogid.html

Adding the Exponents: If b is any positive real number then

bx by = bx+y

for all x and y. This is the single most important identity concerning logs and exponents. Since ex is only a special case of an exponential function, it is also true that

ex ey = ex+y

Multiplying the Exponents: If b is any positive real number then

(bx)y = bxy


for all x and y. Again since ex is a special case of an exponential function, it is also true that

(ex)y = exy


Converting to roots to exponents: The nth root of x is the same as

x1/n

for all positive x. Since square roots are a special case of nth roots, this means that

_
√x = x1/2

In addition:

__
√ex = ex/2

Converting to ex form: If b is any positive real number then

bx = ex ln(b)

for all x. This includes the case where you have xx:

xx = ex ln(x)

or if you have f(x)x:

f(x)x = ex ln(f(x))

or if you have xf(x):

xf(x) = ef(x) ln(x)

or if you have f(x)g(x):

f(x)g(x) = eg(x) ln(f(x))

As an example, suppose you had (x2 + 1)1/x. That would be the same as

e(1/x) ln(x2 + 1)

ex is its own derivative: The derivative of ex is ex. This is the property that makes ex special among all other exponential functions.

ex is always positive: You can put in any x, positive or negative, and ex will always be greater than zero. When x is positive, ex > 1. When x is negative, ex < 1. When x = 0 then ex = 1.

The log of the product is the sum of the logs: Let b, x, and y all be positive real numbers. Then

logb(xy) = logb(x) + logb(y)

This is the most important property of logs. Since ln(x) = loge(x), it is also true that

loge(xy) = ln(xy) = loge(x) + loge(y) = ln(x) + ln(y)

The log of the reciprocal is the negative of the log: For any positive b, x, and y

logb(1/x) = -logb(x)

logb(y/x) = logb(y) - logb(x)

This includes

ln(1/x) = -ln(x)

ln(y/x) = ln(y) - ln(x)

Concerning multiplying a log by something else: Let b and x be positive and k any real number. Then

k logb(x) = logb(xk)

This includes

k ln(x) = ln(xk)

It also means that

_
logb(√x) = (1/2)logb(x)

and

_
ln(√x) = (1/2)ln(x)

Converting log bases to natural log You can compute any base log using the natural log function (that is ln) alone. If b and x are both positive then

logb(x) =

ln(x)

ln(b)

Every log function is the inverse of some exponential function: If b is any positive real number, then

blogb(x) = logb(bx) = x

The right-hand part of this equation is true for all x. The left-hand part is true only for positive x. The functions, ex and ln(x) are also inverses of each other.

eln(x) = ln(ex) = x

The same rules for x apply as above.

The derivative of the natural log is the reciprocal: If x is positive, it is always true that the derivative of ln(x) is 1/x.

To find the derivative of logs of other bases, apply the conversion rule. So for the derivative of logb(x) you end up with

1

x ln(b)

The natural log can be expressed as a limit: For all positive x

xh - 1
ln(x) = lim
h -> 0 h

You can only take the log of positive numbers: If x is negative or zero, you CAN'T take the log of x -- not the natural log or the log of any base. In addition, the base of a log must also be positive. As x approaches zero from above, ln(x) tends to minus infinity. As x goes to positive infinity, so does ln(x). So ln(x) has no limit as x goes to infinity or as x goes to zero.

Natural log is positive or negative depending upon whether x is greater than or less than 1: If x > 1, then ln(x) > 0. If x < 1, then ln(x) < 0. If x = 1 then ln(x) = 0. Indeed the log to any base of 1 is always zero.
Something you Can't Do with Logs

There is no formula for the log of a sum: Don't go saying that log(a+b) is equal to log(a) log(b) because this is NOT TRUE.

 

[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}
= \lim_{h\rightarrow 0}a^x\frac{a^h- a^0}{h}[/itex]<br /> = \left(\lim_{h\rightarrow 0}\frac{a^h- 1}{h}\right)a^x[/itex]&lt;br /&gt; (The hard part is showing that that limit exist- which is why I prefer defining ln(x) by the integral and then e&lt;sup&gt;x&lt;/sup&gt; as &lt;b&gt;its&lt;/b&gt; inverse!)&lt;br /&gt; &lt;br /&gt; Given that the limit exists, we see that the derivative of a&lt;sup&gt;x&lt;/sup&gt; is simply a constant times a&lt;sup&gt;x&lt;/sup&gt;. 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 &amp;quot;e&amp;quot; (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&lt;sup&gt;x&lt;/sup&gt; is precisely e&lt;sup&gt;x&lt;/sup&gt; itself- the world&amp;#039;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
<br /> f(x)=\int_{0.5}^{x} \frac{dx}{x}<br />
Then instead of nice addition formula for log, we get
<br /> f(x)+f(y)=\log2 +f(xy) <br />

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} &lt; 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} &lt; 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,
<br /> \int_{a}^x \frac{dt}{t}=\log x - \log a<br />
Since log 1 = 0 (directly from previously mentioned definition), it is obvious that
<br /> \log x = \int_{1}^x \frac{dt}{t}<br />
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
\int_a^b f(x)dx= -\int_b^a f(x) dx[/itex].<br /> <br /> The only thing starting the integration at x= 1 causes is that ln(1)= 0.<br /> Since, by the fundamental theorem of calculus, the derivative of \int_1^x (1/T)dt is 1/x&gt; 0 for x&gt; 0 so ln(x) is an increasing function and so if x&lt; 1, ln(x)&lt; 0 and if x&gt; 1, ln(x)&gt; 0.<br /> <br /> 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&lt; 0.<br /> That is also why you <b>cannot</b> start the integral at 0- 1/t is ubounded in any neighborhood of t= 0.<br /> <br /> &quot;If you defined the integral to be from 0 to x, you would come up with the same answers wouldn&#039;t you?&quot; No, you wouldn&#039;t! Especially because 1/t is not bounded in any neighborhood of 0 and so that integral is impossible. <br /> <br /> <blockquote data-attributes="" data-quote="Bohrok" data-source="post: 2788254" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> Bohrok said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> You know, that&#039;s pretty much exactly what I was wondering in post #19 here<br /> <a href="https://www.physicsforums.com/showthread.php?t=412403&amp;page=2" class="link link--internal">https://www.physicsforums.com/showthread.php?t=412403&amp;page=2</a><br /> <br /> <br /> <br /> 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? </div> </div> </blockquote> Not &quot;luck&quot; but just because 1 is an easy number (and 0 won&#039;t work as I showed above).<br /> <br /> If we were to define <br /> L_a(x)= \int_a^x \frac{1}{t} dt<br /> then we have<br /> L_a(x)= \int_1^x \frac{1}{t}dt- \int_1^a \frac{1}{t}dt<br /> so that L_a(x)= ln(x)- ln(a)= ln(x/a).<br /> <br /> 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 &quot;a&quot; 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 &quot;e&quot;.<br /> <br /> 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<br /> \lim_{h\to 0}\frac{a^{x+h}- a^x}{h}= \lim_{h\to a}\frac{a^xa^h- a^x}{h}<br /> = \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}<br /> that is, the derivative of a^x, for any positive a, is some constant (which depends on a), C_a, times a^x itself.<br /> <br /> Now, it is easy to see that if a= 2,<br /> C_2= \lim_{h\to 0}\frac{2^h- 2}{h}<br /> is close to .69 (take h= .001) and that if a= 3,<br /> C_3= \lim_{h\to 0}\frac{3^h- 3}{h}<br /> is close to 1.099.<br /> <br /> C_2 is less than 1 and C_3 is larger than 1. It doesn&#039;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 &quot;e&quot; and have the result that (e^x)&amp;#039;= e^x (instead of some weird constant times a^x).<br /> <br /> But either way, our choice of numerical value for &quot;e&quot; and the definition of ln(x) are based on the fact that &quot;1&quot; is a convenient number.<br /> <br /> If <b>you</b> were able to choose a value for a <b>positive</b> constant that would keep popping up in your work, what value would you choose?<br /> <br /> (By the way, I just noticed that alice22 resurrected this from a thread that had been closed three years ago.)

 

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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!